Đề kiểm tra 45 phút chương 2: Phân thức đại số - Đề số 1

$\dfrac{3}{2}$

$\dfrac{3}{{2(x - 4)}}$

$\dfrac{{ - 3}}{{2(x - 4)}}$

$12{x^2}{y^3}$

$6{x^3}{y^2}$

$12{x^4}y$

$A = 3$

$A > 4$

$A = 1$

$\dfrac{{7{x^2}z}}{3}$

$\dfrac{{ - 7xz}}{{3y}}$

$\dfrac{{ - 7{x^2}}}{{3y}}$

$\dfrac{1}{{xy}} = \dfrac{z}{{xyz}},\dfrac{1}{{yz}} = \dfrac{x}{{xyz}},\dfrac{3}{{xz}} = \dfrac{3}{{xyz}}$.

$\dfrac{1}{{xy}} = \dfrac{z}{{xyz}},\dfrac{1}{{yz}} = \dfrac{1}{{xyz}},\dfrac{3}{{xz}} = \dfrac{{3y}}{{xyz}}$.

$\dfrac{1}{{xy}} = \dfrac{1}{{xyz}},\dfrac{1}{{yz}} = \dfrac{x}{{xyz}},\dfrac{3}{{xz}} = \dfrac{{3y}}{{xyz}}$.

$A.B = C.D$

$A.C = B.D$

$AC < B.D$

Mẫu thức chung của các phân thức $\dfrac{{2 - a}}{{3a}};\dfrac{1}{4}$ là $12a$.

Mẫu thức chung của các phân thức $\dfrac{1}{{6a}};\dfrac{{4a + 1}}{{18ab}};\dfrac{{10a}}{{9b}}$ là $18ab$.

Mẫu thức chung của các phân thức $\dfrac{1}{{{{\left( {x - 2y} \right)}^2}}};\dfrac{{5x}}{{{{\left( {x - 2y} \right)}^4}}};\dfrac{1}{{3x}}$ là $3x{\left( {x - 2y} \right)^4}$.

$\dfrac{{{x^2} - 3xy}}{{21{y^2} - 7xy}} = \dfrac{x}{{7y}}$

$\dfrac{{2x + 4}}{{{x^2} - x - 6}} = \dfrac{4}{{x - 3}}$

$\dfrac{{2x - 6y}}{{{x^2} - 9{y^2}}} = \dfrac{2}{{x + 9y}}$

$\dfrac{{A.C}}{B}$

$\dfrac{{A + C}}{{{B^2}}}$

$\dfrac{{A + C}}{{2B}}$

$- x$

$2x$

$\dfrac{x}{2}$          

$\dfrac{A}{B}.\dfrac{B}{A} = 1$.  

$\dfrac{A}{B}.\dfrac{C}{D} = \dfrac{C}{D}.\dfrac{A}{B}$.

$\dfrac{A}{B}.\left( {\dfrac{C}{D}.\dfrac{E}{F}} \right) = \dfrac{E}{F}.\left( {\dfrac{C}{D}.\dfrac{A}{B}} \right)$.

$x = 2$.

$x > 2$.

$x < 2$.

$2022$.

$2016$.

$2024$.

$Q = \dfrac{{x + 1}}{3}$

$Q = \dfrac{{{x^2} + 1}}{{ - 3}}$

$Q = \dfrac{{{x^2} - 1}}{3}$

$\dfrac{4}{3}$

$\dfrac{1}{2}$

$1$

$x = 2$

$x \ne 1$

$x =  - 1$

$1$

$\dfrac{4}{5}$

$2$

Bạn Nam đúng, bạn Minh sai.

Bạn Nam sai, bạn Minh đúng.

Hai bạn đều đúng.

$\dfrac{{2a}}{{a - 1}}$ .

$\dfrac{{2{a^2} + 2a}}{{\left( {a - 1} \right)\left( {a + 1} \right)}}$.

$- \dfrac{{2{a^2}}}{{\left( {a - 1} \right)\left( {a + 1} \right)}}$. 

$\dfrac{{x + 9}}{{x + 10}}$

$\dfrac{1}{{x + 10}}$

$\dfrac{1}{{x(x + 1)...(x + 10)}}$

$\dfrac{1}{{2 - x}}$

$\dfrac{{x + 2}}{{x - 2}}$

$\dfrac{{x + 2}}{{2 - x}}$

$6x;{x^2} + 4$

$x;5\left( {{x^2} + 4} \right)$

$3x;{x^2} + 4$

 $\dfrac{x}{{1 - {x^2}}} + \dfrac{y}{{1 - {y^2}}} + \dfrac{z}{{1 - {z^2}}} = \dfrac{{xyz}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right)}}$

 $\dfrac{x}{{1 - {x^2}}} + \dfrac{y}{{1 - {y^2}}} + \dfrac{z}{{1 - {z^2}}} = \dfrac{{3xyz}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right)}}$

$\dfrac{x}{{1 - {x^2}}} + \dfrac{y}{{1 - {y^2}}} + \dfrac{z}{{1 - {z^2}}} = \dfrac{{xyz\left( {x + y + z} \right)}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right)}}$

$1$

$2$

$- 2$

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

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

$\dfrac{{{{\left( {xy + yz + zx} \right)}^2} - \left( {{x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2}} \right)}}{{{x^2} + {y^2} + {x^2}}}:\dfrac{{{{\left( {x + y + z} \right)}^2}}}{{{x^2} + {y^2} + {z^2}}} = 1$