/x-2/ = 2x-3 giúp ạ
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d) \(\left|2x-3\right|=x-3\)
TH1: \(\left|2x-3\right|=2x-3\) với \(2x-3\ge0\Leftrightarrow x\ge\dfrac{3}{2}\)
Pt trở thành:
\(2x-3=x-3\) (ĐK: \(x\ge\dfrac{3}{2}\) )
\(\Leftrightarrow2x-x=-3+3\)
\(\Leftrightarrow x=0\left(ktm\right)\)
TH2: \(\left|2x-3\right|=-\left(2x-3\right)\) với \(2x-3< 0\Leftrightarrow x< \dfrac{3}{2}\)
Pt trở thành:
\(-\left(2x-3\right)=x-3\)
\(\Leftrightarrow-2x+3=x-3\)
\(\Leftrightarrow-2x-x=-3-3\)
\(\Leftrightarrow-3x=-6\)
\(\Leftrightarrow x=-\dfrac{6}{-3}=2\left(ktm\right)\)
Vậy Pt vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow2x^2+6x-2x^2=30\Rightarrow6x=30\Rightarrow x=5\)
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\(\left(3-2x\right)^2=\left(x-2\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x-2\right)^2-\left(x-2\right)\left(2x-3\right)=0\)
\(\Leftrightarrow9x^2-12x+4-\left(2x^2-7x+6\right)=0\)
\(\Leftrightarrow9x^2-12x+4-2x^2+7x-6=0\)
\(\Leftrightarrow7x^2-5x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{7}\end{matrix}\right.\)
Vậy \(S=\left\{1;-\dfrac{2}{7}\right\}\)
`(3-2x)^2=(x-2)(2x-3)`
`<=>(2x-3)^2 -(x-2)(2x-3)=0`
`<=> (2x-3)(2x-3-x+2)=0`
`<=> (2x-3)(x-1)=0`
\(< =>\left[{}\begin{matrix}2x-3=0\\x-1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(x-3\right)\left(2x^2-3x+4\right)\)
\(=2x^3-3x^2+4x-6x^2+9x-12\)
\(=2x^3-9x^2+13x-12\)
b: \(\left(4x^2y-5xy^2+6xy\right):2xy\)
\(=\dfrac{4x^2y-5xy^2+6xy}{2xy}\)
\(=\dfrac{2xy\cdot2x-2xy\cdot2,5y+2xy\cdot3}{2xy}\)
\(=2x-2,5y+3\)
c: \(\dfrac{x}{2x+4}-\dfrac{2}{x^3+2x}\)
\(=\dfrac{x\left(x^3+2x\right)-2\left(2x+4\right)}{x\left(x^2+2\right)\cdot2\cdot\left(x+2\right)}\)
\(=\dfrac{x^4+2x^2-4x-8}{2x\left(x^2+2\right)\left(x+2\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: (x-3)(x-1)-x(x-2)=0
=>\(x^2-4x+3-x^2+2x=0\)
=>\(-2x+3=0\)
=>-2x=-3
=>\(x=\dfrac{3}{2}\)
b: \(\left(x+2y\right)^2-\left(2x-y\right)^2\)
\(=\left(x+2y+2x-y\right)\left(x+2y-2x+y\right)\)
\(=\left(3x+y\right)\left(-x+3y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-1\right)^2+\left(x-3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-1+x-3\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{2};\dfrac{4}{3}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
んuリ イ Đặt ĐK cho GTTĐ thôi Tú
Với x ≥ 2 pt <=> x - 2 = 2x - 3 <=> -x = -1 <=> x = 1 (ktm)
Với x < 2 pt <=> -x + 2 = 2x - 3 <=> -3x = -5 <=> x = 5/3 (tm)
Vậy ....
\(\left|x-2\right|=2x-3\)
ĐK : \(2x-3\ge0\Leftrightarrow x\ge\frac{3}{2}\)
TH1 : \(x-2=2x-3\Leftrightarrow-x=-1\Leftrightarrow x=1\)( ktm )
TH2 : \(x-2=-2x+3\Leftrightarrow3x=5\Leftrightarrow x=\frac{5}{3}\)( tm )
Vậy tập nghiệm của pt là S = { 5/3 }