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\(-4\left(x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=-3\) \(3\)
<=> \(-4\left(x^2-2x+1\right)+4x^2-1=-3\)
<=> \(-4x^2+8x-4+4x^2-1=-3\)
<=> \(8x-5=-3\)
<=> \(8x=2\)
<=> \(x=\frac{1}{4}\)
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) \(\left(2x-1\right)^2-3\left(2x-1\right)^2=0\)
=> \(\left(2x-1\right)^2\left(1-3\right)=0\)
=> \(\left(2x-1\right)^2.\left(-2\right)=0\)
=> \(\left(2x-1\right)^2=0\)
=> \(2x-1=0\)
=> \(2x=1\)
=> \(x=1:2=\frac{1}{2}\)
(3x-1)2-5(2x+1)2+(6x-3)(2x+1)=(x-1)2
<=> (3x-1)2+2(3x-1)(2x+1)+(2x+1)2-6(2x+1)2=(x-1)2
<=> (5x)2-6(4x2+4x+1)-(x2-2x+1)=0
<=> -22x-7=0
=> x=-7/22
\(\left(3x-1\right)^2-5\left(2x+1\right)^2+\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow9x^2-6x+1+\left(2x+1\right)\left[-5\left(2x+1\right)+6x-3\right]=x^2-1\)
\(\Leftrightarrow9x^2-6x+1+\left(2x+1\right)\left[-10x-5+6x-3\right]=x^2-1\)
\(\Leftrightarrow9x^2-6x+1+\left(2x+1\right)\left[-4x-8\right]=x^2-1\)
\(\Leftrightarrow9x^2-6x+1-4x\left(2x+1\right)-8\left(2x+1\right)=x^2-1\)
\(\Leftrightarrow9x^2-6x+1-8x^2-4x-16x-8=x^2-1\)
\(\Leftrightarrow\left(9x^2-8x^2-x^2\right)-\left(4x+6x+16x\right)+\left(1-8\right)=-1\)
\(\Leftrightarrow0-26x-7=-1\)
\(\Leftrightarrow-26x=-1+7\)
\(\Leftrightarrow-26x=6\)
\(\Leftrightarrow x=\frac{-3}{13}\)
https://lazi.vn/edu/exercise/giai-phuong-trinh-x-1-x-22-x-1-x-4-32x-4-x-42-0-1
chỉ tiềm thấy cái này thôi ~ vì mk k thể giải đc nên nhờ mạng nên thông cảm cho nha
Ta có: \(\left(x-2\right)\left(x-1\right)=x\left(2x+1\right)+2\)
\(\Leftrightarrow x^2-3x+2-2x^2-x-2=0\)
\(\Leftrightarrow-x^2-4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)