Thực hiện phép tính:
a) x^2+y^2/x-y+2xy/y-x
b)5x-7/2(x-1)-4x/x^2-1+9-3x/2(x-1)
Dấu / là mình thay cho dấu phân số ạ
Thực hiện phép tính:
a) x^2+y^2/x-y+2xy/y-x
b)5x-7/2(x-1)-4x/x^2-1+9-3x/2(x-1)
Dấu / là mình thay cho dấu phân số ạ
Đáp án:
$\begin{array}{l}
a)\frac{{{x^2} + {y^2}}}{{x – y}} + \frac{{2xy}}{{y – x}}\\
= \frac{{{x^2} + {y^2}}}{{x – y}} – \frac{{2xy}}{{x – y}}\\
= \frac{{{x^2} – 2xy + {y^2}}}{{x – y}}\\
= \frac{{{{\left( {x – y} \right)}^2}}}{{x – y}}\\
= x – y\\
b)\frac{{5x – 7}}{{2\left( {x – 1} \right)}} – \frac{{4x}}{{{x^2} – 1}} + \frac{{9 – 3x}}{{2\left( {x – 1} \right)}}\\
= \frac{{5x – 7 + 9 – 3x}}{{2\left( {x – 1} \right)}} – \frac{{4x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \frac{{2x + 2}}{{2\left( {x – 1} \right)}} – \frac{{4x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \frac{{2\left( {x + 1} \right)\left( {x + 1} \right) – 4x.2}}{{2.\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \frac{{2\left( {{x^2} + 2x + 1} \right) – 8x}}{{2.\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \frac{{2{x^2} + 4x + 2 – 8x}}{{2.\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \frac{{{x^2} – 2x + 1}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \frac{{{{\left( {x – 1} \right)}^2}}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \frac{{x – 1}}{{x + 1}}
\end{array}$
Đáp án:
a)x-y
b)$\frac{x^{2}-x+1}{2(x-1)(x+1)}$
Giải thích các bước giải:
a)$\frac{x^{2}+y^{2}}{x-y}$ +$\frac{2xy}{y-x}$
=$\frac{x^{2}+y^{2}}{x-y}$-$\frac{2xy}{x-y}$
=$\frac{x^{2}+y^{2}-2xy}{x-y}$
=$\frac{x^{2}-2xy+y^{2}}{x-y}$
=$\frac{(x-y)^{2}}{x-y}$
=x-y
b)$\frac{5x-7}{2(x-1)}$ -$\frac{4x}{x^{2}-1}$ +$\frac{9-3x}{2(x-1)}$
=$\frac{5x-7}{2(x-1)}$-$\frac{4x}{(x-1)(x+1)}$ +$\frac{9-3x}{2(x-1)}$ Mẫu thức chung:2(x-1)(x+1)
=$\frac{5x^{2}-2x-7}{2(x-1)(x+1)}$ -$\frac{8x}{2(x-1)(x+1)}$ +$\frac{6x+9-3x^{2}}{2(x-1)(x+1)}$
=$\frac{5x^{2}-2x-7-8x+6x+9-3x^{2}}{2(x-1)(x+1)}$
=$\frac{2x^{2}-4x+2}{2(x-1)(x+1)}$
=$\frac{2(x^{2}-x+1)}{2(x-1)(x+1)}$
=$\frac{x^{2}-x+1}{(x-1)(x+1)}$