a. \(\displaystyle{A \over {2x - 1}} = {{6{x^2} + 3x} \over {4{x^2} - 1}}\)
\( \Rightarrow A\left( {4{x^2} - 1} \right)\)\(\, = \left( {2x - 1} \right).\left( {6{x^2} + 3x} \right)\)
\( \Rightarrow A\left( {2x - 1} \right)\left( {2x + 1} \right) \)\(\,= \left( {2x - 1} \right).3x\left( {2x + 1} \right)\)
\( \Rightarrow A = 3x\)
Ta có: \(\displaystyle {{3x} \over {2x - 1}} = {{6{x^2} + 3x} \over {4{x^2} - 1}}\)
b. \(\displaystyle{{4{x^2} - 3x - 7} \over A} = {{4x - 7} \over {2x + 3}}\)
\( \Rightarrow \left( {4{x^2} - 3x - 7} \right)\left( {2x + 3} \right) \)\(\,= A\left( {4x - 7} \right) \)
\(\Rightarrow \left( {4{x^2} + 4x - 7x - 7} \right)\left( {2x + 3} \right) \)\(\,= A\left( {4x - 7} \right) \)
\(\Rightarrow \left[ {4x\left( {x + 1} \right) - 7\left( {x + 1} \right)} \right]\left( {2x + 3} \right) \)\(\,= A\left( {4x - 7} \right) \)
\(\Rightarrow \left( {x + 1} \right)\left( {4x - 7} \right)\left( {2x + 3} \right)\)\(\, = A\left( {4x - 7} \right) \)
\(\Rightarrow A = \left( {x + 1} \right)\left( {2x + 3} \right)\)\(\, = 2{x^2} + 3x + 2x + 3 \)\(\,= 2{x^2} + 5x + 3 \)
Ta có: \(\displaystyle {{4{x^2} - 3x - 7} \over {2{x^2} + 5x + 3}} = {{4x - 7} \over {2x + 3}}\)
c. \(\displaystyle {{4{x^2} - 7x + 3} \over {{x^2} - 1}} = {A \over {{x^2} + 2x + 1}}\)
\( \Rightarrow \left( {4{x^2} - 7x + 3} \right).\left( {{x^2} + 2x + 1} \right)\)\(\, = A.\left( {{x^2} - 1} \right) \)
\( \Rightarrow \left( {4{x^2} - 4x - 3x + 3} \right).{\left( {x + 1} \right)^2}\)\(\, = A\left( {x + 1} \right)\left( {x - 1} \right) \)
\( \Rightarrow \left[ {4x\left( {x - 1} \right) - 3\left( {x - 1} \right)} \right].{\left( {x + 1} \right)^2} \)\(\,= A\left( {x + 1} \right)\left( {x - 1} \right) \)
\( \Rightarrow \left( {x - 1} \right)\left( {4x - 3} \right){\left( {x + 1} \right)^2}\)\(\, = A\left( {x + 1} \right)\left( {x - 1} \right) \)
\(\Rightarrow A = \left( {4x - 3} \right)\left( {x + 1} \right) \)\(\,= 4{x^2} + 4x - 3x - 3 = 4{x^2} + x - 3 \)
Ta có: \(\displaystyle\displaystyle{{4{x^2} - 7x + 3} \over {{x^2} - 1}} = {{4{x^2} + x - 3} \over {{x^2} + 2x + 1}}\)
d. \(\displaystyle{{{x^2} - 2x} \over {2{x^2} - 3x - 2}} = {{{x^2} + 2x} \over A}\)
\( \Rightarrow \left( {{x^2} - 2x} \right).A \)\(\,= \left( {2{x^2} - 3x - 2} \right)\left( {{x^2} + 2x} \right) \)
\(\Rightarrow x\left( {x - 2} \right).A \)\(\,= \left( {2{x^2} - 4x + x - 2} \right).x\left( {x + 2} \right) \)
\(\Rightarrow x\left( {x - 2} \right).A \)\(\,= \left[ {2x\left( {x - 2} \right) + \left( {x - 2} \right)} \right].x\left( {x + 2} \right) \)
\( \Rightarrow x\left( {x - 2} \right).A\)\(\, = \left( {2x + 1} \right)\left( {x - 2} \right).x.\left( {x + 2} \right) \)
\( \Rightarrow A = \left( {2x + 1} \right)\left( {x + 2} \right)\)\(\, = 2{x^2} + 4x + x + 2 = 2{x^2} + 5x + 2 \)
Ta có : \(\displaystyle{{{x^2} - 2x} \over {2{x^2} - 3x - 2}} = {{{x^2} + 2x} \over 2{x^2} + 5x + 2}\)
