Các phép tính với số hữu tỉ

$\frac{2}{3} - \frac{{ - 3}}{7} = \frac{2}{3} + \frac{3}{7} = \frac{{14}}{{21}} + \frac{9}{{21}} = \frac{{23}}{{21}}$

$\frac{{ - 2}}{3} + \frac{2}{5}:\frac{{ - 3}}{5} = \frac{{ - 2}}{3} + \frac{2}{5}.\frac{{ - 5}}{3} = \frac{{ - 2}}{3} + \frac{{ - 2}}{3} = \frac{{ - 4}}{3}$

Ta có $\dfrac{{ - 2}}{{13}} + \dfrac{{ - 11}}{{26}} = \dfrac{{ - 4}}{{26}} + \dfrac{{ - 11}}{{26}} = \dfrac{{ - 15}}{{26}}$

Do đó kết quả là số hữu tỉ âm.

Ta có $- \dfrac{6}{7}.\dfrac{{21}}{{12}} =  - \dfrac{6}{7}.\dfrac{7}{4} = \dfrac{{ - 6}}{4} =  - \dfrac{3}{2}$

$\begin{array}{l}3\frac{1}{2} - \frac{2}{3}:\frac{5}{{ - 3}} - 0,3\\ = \frac{7}{2} - \frac{2}{3}.\frac{{ - 3}}{5} - \frac{3}{{10}}\\ = \frac{7}{2} - \frac{{ - 2}}{5} - \frac{3}{{10}}\\ = \frac{7}{2} + \frac{2}{5} - \frac{3}{{10}}\\ = \frac{{35}}{{10}} + \frac{4}{{10}} - \frac{3}{{10}}\\ = \frac{{36}}{{10}}\\ = \frac{{18}}{5}\\ = 3\frac{3}{5}\end{array}$

$\begin{array}{l} - 0,12 - 2x =  - 1\frac{2}{5}\\ \Leftrightarrow \frac{{ - 12}}{{100}} - 2x = \frac{{ - 7}}{5}\\ \Leftrightarrow \frac{{ - 3}}{{25}} - 2x = \frac{{ - 7}}{5}\\ \Leftrightarrow 2x = \frac{{ - 3}}{{25}} - (\frac{{ - 7}}{5})\\ \Leftrightarrow 2x = \frac{{ - 3}}{{25}} + \frac{{35}}{{25}}\\ \Leftrightarrow 2x = \frac{{32}}{{25}}\\ \Leftrightarrow x = \frac{{32}}{{25}}:2\\ \Leftrightarrow x = \frac{{32}}{{25}}.\frac{1}{2}\\ \Leftrightarrow x = \frac{{16}}{{25}}\end{array}$

$\dfrac{{ - 3}}{{14}} = \dfrac{{7 - 10}}{{14}} = \dfrac{7}{{14}} - \dfrac{{10}}{{14}}$$= \dfrac{1}{2}-\dfrac{5}{7}$ nên C đúng

+) Đáp án B: $\dfrac{1}{{14}} - \dfrac{1}{7} = \dfrac{1}{{14}} - \dfrac{2}{{14}} = \dfrac{{ - 1}}{{14}}\ne \dfrac{{ - 3}}{{14}}$ nên loại B.

+) Đáp án A: $\dfrac{2}{3} - \dfrac{5}{7} = \dfrac{{14}}{{21}} - \dfrac{{15}}{{21}} = \dfrac{{ - 1}}{{21}}\ne \dfrac{{ - 3}}{{14}}$ nên loại A.

+) Đáp án D: $\dfrac{3}{{14}} - \dfrac{5}{{14}} = \dfrac{{ - 2}}{{14}} = \dfrac{{ - 1}}{7}\ne \dfrac{{ - 3}}{{14}}$ nên loại D.

Ta có $\dfrac{3}{2}.\dfrac{4}{7} = \dfrac{{3.4}}{{2.7}} = \dfrac{6}{7} > 0$

$M = \dfrac{{11}}{{20}}.68 - 4,2.2022 + 4\dfrac{1}{5}.2022$$- 68.( - 0,45)$$= 0,55.68 - 4,2.2022 + 4,2.2022$$+ 68.0,45$$= (0,55.68 + 68.0,45)$$+ ( - 4,2.2022 + 4,2.2022)$$= 68.(0,55 + 0,45) + 0$$= 68.1\\ = 68$

$\begin{array}{l}\frac{{x + \frac{3}{2}}}{6} = \frac{{ - 5}}{{12}}\\ \Leftrightarrow \frac{{2.(x + \frac{3}{2})}}{{12}} = \frac{{ - 5}}{{12}}\\ \Leftrightarrow \frac{{2x + 3}}{{12}} = \frac{{ - 5}}{{12}}\\ \Leftrightarrow 2x + 3 =  - 5\\ \Leftrightarrow 2x =  - 5 - 3\\ \Leftrightarrow 2x =  - 8\\ \Leftrightarrow x =  - 4\end{array}$

Vậy x = -4

$x + \dfrac{1}{2} = \dfrac{3}{4}$

$x\,\, = \dfrac{3}{4} - \dfrac{1}{2}$

$x = \dfrac{3}{4} - \dfrac{2}{4}$

$x = \dfrac{1}{4}$

Ta có

$A = \dfrac{{ - 5}}{6}.\dfrac{{12}}{{ - 7}}.\left( {\dfrac{{ - 21}}{{15}}} \right) = \dfrac{{\left( { - 5} \right).12.\left( { - 21} \right)}}{{6.\left( { - 7} \right).15}} = \dfrac{{\left( { - 5} \right).2.6.\left( { - 7} \right).3}}{{6.\left( { - 7} \right).5.3}} =  - 2$

$B = \dfrac{1}{6}.\dfrac{9}{{ - 8}}.\left( {\dfrac{{ - 12}}{{11}}} \right) = \dfrac{{9.\left( { - 12} \right)}}{{6.\left( { - 8} \right).11}} = \dfrac{9}{{44}}$

Suy ra $A < B$ .

Ta có:

$\begin{array}{l}\frac{{\frac{3}{{11}} + \frac{3}{{17}} - \frac{3}{{23}} + \frac{3}{{29}}}}{{\frac{7}{{11}} + \frac{7}{{17}} - \frac{7}{{23}} + \frac{7}{{29}}}}\\ = \frac{{3.(\frac{1}{{11}} + \frac{1}{{17}} - \frac{1}{{23}} + \frac{1}{{29}})}}{{7.(\frac{1}{{11}} + \frac{1}{{17}} - \frac{1}{{23}} + \frac{1}{{29}})}}\\ = \frac{3}{7}\end{array}$

Ta xét 2 trường hợp sau:

+ Trường hợp 1:

$\left\{ {_{x - 1 > 0}^{2x + 7 < 0}} \right. \Leftrightarrow \left\{ {_{x > 1}^{2x <  - 7}} \right. \Leftrightarrow \left\{ {_{x > 1}^{x < \frac{{ - 7}}{2}}} \right.$ ( Vô lí)

+ Trường hợp 2:

$\left\{ {_{x - 1 < 0}^{2x + 7 > 0}} \right. \Leftrightarrow \left\{ {_{x < 1}^{2x >  - 7}} \right. \Leftrightarrow \left\{ {_{x < 1}^{x > \frac{{ - 7}}{2}}} \right. \Leftrightarrow \frac{{ - 7}}{2} < x < 1$

Mà x nguyên

$\Rightarrow x \in \{  - 3; - 2; - 1;0\}$

Vậy có 4 giá trị của x thỏa mãn

Ta có $A = \dfrac{1}{3} - \left[ {\left( { - \dfrac{5}{4}} \right) - \left( {\dfrac{1}{4} + \dfrac{3}{8}} \right)} \right]$

$= \dfrac{1}{3} - \left[ {\left( { - \dfrac{5}{4}} \right) - \left( {\dfrac{2}{8} + \dfrac{3}{8}} \right)} \right]$

$= \dfrac{1}{3} - \left[ {\left( { - \dfrac{5}{4}} \right) - \dfrac{5}{8}} \right]$

$= \dfrac{1}{3} - \left[ {\left( { - \dfrac{{10}}{8}} \right) - \dfrac{5}{8}} \right]$

$= \dfrac{1}{3} - \left( { - \dfrac{{15}}{8}} \right)$

$= \dfrac{1}{3} + \dfrac{{15}}{8}$

$= \dfrac{8}{{24}} + \dfrac{{45}}{{24}}$

$= \dfrac{{53}}{{24}}$

Vậy  $A = \dfrac{{53}}{{24}} > \dfrac{{48}}{{24}} = 2$  hay $A > 2$ .

Ta có $x:\left( {\dfrac{2}{5} - 1\dfrac{2}{5}} \right) = 1$

$x:\left( {\dfrac{2}{5} - \dfrac{7}{5}} \right) = 1$

$x:\left( {\dfrac{{ - 5}}{5}} \right) = 1$

$x:\left( { - 1} \right) = 1$

$x = 1.\left( { - 1} \right)$

$x =  - 1$

Vậy $x =  - 1$ .

$\begin{array}{l}(\frac{1}{3} - 1).(\frac{1}{4} - 1)....(\frac{1}{{2022}} - 1)\\ = \frac{{ - 2}}{3}.\frac{{ - 3}}{4}.....\frac{{ - 2021}}{{2022}}\\ = \frac{2}{{2022}}\\ = \frac{1}{{1011}}\end{array}$

Lời giải

Đặt $Q =  P – 3  = 3 + 30 + 33 + 36 +…+ 3300 – 3 = 30 + 33 + 36 +…+ 3300$

Số số hạng của tổng Q là:

$\dfrac{{3300 - 30}}{3} + 1 = 1091$

Tổng Q là: $\dfrac{{(3300 + 30).1091}}{2} = 1816515$

Ta được  $5x = 1816515$

Do đó:$x = 1816515 : 5 = 363 303$

$x + y = \dfrac{a}{b} + \dfrac{c}{d} = \dfrac{{ad}}{{bd}} + \dfrac{{cb}}{{bd}} = \dfrac{{ad + cb}}{{bd}}.$

Ta có $\dfrac{1}{3}x + \dfrac{2}{5}\left( {x - 1} \right) = 0$

$\dfrac{1}{3}x + \dfrac{2}{5}x - \dfrac{2}{5} = 0$

$\dfrac{1}{3}x + \dfrac{2}{5}x = \dfrac{2}{5}$

$x\left( {\dfrac{1}{3} + \dfrac{2}{5}} \right) = \dfrac{2}{5}$

$x.\left( {\dfrac{5}{{15}} + \dfrac{6}{{15}}} \right) = \dfrac{2}{5}$

$x.\dfrac{{11}}{{15}} = \dfrac{2}{5}$

$x = \dfrac{2}{5}:\dfrac{{11}}{{15}}$

$x = \dfrac{2}{5}.\dfrac{{15}}{{11}}$

$x = \dfrac{{2.15}}{{5.11}}$

$x = \dfrac{6}{{11}}$

Vậy có một giá trị của $x$ thoả mãn điều kiện.