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1. A = -4 phần x+2
2. 2x^2 + x = 0 => x = 0 hoặc x = -1/2
Với x = 0 thì A = -2
Với x = -1/2 thì A = -8/3
3. A = 1/2 => -4 phần x + 2 = 1/2
<=> -8 = x + 2
<=> x = -10
4. A nguyên dương => A > 0
=> -4 phần x + 2 > 0
Do -4 < 0 nên -4 phần x + 2 > 0 khi x + 2 < 0
=> x < -2
\(A=\frac{1}{x+2}+\frac{1}{x-2}+\frac{x^2+1}{x^2-4}\)
\(=\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}\)
Với \(\forall x\in\left[-2;2\right]\) thì \(\left(x-2\right)\left(x+2\right)< 0\Rightarrow\frac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}< 0\Rightarrow A< 0\)
1) ĐKXĐ : \(\left\{{}\begin{matrix}x^3-1\ne0\\x^3+x\ne0\\x^2+x\ne0\\3x+\left(x-1\right)^2\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x-1\ne0\\x\left(x^2+1\right)\ne0\\x\left(x+1\right)\ne0\\x^2+x+1\ne0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x-1\ne0\\x\ne0\\x+1\ne0\\\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne0\\x\ne-1\\\left(x+\frac{1}{2}\right)^2\ne-\frac{3}{4}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne\pm1\\x\ne0\end{matrix}\right.\)
2) Ta có : \(P=\left(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}+\frac{1}{x-1}\right):\frac{x^2+x}{x^3+x}\)
=> \(P=\left(\frac{x^2-2x+1}{3x+x^2-2x+1}-\frac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\right):\frac{x^2+x}{x^3+x}\)
=> \(P=\left(\frac{\left(x-1\right)^2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\frac{x^2+x}{x^3+x}\)
=> \(P=\left(\frac{\left(x-1\right)^3-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\frac{x^2+x}{x^3+x}\)
=> \(P=\left(\frac{x^3-3x^2+3x-1-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\frac{x\left(x+1\right)}{x\left(x^2+1\right)}\)
=> \(P=\left(\frac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\frac{x+1}{x^2+1}\)
=> \(P=\left(\frac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right):\frac{x+1}{x^2+1}\)
=> \(P=1:\frac{x+1}{x^2+1}=\frac{x^2+1}{x+1}\)
- Thay P = 0 vào phương trình trên ta được :\(\frac{x^2+1}{x+1}=0\)
=> \(x^2+1=0\)
=> \(x^2=-1\) ( Vô lý )
Vậy phương trình vô nghiệm .
3) Ta có : \(\left|P\right|=1\)
=> \(\left|\frac{x^2+1}{x+1}\right|=1\)
=> \(\frac{x^2+1}{\left|x+1\right|}=1\)
=> \(\left|x+1\right|=x^2+1\)
TH1 : \(x+1\ge0\left(x\ge-1\right)\)
=> \(x+1=x^2+1\)
=> \(x^2=x\)
=> \(x=1\) ( TM )
TH2 : \(x+1< 0\left(x< -1\right)\)
=> \(-x-1=x^2+1\)
=> \(x^2+1+1+x=0\)
=> \(x^2+\frac{1}{2}x.2+\frac{1}{4}+\frac{7}{4}=0\)
=> \(\left(x+\frac{1}{2}\right)^2=-\frac{7}{4}\) ( Vô lý )
Vậy giá trị của x thỏa mãn là x = 1 .
a.)Đkxđ bạn tự tìm nha!!!
A=\(\left(\frac{1}{x-1}-\frac{x}{1-x^3}.\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)
\(\Leftrightarrow\)\(\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)
\(\Leftrightarrow\)\(\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{x^2+x+1}\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{x^2+x+1}\)
\(\Leftrightarrow\)\(\frac{2x+1}{\left(x-1\right)\left(x+1\right)}:\frac{2x+1}{x^2+2x+1}\)
\(\Leftrightarrow\)\(\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x+1\right)^2}{2x+1}\)
\(\Leftrightarrow\)\(\frac{x+1}{x-1}\left(tm\text{đ}k\right)\)
b.)Thay \(x=\frac{1}{2}\)vào A \(\Rightarrow\)\(A=-3\)
a) Đk: x > 0 và x khác +-1
Ta có: A = \(\left(\frac{x+1}{x}-\frac{1}{1-x}-\frac{x^2-2}{x^2-x}\right):\frac{x^2+x}{x^2-2x+1}\)
A = \(\left[\frac{\left(x-1\right)\left(x+1\right)+x-x^2+2}{x\left(x-1\right)}\right]:\frac{x\left(x+1\right)}{\left(x-1\right)^2}\)
A = \(\frac{x^2-1+x-x^2+2}{x\left(x-1\right)}\cdot\frac{\left(x-1\right)^2}{x\left(x+1\right)}\)
A = \(\frac{x+1}{x}\cdot\frac{x-1}{x\left(x+1\right)}=\frac{x-1}{x^2}\)
b) Ta có: A = \(\frac{x-1}{x^2}=\frac{1}{x}-\frac{1}{x^2}=-\left(\frac{1}{x^2}-\frac{1}{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\frac{1}{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
Dấu "=" xảy ra <=> 1/x - 1/2 = 0 <=> x = 2 (tm)
Vậy MaxA = 1/4 <=> x = 2
M xác định
\(\Leftrightarrow\hept{\begin{cases}x-1\ne0\\x^2-x\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\left(x-1\right)\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne0;x\ne1\end{cases}}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\ne0\end{cases}}\)
Vậy ĐKXĐ của M là \(\hept{\begin{cases}x\ne1\\x\ne0\end{cases}}\)
\(M=\frac{3}{x-1}+\frac{1}{x^2-x}=\frac{3}{x-1}+\frac{1}{x\left(x-1\right)}=\frac{3x}{x\left(x-1\right)}+\frac{1}{x\left(x-1\right)}=\frac{3x+1}{x\left(x-1\right)}\)
Thay x=5 ta có:
\(M=\frac{3.5+1}{5\left(5-1\right)}=\frac{15+1}{5.4}=\frac{16}{20}=\frac{4}{5}\)
Vậy \(M=5\)tại x=5
\(M=0\)
\(\Leftrightarrow\frac{3x+1}{x\left(x-1\right)}=0\Leftrightarrow3x+1=0\Leftrightarrow x=-\frac{1}{3}\)( thỏa mãn đkxđ)
Vậy với \(x=-\frac{1}{3}\)thì \(M=0\)
\(M=-1\)
\(\Leftrightarrow\frac{3x+1}{x\left(x-1\right)}=-1\Leftrightarrow3x+1=-x^2+x\Leftrightarrow x^2+2x+1=0\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)
Vậy với \(x=-1\)thì \(M=-1\)
Bài làm:
a) đkxđ: \(x\ne\pm1\)
Ta có:
\(M=\frac{x+1}{x^2-1}-\frac{x^2+2}{x^3-1}-\frac{x+1}{x^2+x+1}\)
\(M=\frac{1}{x-1}-\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x+1}{x^2+x+1}\)
\(M=\frac{x^2+x+1-x^2-2-\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(M=\frac{x-1-x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(M=\frac{x\left(1-x\right)}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{x}{x^2+x+1}\)
b) Mà x khác 1
=> x = -2, khi đó:
\(M=-\frac{-2}{4-2+1}=\frac{2}{3}\)