Sử dụng quy tắc đổi dấu để thực hiện các phép tính sau:
a) 3 x 2 − x x − 1 + x + 2 1 − x + 3 − 2 x 2 x − 1 với x ≠ 1 ;
b) 2 y + 2 + 4 y − 2 + 5 y + 2 4 − y 2 với y ≠ ± 2 .
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\frac{16+x}{x^2-2x}-\frac{18}{x^2-2x}\)
\(=\frac{16+x-18}{x\left(x-2\right)}\)
\(=\frac{-2+x}{x\left(x-2\right)}\)
a) \(\frac{16+x}{x^2-2x}+\frac{18}{2x-x^2}=\frac{16+x-18}{x^2-2x}=\frac{x-2}{x\left(x-2\right)}=\frac{1}{x}\)
b) \(\frac{2y}{2x^2-xy}+\frac{4x}{xy-2x^2}=\frac{2y-4x}{2x^2-xy}=\frac{-2\left(2x-y\right)}{x\left(2x-y\right)}=\frac{-2}{x}\)
c) \(\frac{4-x^2}{x-3}+\frac{2x-2x^2}{3-x}+\frac{5-4x}{x-3}=\frac{4-x^2+2x^2-2x+5-4x}{x-3}=\frac{x^2-6x+9}{x-3}=\frac{\left(x-3\right)^2}{x-3}=x-3\)
a: \(\dfrac{6}{x^2+4x}+\dfrac{3}{2x+8}\)
\(=\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\)
\(=\dfrac{12+3x}{2x\left(x+4\right)}=\dfrac{3\left(x+4\right)}{2x\left(x+4\right)}=\dfrac{3}{2x}\)
b: \(\dfrac{x+1}{2x-2}+\dfrac{x-1}{2x+2}+\dfrac{x^2}{1-x^2}\)
\(=\dfrac{x+1}{2\left(x-1\right)}+\dfrac{x-1}{2\left(x+1\right)}-\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2+\left(x-1\right)^2-2x^2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+2x+1+x^2-2x+1-2x^2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x^2-1}\)
c: \(\dfrac{1}{x^2+xy}+\dfrac{2}{y^2-x^2}+\dfrac{1}{xy-x^2}\)
\(=\dfrac{1}{x\left(x+y\right)}-\dfrac{2}{\left(x-y\right)\left(x+y\right)}-\dfrac{1}{x\left(x-y\right)}\)
\(=\dfrac{x-y-2x-x-y}{x\left(x-y\right)\left(x+y\right)}=\dfrac{-2x-2y}{x\left(x-y\right)\left(x+y\right)}\)
\(=-\dfrac{2}{x\left(x-y\right)}\)
a/ \(\left(2x+3\right)\left(x-5\right)-\left(x-1\right)^2+x\left(7-x\right)\)
\(=2x^2-2x-15-x^2+2x-1+7x-x^2\)
\(=7x-16\)