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\(2x-2=8-3x\)
\(\Leftrightarrow\)\(2x+3x=8+2\)
\(\Leftrightarrow\)\(5x=10\)
\(\Leftrightarrow\)\(x=2\)
Vậy...
\(x^2-3x+1=x+x^2\)
\(\Leftrightarrow\)\(x^2-3x-x-x^2=-1\)
\(\Leftrightarrow\)\(-4x=-1\)
\(\Leftrightarrow\)\(x=\frac{1}{4}\)
Vậy...
mấy cái này bấm máy tính là đc òi. giải mất thời gian lắm :))
a) \(\frac{x-5}{4}-2x+1=\frac{x}{3}-\frac{2-x}{6}\)
<=> \(\frac{1}{4}x-\frac{5}{4}-2x+1=\frac{1}{3}x-\frac{1}{3}+\frac{1}{6}x\)
<=> \(-\frac{7}{4}x-\frac{1}{2}x=-\frac{1}{3}+\frac{1}{4}\)
<=> \(-\frac{9}{4}x=-\frac{1}{12}\)
<=> \(x=\frac{1}{27}\)
Vậy ...
b) ( x2 - 4 ) - ( x - 2 )( 3 - 2x ) = 0
<=> ( x - 2 )( x + 2 ) - ( x - 2 )( 3 - 2x ) = 0
<=> ( x - 2 )( x + 2 - 3 + 2x ) = 0
<=> ( x - 2 )( 3x - 1 ) = 0
<=> x = 2 hoặc x = 1/3
Vậy ...
\(\frac{x+1}{2011}+\frac{x+2}{2010}=\frac{x+3}{2009}+\frac{x+4}{2008}\Leftrightarrow\frac{x+1}{2011}+1+\frac{x+2}{2010}+1=\frac{x+3}{2009}+1+\frac{x+4}{2008}+1\)
\(\Leftrightarrow\frac{x+1}{2011}+\frac{2011}{2011}+\frac{x+2}{2010}+\frac{2010}{2010}=\frac{x+3}{2009}+\frac{2009}{2009}+\frac{x+4}{2008}+\frac{2008}{2008}\)
\(\Leftrightarrow\frac{x+1+2011}{2011}+\frac{x+2+2010}{2010}=\frac{x+3+2009}{2009}+\frac{x+4+2008}{2008}\)
\(\Leftrightarrow\frac{x+2012}{2011}+\frac{x+2012}{2010}=\frac{x+2012}{2009}+\frac{x+2012}{2008}\)
\(\Leftrightarrow\left(x+2012\right)\left(\frac{1}{2011}+\frac{1}{2010}\right)=\left(x+2012\right)\left(\frac{1}{2009}+\frac{1}{2008}\right)\)
\(\Leftrightarrow\left(x+2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}=0\right)\)
mà 1/2011+1/2010-1/2009-1/2008 khác 0
\(\Rightarrow x+2012=0\Rightarrow x=-2012\)
\(\left(3x-2\right)^2-x\left(9x-2\right)=24\Leftrightarrow9x^2-12x+4-9x^2+2x=24\)
\(\Leftrightarrow-10x+4=24\Leftrightarrow-10x=20\Leftrightarrow x=-2\)
1; Ta có : x+1/2011 + x+2/2010 = x+3/2009 + x+4/ 2008
Suy ra: 2+(x+1/2011 + x+2/2010 ) = 2+( x+3/2009 + x+4/2008)
suy ra ban tach 2=1+1 roi cong 1 voi tưng phân số trên nha sẽ ra kết quả ngay thôi
2; gợi ý nè : (3x-2)^2 =(3x)^2 + 2*3x*2+2^2
a) Ta có: \(x^3-6x^2+11x-6=0\)
\(\Leftrightarrow x^3-x^2-5x^2+5x+6x-6=0\)
\(\Leftrightarrow x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\x-2=0\\x-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\x=2\\x=3\end{cases}}\)
Vậy nghiệm của phương trình là {1;2;3}
Mình đang bận. Câu 2 tí nữa giải quyết sau...
c) \(\left|2x-3\right|=4\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=4\\2x-3=-4\end{cases}}\)
\(TH:2x-3=4\)
\(\Leftrightarrow2x=4+3\)
\(\Leftrightarrow2x=7\)
\(\Leftrightarrow x=\frac{7}{2}\)
\(TH:2x-3=-4\)
\(\Leftrightarrow2x=-4+3\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy \(x\in\left\{\frac{7}{2};\frac{-1}{2}\right\}\)
e) \(\frac{x-1}{x-3}>1\)
\(ĐKXĐ:x\ne3\)
\(\Leftrightarrow\frac{x-3+2}{x-3}>1\)
\(\Leftrightarrow\frac{x-3}{x-3}+\frac{2}{x-3}>1\)
\(\Leftrightarrow1+\frac{2}{x-3}>1\)
\(\Leftrightarrow\frac{2}{x-3}>0\)
\(\Leftrightarrow x-3>0\)
\(\Leftrightarrow x>3\)
ĐKXĐ:...
Đặt \(x+\frac{1}{x}=a\Rightarrow a^3=x^3+\frac{1}{x^3}+3x.\frac{1}{x}\left(x+\frac{1}{x}\right)\)
\(\Rightarrow a^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=x^3+\frac{1}{x^3}+3a\)
\(\Rightarrow x^3+\frac{1}{x^3}=a^3-3a\)
Thay vào pt ta được:
\(4\left(a^3-3a\right)=13a\)
\(\Leftrightarrow4a^3-25a=0\Leftrightarrow a\left(4a^2-25\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\a=\frac{5}{2}\\a=-\frac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=0\\x+\frac{1}{x}=\frac{5}{2}\\x+\frac{1}{x}=-\frac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\left(vn\right)\\2x^2-5x+2=0\\2x^2+5x+2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=\pm\frac{1}{2}\end{matrix}\right.\)