2=0+... giúp em với
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(1-2x)2-(3x-2)2=0
(1-2x-3x+2)(1-2x+3x-2)=0
(-5x+3)(x-1)=0
Th1: -5x+3=0 Th2: x-1=0
-5x=-3 x=1
x=3/5
vậy ...
Ta có: \(6x^2-x-2=0\)
\(\Leftrightarrow6x^2-4x+3x-2=0\)
\(\Leftrightarrow2x.\left(3x-2\right)+\left(3x-2\right)=0\)
\(\Leftrightarrow\left(2x+1\right).\left(3x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\3x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{2}{3}\end{cases}}\)
6x2 - 2 - x = 0
=> 6x2 + 3x - 4x - 2 = 0
=> 3x(2x + 1) - 2(2x + 1) = 0
=> (3x - 2)(2x + 1) = 0
=> \(\orbr{\begin{cases}3x-2=0\\2x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=2\\2x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{1}{2}\end{cases}}\)
\(x^2-8x+20=\left(x^2-8x+16\right)+4=\left(x-4\right)^2+4\ge4>0\forall x\)
2x . ( x-2) - x+2 = 0
\(\Leftrightarrow2x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}}\)
\(\dfrac{1}{2}-3x+\left|x-1\right|=0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}-0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}\\ \Rightarrow\left|x-1\right|=\dfrac{1}{2}-3x\\ \Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}-3x\\x-1=-\dfrac{1}{2}+3x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+3x=\dfrac{1}{2}+1\\x-3x=-\dfrac{1}{2}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=\dfrac{3}{2}\\2x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)
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\(\dfrac{1}{2}\left|2x-1\right|+\left|2x-1\right|=x+1\\ \Rightarrow\left|2x-1\right|\cdot\left(\dfrac{1}{2}+1\right)=x+1\\ \Rightarrow\left|2x-1\right|\cdot\dfrac{3}{2}=x+1\\ \Rightarrow\left|2x-1\right|=x+1:\dfrac{3}{2}\\ \Rightarrow\left|2x-1\right|=x+\dfrac{2}{3}\\ \Rightarrow\left[{}\begin{matrix}2x-1=x+\dfrac{2}{3}\\2x-1=-x-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=\dfrac{2}{3}+1\\2x+x=-\dfrac{2}{3}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\3x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{1}{9}\end{matrix}\right.\)
TA CÓ 0=02
⇒X-11+Y+X+4-Y=0
⇒(X+X)+(-11+4)+(Y-Y)=0
⇒2X+(-7)+0=0
⇒2X=0-(-7)
⇒2X=7
⇒X=7:2
⇒X=3,5
VẬY X =3,5
2 = 0 + 2 nhé
2 = 0 + 2 nhé !