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Lời giải:
a) ĐKXĐ: \(\left\{\begin{matrix} x+1\neq 0\\ x-1\neq 0\\ 2-2x^2\neq 0\end{matrix}\right.\Leftrightarrow x\neq \pm 1\)
b)
\(A=\left[\frac{x(x-1)}{(x-1)(x+1)}+\frac{x+1}{(x+1)(x-1)}+\frac{2x}{(x-1)(x+1)}\right].\frac{1}{x+1}=\frac{x^2+2x+1}{(x-1)(x+1)}.\frac{1}{x+1}\)
\(=\frac{(x+1)^2}{(x-1)(x+1)}.\frac{1}{x+1}=\frac{1}{x-1}\)
Để $A$ nguyên thì $1\vdots x-1$
$\Rightarrow x-1\in\left\{\pm 1\right\}$
$\Rightarrow x\in\left\{0;2\right\}$ (đều thỏa mãn đkxđ)
a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(A=\left(\dfrac{x}{x+1}+\dfrac{1}{x-1}-\dfrac{4x}{2-2x^2}\right):\left(x+1\right)\)
\(=\left(\dfrac{2x\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}+\dfrac{2\left(x+1\right)}{2\left(x+1\right)\left(x-1\right)}+\dfrac{4x}{2\left(x+1\right)\left(x-1\right)}\right)\cdot\dfrac{1}{x+1}\)
\(=\dfrac{2x^2-2x+2x+2+4x}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{2x^2+4x+2}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{2\left(x^2+2x+1\right)}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{2\left(x+1\right)^2}{2\left(x+1\right)^2\cdot\left(x-1\right)}\)
\(=\dfrac{1}{x-1}\)
b) Để A nguyên thì \(1⋮x-1\)
\(\Leftrightarrow x-1\inƯ\left(1\right)\)
\(\Leftrightarrow x-1\in\left\{1;-1\right\}\)
hay \(x\in\left\{2;0\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{2;0\right\}\)
Vậy: Để A nguyên thì \(x\in\left\{2;0\right\}\)
a: |x-1|=3
=>x-1=3 hoặc x-1=-3
=>x=-2(nhận) hoặc x=4(loại)
Khi x=-2 thì \(A=\dfrac{4+4}{-2-4}=\dfrac{8}{-6}=\dfrac{-4}{3}\)
b: ĐKXĐ: x<>4; x<>-4
\(B=\dfrac{-\left(x+4\right)}{x-4}+\dfrac{x-4}{x+4}-\dfrac{4x^2}{\left(x-4\right)\left(x+4\right)}\)
\(=\dfrac{-x^2-8x-16+x^2-8x+16-4x^2}{\left(x-4\right)\left(x+4\right)}=\dfrac{-4x^2-16x}{\left(x-4\right)\left(x+4\right)}\)
=-4x/x-4
c: A+B
=-4x/x-4+x^2+4/x-4
=(x-2)^2/(x-4)
A+B>0
=>x-4>0
=>x>4
a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}
Thay x = 2, ta có B không tồn tại
Thay x = -1, ta có B = \(\dfrac{1}{3}\)
b)ĐKXĐ:x ≠ 2,-2
Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x
Do đó không tồn tại x thỏa mãn đề bài
em c.ơn trước
a) Phân thức B xác định \(\Leftrightarrow\hept{\begin{cases}2x-2\ne0\\x^2-1\ne0\\2x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne\left\{\pm1\right\}\\x\ne-1\end{cases}\Leftrightarrow}x\ne\left\{\pm1\right\}}\)
b) \(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\cdot\frac{4x^2-4}{5}\)
\(B=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{3\cdot2}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{\left(2x\right)^2-2^2}{5}\)
\(B=\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(2x-2\right)\left(2x+2\right)}{5}\)
\(B=\frac{10\cdot2\left(x-1\right)\cdot2\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)\cdot5}\)
\(B=\frac{40\left(x-1\right)\left(x+1\right)}{10\left(x-1\right)\left(x+1\right)}\)
\(B=4\)
Vậy với mọi giá trị của x thì B luôn bằng 4
Vậy giá trị của B không phụ thuộc vào biến ( đpcm )
\(Giải:\)
\(ĐKXĐ:x\ne\pm1\)
\(B=\left[\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right]=\left[\frac{x+1}{2x-2}+\frac{12}{4x^2-4}-\frac{x+3}{2x+2}\right]\)
\(=\left[\frac{x+1}{2x-2}+\frac{12}{\left(2x+2\right)\left(2x-2\right)}-\frac{x+3}{2x+2}\right]\)
\(=\left[\frac{\left(x+1\right)\left(2x+2\right)}{\left(2x+2\right)\left(2x-2\right)}+\frac{12}{\left(2x+2\right)\left(2x-2\right)}-\frac{\left(x+3\right)\left(2x-2\right)}{\left(2x-2\right)\left(2x+2\right)}\right]\)
\(=\frac{2x^2+4x+14-2x^2+2x-6x+6}{\left(2x-2\right)\left(2x+2\right)}\)
\(=\frac{6}{\left(2x-2\right)\left(2x+2\right)}\)
Các bạn giải chi tiết hộ mình với!mình cảm ơn nhìu.
a: ĐKXĐ: \(x\notin\left\{0;-4;-2;2\right\}\)
b: \(B=\dfrac{1}{x+2}-\dfrac{x^2-4}{x+4}\cdot\left(\dfrac{4x^2+x^2+4x+4}{4x^2\left(x+2\right)^2}\right)\)
\(=\dfrac{1}{x+2}-\dfrac{\left(x-2\right)}{x+4}\cdot\dfrac{5x^2+4x+4}{4x^2\left(x+2\right)}\)
\(=\dfrac{4x^3+16x^2-\left(x-2\right)\left(5x^2+4x+4\right)}{4x^2\left(x+4\right)\left(x+2\right)}\)
\(=\dfrac{4x^3+16x^2-5x^3-4x^2-4x+10x^2+8x+8}{4x^2\left(x+4\right)\left(x+2\right)}\)
\(=\dfrac{-x^3+22x^2+4x+8}{4x^2\left(x+4\right)\left(x+2\right)}\)