Đề thi Giữa kì 1 Toán 8 có đáp án (Đề 4)

Thực hiện phép tính:1) \(\frac{{x - 5}}{{x - 2}} - \frac{{x + 4}}{{2x - {x^2}}}\);2) \(\frac{{x - 3}}{{x + 2}} + \frac{{4x}}{{x - 3}} - \frac{{8x + 4{x^2}}}{{{x^2} - x - 6}}\).

3/4

Thực hiện phép tính:

1) \(\frac{{x - 5}}{{x - 2}} - \frac{{x + 4}}{{2x - {x^2}}}\);

2) \(\frac{{x - 3}}{{x + 2}} + \frac{{4x}}{{x - 3}} - \frac{{8x + 4{x^2}}}{{{x^2} - x - 6}}\).

0/3000 ký tự
Giải thích

Hướng dẫn giải

1) \(\frac{{x - 5}}{{x - 2}} - \frac{{x + 4}}{{2x - {x^2}}}\)

\( = \frac{{x - 5}}{{x - 2}} + \frac{{x + 4}}{{{x^2} - 2x}}\)

\( = \frac{{x - 5}}{{x - 2}} + \frac{{x + 4}}{{x\left( {x - 2} \right)}}\)

\( = \frac{{x\left( {x - 5} \right)}}{{x\left( {x - 2} \right)}} + \frac{{x + 4}}{{x\left( {x - 2} \right)}}\)

\( = \frac{{{x^2} - 5x + x + 4}}{{x\left( {x - 2} \right)}}\)

\( = \frac{{{x^2} - 4x + 4}}{{x\left( {x - 2} \right)}}\)

\( = \frac{{{{\left( {x - 2} \right)}^2}}}{{x\left( {x - 2} \right)}}\)

\( = \frac{{x - 2}}{x}\)

2) \(\frac{{x - 3}}{{x + 2}} + \frac{{4x}}{{x - 3}} - \frac{{8x + 4{x^2}}}{{{x^2} - x - 6}}\)

\( = \frac{{x - 3}}{{x + 2}} + \frac{{4x}}{{x - 3}} - \frac{{8x + 4{x^2}}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\)

\( = \frac{{{{\left( {x - 3} \right)}^2}}}{{x + 2}} + \frac{{4x\left( {x + 2} \right)}}{{x - 3}} - \frac{{8x + 4{x^2}}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\)

\( = \frac{{{{\left( {x - 3} \right)}^2} + 4{x^2} + 8x - 8x - 4{x^2}}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\)

\( = \frac{{{{\left( {x - 3} \right)}^2}}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\)

\( = \frac{{x - 3}}{{x + 2}}\)