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Bài Làm:
\(1,\left(2x-1\right)^2-3\left(2x-1\right)^2=0\)
\(\Leftrightarrow-2\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy ...
\(2,\left(x-1\right)^2\left(x+1\right)=x+1\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2-2x+1-1\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=2\end{matrix}\right.\)
Vậy ...
\(3,x^4-3x^2=x^2\)
\(\Leftrightarrow x^4-3x^2-x^2=0\)
\(\Leftrightarrow x^4-4x^2=0\)
\(\Leftrightarrow x^2\left(x^2-4\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy ...
Chúc pạn hok tốt!!!
a: =>6x-3x^2-5=4-3x^2-2
=>6x-5=2
=>6x=7
=>x=7/6
b: =>20x+5-12x^2-3x=6x^2-10x+3x-5
=>-12x^2+17x+5-6x^2+7x+5=0
=>-18x^2+24x+10=0
=>x=5/3 hoặc x=-1/3
a: \(\dfrac{3x+2}{4}-\dfrac{3x+1}{3}=\dfrac{5}{6}\)
=>3(3x+2)-4(3x+1)=10
=>9x+6-12x-4=10
=>-3x+2=10
=>-3x=8
=>x=-8/3
b: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{9x-10}{4-x^2}\)
=>(x-1)(x-2)-x(x+2)=-9x+10
=>x^2-3x+2-x^2-2x=-9x+10
=>-5x+2=-9x+10
=>x=2(loại)
1/ \(\left(2x-1\right)^2-3\left(2x-1\right)^2=0\)
\(\left(2x-1\right)^2\left(1-3\right)=0\)
\(\left(2x-1\right)^2\cdot\left(-2\right)=0\)
\(\Rightarrow\text{ }\left(2x-1\right)^2=0\)
\(2x-1=0\)
\(2x=0+1=1\)
\(x=\frac{1}{2}\)
1) \(\left(2x-1\right)^2-3\left(2x-1\right)^2=0\)
=> \(\left(2x-1\right)^2\left(1-3\right)=0\)
=> \(\left(2x-1\right)^2.\left(-2\right)=0\)
=> \(\left(2x-1\right)^2=0\)
=> \(2x-1=0\)
=> \(2x=1\)
=> \(x=1:2=\frac{1}{2}\)