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\(3x-x\left(x-2\right)=-\left(x+1\right)^2\)
\(\Leftrightarrow3x-x^2+2x=-\left(x^2+2x+1\right)\)
\(\Leftrightarrow5x-x^2=-x^2-2x-1\)
\(\Leftrightarrow-x^2+x^2+5x+2x=-1\)
\(\Leftrightarrow7x=-1\)
\(\Leftrightarrow x=\left(-1\right)\div7\)
\(\Leftrightarrow x=-\dfrac{1}{7}\)
Ko bt đúng or sai :>
3x -x(x-2)= -(x+1)^2
<=>3x -x^2 +2x= -x^2-2x -1
<=> -x^2 +x^2 +5x +2x=-1
<=>7x= -1
<=>x= -1/7
![](https://rs.olm.vn/images/avt/0.png?1311)
=>3x^3-6x^2+x^2-2x+x-2>0
=>(x-2)(3x^2+x+1)>0
=>x-2>0
=>x>2
\(3x^3-5x^2-x-2>0\)
\(\Leftrightarrow3x^3-6x^2+x^2-2x+x-2>0\)
\(\Leftrightarrow3x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)>0\)
\(\Leftrightarrow\left(x-2\right)\left(3x^2+x+1\right)>0\)
Mặt khác: \(3x^2+x+1=2x^2+\left(x^2+x+1\right)\)
Ta lại có: \(x^2+x+1=x^2+2x\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(\Rightarrow3x^2+x+1>0\)
\(\Rightarrow x-2>0\)
\(\Leftrightarrow x>2\)
Vậy bpt có nghiệm là \(x>2\)
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ta có :
\(\left|x+1\right|+\left|x-1\right|=1+\left|\left(x-1\right)\left(x+1\right)\right|\)
\(\Leftrightarrow\left|x-1\right|\left|x+1\right|-\left|x-1\right|-\left|x+1\right|+1=0\)
\(\Leftrightarrow\left(\left|x-1\right|-1\right)\left(\left|x+1\right|-1\right)=0\Leftrightarrow\orbr{\begin{cases}\left|x-1\right|=1\\\left|x+1\right|=1\end{cases}}\)
\(\Leftrightarrow x\in\left\{-2,0,2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)
hay \(x\in\left\{0;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)
=>x-1=0
hay x=1
d: \(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)
aGiải phương trình |x-1|+|x-2|=|2x-3|
b)Giải phương trình 1/(x−2 )+ 2/(x−3) − 3/(x−5) = 1/(x^2 −5x+6)
![](https://rs.olm.vn/images/avt/0.png?1311)
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1. a = 3 thì phương trình trở thành:
\(\frac{x+3}{3-x}-\frac{x-3}{3+x}=\frac{-3\left[3.\left(-3\right)+1\right]}{\left(-3\right)^2}-x^2\)
\(\Leftrightarrow\frac{\left(x+3\right)^2+\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}=\frac{-3\left[-9+1\right]}{9}-x^2\)
\(\Leftrightarrow\frac{x^2+6x+9+x^2-6x+9}{\left(3-x\right)\left(3+x\right)}=\frac{-3.\left(-8\right)}{9}-x^2\)
\(\Leftrightarrow\frac{2x^2+18}{9-x^2}=\frac{24}{9}-x^2\)
\(\Leftrightarrow\frac{2x^2+18}{9-x^2}+x^2=\frac{24}{9}\)
\(\Leftrightarrow\frac{2x^2+18+9x^2-x^4}{9-x^2}=\frac{24}{9}\)
\(\Leftrightarrow\frac{11x^2+18-x^4}{9-x^2}=\frac{24}{9}\)
\(\Leftrightarrow99x^2+18-9x^4=216-24x^2\)
\(\Leftrightarrow9x^4-123x^2+198=0\)
Đặt \(x^2=t\left(t\ge0\right)\)
Phương trình trở thành \(9t^2-123t+198=0\)
Ta có \(\Delta=123^2-4.9.198=8001,\sqrt{\Delta}=3\sqrt{889}\)
\(\Rightarrow\orbr{\begin{cases}t=\frac{123+3\sqrt{889}}{18}=\frac{41+\sqrt{889}}{6}\\t=\frac{123-3\sqrt{889}}{18}=\frac{41-\sqrt{889}}{6}\end{cases}}\)
Lúc đó \(\orbr{\begin{cases}x^2=\frac{41+\sqrt{889}}{6}\\x^2=\frac{41-\sqrt{889}}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{\frac{41+\sqrt{889}}{6}}\\x=\pm\sqrt{\frac{41-\sqrt{889}}{6}}\end{cases}}\)
Vậy pt có 4 nghiệm \(S=\left\{\pm\sqrt{\frac{41+\sqrt{889}}{6}};\pm\sqrt{\frac{41-\sqrt{889}}{6}}\right\}\)
\(3x-x\left(x-2\right)=-x\left(x+1\right)^2\)
\(\Leftrightarrow3x-x^2+2x=-x^2-2x-1\)
\(\Leftrightarrow-x^2+x^2+3x+2x+2x+1=0\)
\(\Leftrightarrow7x+1=0\)
\(\Leftrightarrow x=-\dfrac{1}{7}\)
Vậy \(S=\left\{-\dfrac{1}{7}\right\}\)