cho biết x>1 . Hãy rút gọn biểu thức A= \(|x-1|\)+2-x
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a, ĐKXĐ: x≠±2
A=\(\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right)\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
A=\(\left(\dfrac{x}{x^2-4}-\dfrac{2x+4}{x^2-4}+\dfrac{x-2}{x^2-4}\right)\left(\dfrac{x^2+2x}{x+2}-\dfrac{2x+4}{x+2}+\dfrac{10-x^2}{x+2}\right)\)
A=\(\left(\dfrac{-6}{x^2-4}\right)\left(\dfrac{6}{x+2}\right)\)
A=\(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\)
b, |x|=\(\dfrac{1}{2}\)
TH1z: x≥0 ⇔ x=\(\dfrac{1}{2}\) (TMĐKXĐ)
TH2: x<0 ⇔ x=\(\dfrac{-1}{2}\) (TMĐXĐ)
Thay \(\dfrac{1}{2}\), \(\dfrac{-1}{2}\) vào A ta có:
\(\dfrac{-36}{\left(\dfrac{1}{2}-2\right)\left(\dfrac{1}{2}+2\right)^2}\)=\(\dfrac{96}{25}\)
\(\dfrac{-36}{\left(\dfrac{-1}{2}-2\right)\left(\dfrac{-1}{2}+2\right)^2}\)=\(\dfrac{32}{5}\)
c, A<0 ⇔ \(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\) ⇔ (x-2)(x+2)2 < 0
⇔ {x-2>0 ⇔ {x>2
[ [
{x+2<0 {x<2
⇔ {x-2<0 ⇔ {x<2
[ [
{x+2>0 {x>2
⇔ x<2
Vậy x<2 (trừ -2)
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a, ĐKXĐ: x≠±3
A=\(\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{3-x}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{3-x}{x-3}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{9-x^2}{x^2-9}+\dfrac{x^2-3x}{x^2-9}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{-3}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\dfrac{-1}{x^2}\)
b, Thay x=\(-\dfrac{1}{2}\) (TMĐKXĐ) vào A ta có:
\(\dfrac{-1}{\left(-\dfrac{1}{2}\right)^2}\)=-4
c, A<0 ⇔ \(\dfrac{-1}{x^2}< 0\) ⇔ x2>0 (Đúng với mọi x)
Vậy để A<0 thì x đúng với mọi giá trị (trừ ±3)
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a, Với x khác 1
\(A=\dfrac{x^2+x+1-3x^2+2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1-x}{\left(x-1\right)\left(x^2+x+1\right)}=-\dfrac{1}{x^2+x+1}\)
b, Ta có \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\Rightarrow\dfrac{-1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< 0\)
Vậy với x khác 1 thì bth A luôn nhận gtri âm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=x^2-3x-10-x^2+3x=-10\\ b,=\left(x+1\right)\left(x+1-x+1\right)=2\left(x+1\right)=2x+2\)
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a) \(A=\left(x-1\right).\left(x+1\right)+\left(x+2\right).\left(x^2+2x+4\right)-x.\left(x^2+x+2\right)\)
\(=x^2-1+x^3+2x^2+4x+2x^2+4x+8-x^3-x^2-2x\)
\(=\left(x^3-x^3\right)+\left(x^2+2x^2+2x^2-x^2\right)+\left(4x+4x-2x\right)+\left(-1+8\right)\)
\(=4x^2+6x+7\)
b) Thay vào ta được
\(A=4.\left(\frac{1}{2}\right)^2+6.\frac{1}{2}+7=1+3+7=11\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(C=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(\dfrac{1-x^3}{1-x}+x\right)\left(\dfrac{1+x^3}{1+x}-x\right)\right]\)
\(=\dfrac{x\left(x^2-1\right)^2}{x^2+1}:\left[\left(\dfrac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}+x\right)\left(\dfrac{\left(1+x\right)\left(1-x+x^2\right)}{\left(1+x\right)}-x\right)\right]\)
\(=\dfrac{x\left(x^2-1\right)^2}{x^2+1}:\left[\left(x^2+2x+1\right)\left(x^2-2x+1\right)\right]\)
\(=\dfrac{x\left(x-1\right)^2\cdot\left(x+1\right)^2}{\left(x^2+1\right)}\cdot\dfrac{1}{\left(x+1\right)^2\cdot\left(x-1\right)^2}\)
\(=\dfrac{x}{x^2+1}\)
b) Thay \(x=-\dfrac{3}{2}\) vào C, ta được:
\(C=\dfrac{-3}{2}:\left(\dfrac{9}{4}+1\right)=\dfrac{-3}{2}:\dfrac{13}{4}=\dfrac{-3}{2}\cdot\dfrac{4}{13}=\dfrac{-6}{13}\)
c) Ta có: \(C=\dfrac{1}{2}\)
nên \(\dfrac{x}{x^2+1}=\dfrac{1}{2}\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\)(Loại)