Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ta có : \(\left(a-\dfrac{x^2+a^2}{x+a}\right).\left(\dfrac{2a}{x}-\dfrac{4a}{x-a}\right)\)
\(=\dfrac{-x^2-a^2+ax+a^2}{x+a}.\dfrac{2a\left(x-a\right)-4ax}{x\left(x-a\right)}\)
\(=\dfrac{-x^2+ax}{x+a}.\dfrac{2ax-2a^2-4ax}{x\left(x-a\right)}\)
\(=\dfrac{-x\left(x-a\right)}{x+a}.\dfrac{-2a^2-2ax}{x\left(x-a\right)}\)
\(=\dfrac{-x\left(x-a\right)}{x+a}.\dfrac{-2a\left(a+x\right)}{x\left(x-a\right)}=\dfrac{2a}{1}=2a\) vì a nguyên \(\Rightarrow2a\) nguyên (đpcm)
Bài 3:
a: ĐKXĐ: x<>2
b: \(M=\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}\)
c: Khi x=4001/2000 thì \(M=\dfrac{3}{\dfrac{4001}{2000}-2}=3:\dfrac{1}{2000}=6000\)
a)
\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)
c)
\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)
d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)
f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)
g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)
Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
a) Rút gọn A
\(A=\left(\dfrac{1}{2-x}+\dfrac{1}{2+x}\right):\left(\dfrac{1}{2-x}-\dfrac{1}{2+x}\right)+\dfrac{2}{2+x}\)
ĐKXĐ : \(\left\{{}\begin{matrix}2-x\ne0\\2+x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.andx\ne0\)
Ta có : \(A=\left(\dfrac{2+x}{\left(2-x\right)\left(2+x\right)}+\dfrac{2-x}{\left(2-x\right)\left(2+x\right)}\right):\left(\dfrac{2+x}{\left(2-x\right)\left(2+x\right)}-\dfrac{2-x}{\left(2-x\right)\left(2+x\right)}\right)+\dfrac{2}{2+x}\)
\(A=\dfrac{4}{\left(2-x\right)\left(2+x\right)}.\dfrac{\left(2-x\right)\left(2+x\right)}{2+x}+\dfrac{2}{2+x}\)
\(A=\dfrac{4}{2+x}+\dfrac{2}{2+x}\)
\(A=\dfrac{6}{2+x}\)
