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.

\(\dfrac{{2a}}{{a - 1}}\) .

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

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

\(\dfrac{{x + 9}}{{x + 10}}\)

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

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

\(\dfrac{1}{{2 - x}}\)

\(\dfrac{{x + 2}}{{x - 2}}\)

\(\dfrac{{x + 2}}{{2 - x}}\)

\(6x;{x^2} + 4\)

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

\(3x;{x^2} + 4\)

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

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

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

\(1\)

\(2\)

\( - 2\)

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

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

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