Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
a) Ta có: \(A=\left(1+\dfrac{x^2}{x^2+1}\right):\left(\dfrac{1}{x-1}-\dfrac{2x}{x^3+x-x^2-1}\right)\)
\(=\dfrac{2x^2+1}{x^2+1}:\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}\)
\(=\dfrac{2x^2+1}{x^2+1}\cdot\dfrac{\left(x-1\right)\left(x^2+1\right)}{\left(x-1\right)^2}\)
\(=\dfrac{2x^2+1}{x-1}\)
b) Thay \(x=-\dfrac{1}{2}\) vào A, ta được:
\(A=\left(2\cdot\dfrac{1}{4}+1\right):\left(\dfrac{-1}{2}-1\right)\)
\(=\dfrac{3}{2}:\dfrac{-3}{2}=-1\)
c) Để A<1 thì A-1<0
\(\Leftrightarrow\dfrac{2x^2+1}{x-1}-1< 0\)
\(\Leftrightarrow\dfrac{2x^2+1-x+1}{x-1}< 0\)
\(\Leftrightarrow\dfrac{2x^2-x+2}{x-1}< 0\)
\(\Leftrightarrow x-1< 0\)
hay x<1
a: \(B=\dfrac{x^2-1-2x+3x+1}{x\left(x-1\right)}=\dfrac{x^2+x}{x\left(x-1\right)}=\dfrac{x+1}{x-1}\)
a) B = \(\dfrac{x+1}{x}-\dfrac{2}{x-1}+\dfrac{3x+1}{x\left(x-1\right)}\) (ĐK: \(x\ne0;1\))
= \(\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}-\dfrac{2x}{x\left(x-1\right)}+\dfrac{3x+1}{x\left(x-1\right)}\)
= \(\dfrac{x^2-1-2x+3x+1}{x\left(x-1\right)}=\dfrac{x^2+x}{x\left(x-1\right)}=\dfrac{x+1}{x-1}\)
b) \(\left|x\right|=1< =>\left[{}\begin{matrix}x=1\left(L\right)\\x=-1\left(C\right)\end{matrix}\right.\)
Thay x = -1 vào B, ta có:
\(\dfrac{-1+1}{-1-1}=0\)
c) B nguyên <=> \(\dfrac{x+1}{x-1}\) nguyên <=> \(1+\dfrac{2}{x-1}\) nguyên
<=> 2\(⋮x-1\)
<=> x-1 \(\in\left\{-2;-1;1;2\right\}\)
| x-1 | -2 | -1 | 1 | 2 |
| x | -1 | 0 | 2 | 3 |
| C | L | C | C |
KL: x \(\in\left\{-1;2;3\right\}\)
a: ĐKXĐ: x<>-1
b: \(P=\left(1-\dfrac{x+1}{x^2-x+1}\right)\cdot\dfrac{x^2-x+1}{x+1}\)
\(=\dfrac{x^2-x+1-x-1}{x^2-x+1}\cdot\dfrac{x^2-x+1}{x+1}=\dfrac{x^2-2x}{x+1}\)
c: P=2
=>x^2-2x=2x+2
=>x^2-4x-2=0
=>\(x=2\pm\sqrt{6}\)
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: \(A=\dfrac{x-1+2x^2+2x+2-x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x-1}\)
