\(2\left(x^2+\frac{1}{x^2}\right)-x-\frac{1}{x}-6=0\)
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MN
24 tháng 1 2020
a) \(\left(x+1\right)-\frac{x+1}{3}=\frac{5\left(x+1\right)-1}{6}\)
\(\Leftrightarrow6\left(x+1\right)-2\left(x+1\right)=5\left(x+1\right)-1\)
\(\Leftrightarrow6x+6-2x-2=5x+5-1\)
\(\Leftrightarrow6x-2x-5x=5-1-6+2\)
\(\Leftrightarrow-x=0\)
\(\Leftrightarrow x=0\)
b) \(\left(1-x\right)^2+\left(x+2\right)^2=2x\left(x-3\right)-7\)
\(\Leftrightarrow1-2x+x^2+x^2+4x+4=2x^2-6x-7\)
\(\Leftrightarrow2x^2+2x+5=2x^2-6x-7\)
\(\Leftrightarrow2x+6x=-7-5\)
\(\Leftrightarrow8x=-12\)
\(\Leftrightarrow x=-\frac{3}{2}\)
c) \(2+\frac{x-2}{2}-\frac{2x-4}{3}-\frac{5}{6}\left(2-x\right)=0\)
\(\Leftrightarrow2+\frac{x}{2}-1-\frac{2}{3}x+\frac{4}{3}-\frac{5}{3}+\frac{5}{6}x=0\)
\(\Leftrightarrow\frac{x}{2}-\frac{2}{3}x+\frac{5}{6}x=-2+1-\frac{4}{3}+\frac{5}{3}\)
\(\Leftrightarrow\frac{2}{3}x=-\frac{2}{3}\)
\(\Leftrightarrow x=-1\)
VM
16 tháng 3 2020
a) \(\left(\frac{2}{3}x-1\right).\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x-1=0\\\frac{3}{4}x+\frac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=1\\\frac{3}{4}x=-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1:\frac{2}{3}\\x=\left(-\frac{1}{2}\right):\frac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{3}{2};-\frac{2}{3}\right\}.\)
b) \(\left(x-1\right)^{x-2}=\left(x-1\right)^{x+6}\)
\(\Rightarrow\left(x-1\right)^{x-2}-\left(x-1\right)^{x+6}=0\)
\(\Rightarrow\left(x-1\right)^{x-2}.\left[1-\left(x-1\right)^8\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^{x-2}=0\\1-\left(x-1\right)^8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^8=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=1+1\\x=\left(-1\right)+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{1;2;0\right\}.\)
Chúc bạn học tốt!
\(2\left(x^2+\frac{1}{x^2}\right)-x-\frac{1}{x}-6=0\)( ĐKXĐ : x ≠ 0 )
<=> \(2\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)-6=0\)
Đặt \(x+\frac{1}{x}=t\)=> \(t^2=x^2+\frac{1}{x^2}+2\)=> \(x^2+\frac{1}{x^2}=t^2-2\)
Khi đó pt đã cho trở thành 2( t2 - 2 ) - t - 6 = 0
<=> 2t2 - 4 - t - 6 = 0
<=> 2t2 + 4t - 5t - 10 = 0
<=> 2t( t + 2 ) - 5( t + 2 ) = 0
<=> ( t + 2 )( 2t - 5 ) = 0
<=> t = -2 hoặc t = 5/2
Với t = -2 => \(x+\frac{1}{x}=-2\)<=> \(\frac{x^2+1}{x}=-2\)=> x2 + 1 = -2x <=> ( x + 1 )2 = 0 <=> x = -1 (tm)
Với t = 5/2 => \(x+\frac{1}{x}=\frac{5}{2}\)<=> \(\frac{x^2+1}{x}=\frac{5}{2}\)=> 2x2 + 2 = 5x <=> ( 2x - 1 )( x - 2 ) = 0 <=> \(\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}\left(tm\right)}\)
Vậy ...