Câu 2 trang 24 Sách bài tập (SBT) Toán 8 tập 1

Dùng định nghĩa hai phân thức bằng nhau, hãy tìm đa thức A trong mỗi đẳng thức sau:

a. \({A \over {2x – 1}} = {{6{x^2} + 3x} \over {4{x^2} – 1}}\)

b. \({{4{x^2} – 3x – 7} \over A} = {{4x – 7} \over {2x + 3}}\)

c. \({{4{x^2} – 7x + 3} \over {{x^2} – 1}} = {A \over {{x^2} + 2x + 1}}\)

d. \({{{x^2} – 2x} \over {2{x^2} – 3x – 2}} = {{{x^2} + 2x} \over A}\)

Giải:

a. \({A \over {2x – 1}} = {{6{x^2} + 3x} \over {4{x^2} – 1}}\)

\( \Rightarrow A\left( {4{x^2} – 1} \right) = \left( {2x – 1} \right).\left( {6{x^2} + 3x} \right)\)

\( \Rightarrow A\left( {2x – 1} \right)\left( {2x + 1} \right) = \left( {2x – 1} \right).3x\left( {2x + 1} \right)\)

\( \Rightarrow A = 3x\)

Ta có: \({{3x} \over {2x – 1}} = {{6{x^2} + 3x} \over {4{x^2} – 1}}\)

b. \({{4{x^2} – 3x – 7} \over A} = {{4x – 7} \over {2x + 3}}\)

 \(\eqalign{  &  \Rightarrow \left( {4{x^2} – 3x – 7} \right)\left( {2x + 3} \right) = A\left( {4x – 7} \right)  \cr  &  \Rightarrow \left( {4{x^2} + 4x – 7x – 7} \right)\left( {2x + 3} \right) = A\left( {4x – 7} \right)  \cr  &  \Rightarrow \left[ {4x\left( {x + 1} \right) – 7\left( {x + 1} \right)} \right]\left( {2x + 3} \right) = A\left( {4x – 7} \right)  \cr  &  \Rightarrow \left( {x – 1} \right)\left( {4x – 7} \right)\left( {2x + 3} \right) = A\left( {4x – 7} \right)  \cr  &  \Rightarrow A = \left( {x + 1} \right)\left( {2x + 3} \right) = 2{x^2} + 3x + 2x + 3 = 2{x^2} + 5x + 3 \cr} \)

Ta có: \({{4{x^2} – 3x – 7} \over {2{x^2} + 5x + 3}} = {{4x – 7} \over {2x + 3}}\)

c. \({{4{x^2} – 7x + 3} \over {{x^2} – 1}} = {A \over {{x^2} + 2x + 1}}\)    

\(\eqalign{  &  \Rightarrow \left( {4{x^2} – 7x + 3} \right).\left( {{x^2} + 2x + 1} \right) = A.\left( {{x^2} – 1} \right)\left( {{\pi  \over 2} – \theta } \right)  \cr  &  \Rightarrow \left( {4{x^2} – 4x – 3x + 3} \right).{\left( {x + 1} \right)^2} = A\left( {x + 1} \right)\left( {x – 1} \right)  \cr  &  \Rightarrow \left[ {4x\left( {x – 1} \right) – 3\left( {x – 1} \right)} \right].{\left( {x + 1} \right)^2} = A\left( {x + 1} \right)\left( {x – 1} \right)  \cr  &  \Rightarrow \left( {x – 1} \right)\left( {4x – 3} \right){\left( {x + 1} \right)^2} = A\left( {x + 1} \right)\left( {x – 1} \right)  \cr  &  \Rightarrow A = \left( {4x – 3} \right)\left( {x + 1} \right) = 4{x^2} + 4x – 3x – 3 = 4{x^2} + x – 3 \cr} \)

Ta có:    \({{4{x^2} – 7x + 3} \over {{x^2} – 1}} = {{4{x^2} + x – 3} \over {{x^2} + 2x + 1}}\)

d. \({{{x^2} – 2x} \over {2{x^2} – 3x – 2}} = {{{x^2} + 2x} \over A}\)    

\(\eqalign{  &  \Rightarrow \left( {{x^2} – 2x} \right).A = \left( {2{x^2} – 3x – 2} \right)\left( {{x^2} + 2x} \right)  \cr  &  \Rightarrow x\left( {x – 2} \right).A = \left( {2{x^2} – 4x + x – 2} \right).x\left( {x + 2} \right)  \cr  &  \Rightarrow x\left( {x – 2} \right).A = \left[ {2x\left( {x – 2} \right) + \left( {x – 2} \right)} \right].x\left( {x + 2} \right)  \cr  &  \Rightarrow x\left( {x – 2} \right).A = \left( {2x + 1} \right)\left( {x – 2} \right).x.\left( {x + 2} \right)  \cr &  \Rightarrow A = \left( {2x + 1} \right)\left( {x + 2} \right) = 2{x^2} + 4x + x + 2 = 2{x^2} + 5x + 2 \cr} \)

Ta có: \({{{x^2} – 2x} \over {2{x^2} – 3x – 2}} = {{{x^2} + 2x} \over {{x^2} + 2x + 1}}\)