Nhân, chia các phân thức

Ta có  $\dfrac{{24x{y^2}{z^2}}}{{12{x^2}z}}.\dfrac{{4{x^2}y}}{{6x{y^4}}}$$= \dfrac{{24x{y^2}{z^2}.4{x^2}y}}{{12{x^2}z.6x{y^4}}} = \dfrac{{96{x^3}{y^3}{z^2}}}{{72{x^3}{y^4}z}} = \dfrac{{4z}}{{3y}}$ .

Ta có   $\dfrac{{5\left( {x + 1} \right)}}{{x{y^2}}}:\dfrac{{10\left( {x + 1} \right)}}{{3{x^2}y}}$$= \dfrac{{5\left( {x + 1} \right)}}{{x{y^2}}}.\dfrac{{3{x^2}y}}{{10\left( {x + 1} \right)}} = \dfrac{{15\left( {x + 1} \right){x^2}y}}{{10\left( {x + 1} \right)x{y^2}}} = \dfrac{{3x}}{{2y}}$ .

Ta có  $\dfrac{{5x + 2}}{{3x{y^2}}}:\dfrac{{10x + 4}}{{{x^2}y}} = \dfrac{{5x + 2}}{{3x{y^2}}}:\dfrac{{2\left( {5x + 2} \right)}}{{{x^2}y}} = \dfrac{{5x + 2}}{{3x{y^2}}}.\dfrac{{{x^2}y}}{{2\left( {5x + 2} \right)}}$$= \dfrac{{\left( {5x + 2} \right){x^2}y}}{{6x{y^2}\left( {5x + 2} \right)}} = \dfrac{x}{{6y}}$

Vậy đa thức cần điền là $x$ .

Ta có $\dfrac{{ - 27{z^4}}}{{6{y^3}z}}.\dfrac{{2{y^2}}}{{ - 45{x^2}z}}$$= \dfrac{{ - 27{z^4}.2{y^2}}}{{6{y^3}z.\left( { - 45{x^2}z} \right)}} = \dfrac{{ - 54.{z^4}{y^2}}}{{ - 270.{x^2}{y^3}{z^2}}} = \dfrac{{{z^2}}}{{5{x^2}y}}$ nên A sai.

* $\dfrac{{ - 9x{z^4}}}{{18{y^3}z}}.\dfrac{{8x{y^2}}}{{ - 45{x^2}z}}$$= \dfrac{{ - 72.{x^2}{y^2}{z^4}}}{{ - 810{x^2}{y^3}{z^2}}} = \dfrac{{4{z^2}}}{{45y}}$  nên B sai.

* $\dfrac{{ - 27x{z^4}}}{{6{y^3}{z^2}}}.\dfrac{{4x{y^2}}}{{ - 45{x^2}}}$$= \dfrac{{ - 108{x^2}{y^2}{z^4}}}{{ - 270{x^2}{y^3}{z^2}}} = \dfrac{{2{z^2}}}{{5y}}$  nên C sai.

* $\dfrac{{ - 27x{z^4}}}{{18{y^3}z}}.\dfrac{{4x{y^2}}}{{15{x^2}z}}$$= \dfrac{{ - 108{x^2}{y^2}{z^4}}}{{270{x^2}{y^3}{z^2}}} = \dfrac{{ - 2{z^2}}}{{5y}}$  nên D đúng.

Ta có: $\dfrac{{{{\left( {x - y} \right)}^2}}}{{{{\left( {x + y} \right)}^2}}}:\dfrac{{{{\left( {x - y} \right)}^4}}}{{{{\left( {x + y} \right)}^3}}}$$= \dfrac{{{{\left( {x - y} \right)}^2}}}{{{{\left( {x + y} \right)}^2}}}.\dfrac{{{{\left( {x + y} \right)}^3}}}{{{{\left( {x - y} \right)}^4}}} = \dfrac{{x + y}}{{{{\left( {x - y} \right)}^2}}}$  nên A đúng.

*) $\dfrac{{{{\left( {x - y} \right)}^2}}}{{{{\left( {x + y} \right)}^2}}}:\dfrac{{{{\left( {x - y} \right)}^3}}}{{{{\left( {x + y} \right)}^3}}}$$= \dfrac{{{{\left( {x - y} \right)}^2}}}{{{{\left( {x + y} \right)}^2}}}.\dfrac{{{{\left( {x + y} \right)}^3}}}{{{{\left( {x - y} \right)}^3}}} = \dfrac{{x + y}}{{x - y}}$ nên B sai.

*) $\dfrac{{x - y}}{{{{\left( {x + y} \right)}^2}}}:\dfrac{{{{\left( {x - y} \right)}^4}}}{{{{\left( {x + y} \right)}^3}}}$$= \dfrac{{x - y}}{{{{\left( {x + y} \right)}^2}}}.\dfrac{{{{\left( {x + y} \right)}^3}}}{{{{\left( {x - y} \right)}^4}}} = \dfrac{{x + y}}{{{{\left( {x - y} \right)}^3}}}$ nên C sai.

*) $\dfrac{{ - {{\left( {x - y} \right)}^2}}}{{{{\left( {x + y} \right)}^2}}}:\dfrac{{{{\left( {x - y} \right)}^4}}}{{{{\left( {x + y} \right)}^3}}}$$= \dfrac{{ - {{\left( {x - y} \right)}^2}}}{{{{\left( {x + y} \right)}^2}}}.\dfrac{{{{\left( {x + y} \right)}^3}}}{{{{\left( {x - y} \right)}^4}}} = \dfrac{{ - \left( {x + y} \right)}}{{{{\left( {x - y} \right)}^2}}}$ nên D sai.

Ta có $\dfrac{{{x^2} - 36}}{{2x + 10}}\,.\,\dfrac{3}{{6 - x}}$$= \dfrac{{\left( {x - 6} \right)\left( {x + 6} \right)}}{{2\left( {x + 5} \right)}}.\dfrac{{ - 3}}{{x - 6}} = \dfrac{{ - 3.\left( {x - 6} \right)\left( {x + 6} \right)}}{{2\left( {x + 5} \right)\left( {x - 6} \right)}} = \dfrac{{ - 3\left( {x + 6} \right)}}{{2\left( {x + 5} \right)}}$

Vậy mẫu thức cần tìm là $2\left( {x + 5} \right)$ .

Ta có $\dfrac{{x + 3}}{{{x^2} - 4}}.\dfrac{{8 - 12x + 6{x^2} - {x^3}}}{{9x + 27}}$$= \dfrac{{x + 3}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{{{\left( {2 - x} \right)}^3}}}{{9\left( {x + 3} \right)}} = \dfrac{{ - \left( {x + 3} \right){{\left( {x - 2} \right)}^3}}}{{\left( {x - 2} \right)\left( {x + 2} \right)9\left( {x + 3} \right)}}$ $= \dfrac{{ - {{\left( {x - 2} \right)}^2}}}{{9\left( {x + 2} \right)}} = \dfrac{{{{\left( {x - 2} \right)}^2}}}{{ - 9\left( {x + 2} \right)}}$.

Vậy  các đa thức thích hợp điền vào chỗ trống ở  tử và mẫu lần lượt là ${\left( {x - 2} \right)^2};\,x + 2$

Ta có $B = \dfrac{{x + y}}{x}.\dfrac{{{x^2} + xy}}{6}.\dfrac{{3x}}{{{x^2} - {y^2}}}$$= \dfrac{{x + y}}{x}.\dfrac{{x\left( {x + y} \right)}}{6}.\dfrac{{3x}}{{\left( {x - y} \right)\left( {x + y} \right)}}$ $= \dfrac{{\left( {x + y} \right).x.\left( {x + y} \right).3x}}{{x.6.\left( {x - y} \right)\left( {x + y} \right)}} = \dfrac{{x\left( {x + y} \right)}}{{2\left( {x - y} \right)}}$ .

Ta có ${x^2} + 5x + 4 = {x^2} + x + 4x + 4 = x\left( {x + 1} \right) + 4\left( {x + 1} \right) = \left( {x + 1} \right)\left( {x + 4} \right)$  nên

$A = \dfrac{{x + 4}}{5}.\dfrac{{x + 1}}{{2x}}.\dfrac{{100x}}{{{x^2} + 5x + 4}}$$= \dfrac{{x + 4}}{5}.\dfrac{{x + 1}}{{2x}}.\dfrac{{100x}}{{\left( {x + 1} \right)\left( {x + 4} \right)}} = \dfrac{{\left( {x + 4} \right)\left( {x + 1} \right).100x}}{{10x\left( {x + 1} \right)\left( {x + 4} \right)}} = 10$

Ta có  $C = \dfrac{{2{x^3}{y^2}}}{{{x^2}{y^5}{z^2}}}:\dfrac{{5{x^2}y}}{{4{x^2}{y^5}}}:\dfrac{{ - 8{x^3}{y^2}{z^3}}}{{15{x^5}{y^2}}}$$= \dfrac{{2{x^3}{y^2}}}{{{x^2}{y^5}{z^2}}}.\dfrac{{4{x^2}{y^5}}}{{5{x^2}y}}:\dfrac{{ - 8{x^3}{y^2}{z^3}}}{{15{x^5}{y^2}}} = \dfrac{{8{x^5}{y^7}}}{{5{x^4}{y^6}{z^2}}}:\dfrac{{ - 8{x^3}{y^2}{z^3}}}{{15{x^5}{y^2}}}$

$= \dfrac{{8xy}}{{5{z^2}}}:\dfrac{{ - 8{x^3}{y^2}{z^3}}}{{15{x^5}{y^2}}}$$= \dfrac{{8xy}}{{5{z^2}}}.\dfrac{{15{x^5}{y^2}}}{{ - 8{x^3}{y^2}{z^3}}}$$= \dfrac{{120{x^6}{y^3}}}{{ - 40{x^3}{y^2}{z^5}}}$$= \dfrac{{ - 3{x^3}y}}{{{z^5}}}$ . Vậy $C = \dfrac{{ - 3{x^3}y}}{{{z^5}}}.$ 

Thay $x = 4;y = 1;z =  - 2$ vào $C = \dfrac{{ - 3{x^3}y}}{{{z^5}}}$ ta được $C = \dfrac{{ - {{3.4}^3}.1}}{{{{\left( { - 2} \right)}^5}}} = 6.$ 

Ta có $M = \dfrac{{{x^2} + {y^2} + xy}}{{{x^2} - {y^2}}}:\dfrac{{{x^3} - {y^3}}}{{{x^2} + {y^2} - 2xy}}$$= \dfrac{{{x^2} + xy + {y^2}}}{{{x^2} - {y^2}}}.\dfrac{{{x^2} - 2xy + {y^2}}}{{{x^3} - {y^3}}}$

$= \dfrac{{\left( {{x^2} + xy + {y^2}} \right){{\left( {x - y} \right)}^2}}}{{\left( {x - y} \right)\left( {x + y} \right)\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)}}$ $= \dfrac{1}{{x + y}} \Rightarrow M = \dfrac{1}{{x + y}}$ .

Và $N = \dfrac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}:\dfrac{{{x^2} - 2xy + {y^2}}}{{{x^4} - {y^4}}}$$= \dfrac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}.\dfrac{{{x^4} - {y^4}}}{{{x^2} - 2xy + {y^2}}}$ $= \dfrac{{\left( {x - y} \right)\left( {x + y} \right)\left( {{x^2} + {y^2}} \right)\left( {{x^2} - {y^2}} \right)}}{{\left( {{x^2} + {y^2}} \right){{\left( {x - y} \right)}^2}}}$

$= \dfrac{{\left( {x + y} \right)\left( {{x^2} - {y^2}} \right)}}{{x - y}} = \dfrac{{\left( {x + y} \right)\left( {x - y} \right)\left( {x + y} \right)}}{{x - y}} = {\left( {x + y} \right)^2}$ $\Rightarrow N = {\left( {x + y} \right)^2}$.

Với $x + y = 6$ thì $M = \dfrac{1}{{{{\left( {x + y} \right)}^2}}} = \dfrac{1}{{{6^2}}} = \dfrac{1}{{36}}$  và $N = {\left( {x + y} \right)^2} = {6^2} = 36$ . Nên $M < N$ .

Ta có  $P = \dfrac{{{x^4} + 3{x^3} + 5}}{{{x^3} + 1}}.\dfrac{{x + 2}}{{x + 1}}.\dfrac{{{x^2} - x + 1}}{{{x^4} + 3{x^3} + 5}}$$= \dfrac{{\left( {{x^4} + 3{x^3} + 5} \right)\left( {x + 2} \right)\left( {{x^2} - x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)\left( {x + 1} \right)\left( {{x^4} + 3{x^3} + 5} \right)}}$ $= \dfrac{{x + 2}}{{{{\left( {x + 1} \right)}^2}}}$ .

Vậy cả hai bạn Mai và Đào đều làm sai.

Ta có $\dfrac{{{x^2} + 5x}}{{x - 2}}.Q = \dfrac{{{x^2} - 25}}{{{x^2} - 2x}}$$\Leftrightarrow Q = \dfrac{{{x^2} - 25}}{{{x^2} - 2x}}:\dfrac{{{x^2} + 5x}}{{x - 2}}$ $\Leftrightarrow Q = \dfrac{{{x^2} - 25}}{{{x^2} - 2x}}.\dfrac{{x - 2}}{{{x^2} + 5x}}$ $\Leftrightarrow Q = \dfrac{{\left( {x - 5} \right)\left( {x + 5} \right).\left( {x - 2} \right)}}{{x\left( {x - 2} \right)x\left( {x + 5} \right)}} \Leftrightarrow Q = \dfrac{{x - 5}}{{{x^2}}}$ .

Ta có:

$\begin{array}{l}\,N:\dfrac{{{x^2} + x + 1}}{{2x + 2}} = \dfrac{{x + 1}}{{{x^3} - 1}}\\N = \dfrac{{x + 1}}{{{x^3} - 1}} \cdot \dfrac{{{x^2} + x + 1}}{{2x + 2}}\\N = \dfrac{{x + 1}}{{(x - 1)({x^2} + x + 1)}} \cdot \dfrac{{{x^2} + x + 1}}{{2(x + 1)}}\\N = \dfrac{1}{{2(x - 1)}}.\end{array}$.

$\begin{array}{l}\,\,\,\,\,\,\dfrac{1}{x} \cdot \dfrac{x}{{x + 1}} \cdot \dfrac{{x + 1}}{{x + 2}} \cdot \dfrac{{x + 2}}{{x + 3}} \cdot \dfrac{{x + 3}}{{x + 4}} \cdot \dfrac{{x + 4}}{{x + 5}} \cdot \dfrac{{x + 5}}{{x + 6}} = 1\\ \Leftrightarrow \dfrac{1}{{x + 6}} = 1\\ \Leftrightarrow x + 6 = 1\\ \Leftrightarrow x =  - 5\end{array}$.

$\begin{array}{l}A = \dfrac{{{5^2} - 1}}{{{3^2} - 1}}:\dfrac{{{9^2} - 1}}{{{7^2} - 1}}:\dfrac{{{{13}^2} - 1}}{{{{11}^2} - 1}}:...:\dfrac{{{{55}^2} - 1}}{{{{53}^2} - 1}}\\A = \dfrac{{{5^2} - 1}}{{{3^2} - 1}} \cdot \dfrac{{{7^2} - 1}}{{{9^2} - 1}} \cdot \dfrac{{{{11}^2} - 1}}{{{{13}^2} - 1}} \cdot  \cdot  \cdot \dfrac{{{{53}^2} - 1}}{{{{55}^2} - 1}}\\A = \dfrac{{4.6}}{{2.4}} \cdot \dfrac{{6.8}}{{8.10}} \cdot \dfrac{{10.12}}{{12.14}} \cdot  \cdot  \cdot \dfrac{{52.54}}{{54.56}}\\A = \dfrac{6}{2} \cdot \dfrac{6}{{10}} \cdot \dfrac{{10}}{{14}}.....\dfrac{{52}}{{56}}\\A = 3 \cdot \dfrac{6}{{56}} = \dfrac{9}{{28}}.\end{array}$

Ta có:

$\dfrac{{{{\left( {xy + yz + zx} \right)}^2} - \left( {{x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2}} \right)}}{{{x^2} + {y^2} + {x^2}}}:\dfrac{{{{\left( {x + y + z} \right)}^2}}}{{{x^2} + {y^2} + {z^2}}}$

$= \dfrac{{{x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2} + 2\left( {x{y^2}z + {z^2}yz + {y^2}zx} \right) - \left( {{x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2}} \right)}}{{{x^2} + {y^2} + {z^2}}}.\dfrac{{{x^2} + {y^2} + {z^2}}}{{{{\left( {x + y + z} \right)}^2}}}$

$= \dfrac{{2xyz\left( {x + y + z} \right)}}{{{{\left( {x + y + z} \right)}^2}}} = \dfrac{{2xyz}}{{\left( {x + y + z} \right)}} = \dfrac{{2xyz}}{{2x}} = yz$ (vì $x = y + z$)

Ta có: ${x^2} + \left( {a - b} \right)x - ab = {x^2} + ax - bx - ab$$= x\left( {x + a} \right) - b\left( {x + a} \right) = \left( {x - b} \right)\left( {x + a} \right)$

${x^2} - \left( {a - b} \right)x - ab = {x^2} - ax + bx - ab$$= x\left( {x - a} \right) + b\left( {x - a} \right) = \left( {x - a} \right)\left( {x + b} \right)$

${x^2} - \left( {a + b} \right)x + ab = {x^2} - ax - bx + ab$$= x\left( {x - a} \right) - b\left( {x - a} \right) = \left( {x - b} \right)\left( {x - a} \right)$

${x^2} + \left( {a + b} \right)x + ab = {x^2} + ax + bx + ab$$= x\left( {x + a} \right) + b\left( {x + a} \right) = \left( {x + a} \right)\left( {x + b} \right)$

${x^2} - \left( {b - 1} \right)x - b = {x^2} - bx + x - b$$= x\left( {x - b} \right) + x - b = \left( {x - b} \right)\left( {x + 1} \right)$

${x^2} + \left( {b + 1} \right)x + b = {x^2} + bx + x + b$$= x\left( {x + b} \right) + x + b = \left( {x + b} \right)\left( {x + 1} \right)$

${x^2} - \left( {b + 1} \right)x + b = {x^2} - bx - x + b$$= x\left( {x - b} \right) - \left( {x - b} \right) = \left( {x - 1} \right)\left( {x - b} \right)$

${x^2} - \left( {1 - b} \right)x - b = {x^2} - x + bx - b$$= x\left( {x - 1} \right) + b\left( {x - 1} \right) = \left( {x + b} \right)\left( {x - 1} \right)$

Khi đó $T = \left[ {\dfrac{{{x^2} + \left( {a - b} \right)x - ab}}{{{x^2} - \left( {a - b} \right)x - ab}}.\dfrac{{{x^2} - \left( {a + b} \right)x + ab}}{{{x^2} + \left( {a + b} \right)x + ab}}} \right]:\left[ {\dfrac{{{x^2} - \left( {b - 1} \right)x - b}}{{{x^2} + \left( {b + 1} \right)x + b}}.\dfrac{{{x^2} - \left( {b + 1} \right)x + b}}{{{x^2} - \left( {1 - b} \right)x - b}}} \right]$

$= \left[ {\dfrac{{\left( {x - b} \right)\left( {x + a} \right)}}{{\left( {x - a} \right)\left( {x + b} \right)}}.\dfrac{{\left( {x - a} \right)\left( {x - b} \right)}}{{\left( {x + a} \right)\left( {x + b} \right)}}} \right]$$:\left[ {\dfrac{{\left( {x - b} \right)\left( {x + 1} \right)}}{{\left( {x + b} \right)\left( {x + 1} \right)}}.\dfrac{{\left( {x - 1} \right)\left( {x - b} \right)}}{{\left( {x + b} \right)\left( {x - 1} \right)}}} \right]$

$= \dfrac{{{{\left( {x - b} \right)}^2}}}{{{{\left( {x + b} \right)}^2}}}:\dfrac{{{{\left( {x - b} \right)}^2}}}{{{{\left( {x + b} \right)}^2}}} = 1$

Vậy $T = 1.$