tim x biet :
2 (x +5 ) - x^2 - 5x =0
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a)\(x^2+2y^2+2xy-2y+1=0\)
\(\Leftrightarrow x^2+2xy+y^2+y^2-2y+1=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}y-1=0\\x+y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=1\\x=-y=-1\end{cases}}\)
Vậy x=-1 y=1
a) \(x^2+2y^2+2xy-2y+1=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+y=0\\y-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-y\\y=1\end{cases}\Rightarrow}x=-1;y=1}\)
b) \(5x^2+3y^2+z^2-4x+6xy+4z+6=0\)
\(\Leftrightarrow\left(2x^2-4x+2\right)+\left(3x^2+6xy+3y^2\right)+\left(z^2+4z+4\right)=0\)
\(\Leftrightarrow2.\left(x-1\right)^2+3.\left(x+y\right)^2+\left(z+2\right)^2=0\)
\(\Rightarrow\) \(\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)
\(\left(x+y\right)^2=0\Rightarrow x+y=0\Rightarrow y=-x=-1\)
\(\left(z+2\right)^2=0\Rightarrow z+2=0\Rightarrow z=-2\)
3x^2-5x+2=0
<=>3x2-3x-2x+2=0
<=>3x.(x-1)-2.(x-1)=0
<=>(x-1)(3x-2)=0
<=>x-1=0 hoặc 3x-2=0
<=>x=1 hoặc 3x=2
<=>x=1 hoặc x=2/3
\(5x\left(x-3\right)-x+3=0\)
<=> \(\left(x-3\right)\left(5x-1\right)=0\)
<=> \(\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}\)
Vay..........
a,Để \(|2x+1|+|x-2|=0\Leftrightarrow\hept{\begin{cases}2x+1=0\\x-2=0\end{cases}}\)(vô lý)
=> ko có x thỏa mãn
b,\(|x+5|=2x-1\Leftrightarrow1-2x< x+5< 2x-1\)
\(16-5x^2-3=0\)
\(\Leftrightarrow16-5x^2=0+3\)
\(\Leftrightarrow16-5x^2=3\)
\(\Leftrightarrow5x^2=16-3\)
\(\Leftrightarrow5x^2=13\)
\(\Leftrightarrow x^2=\frac{13}{5}\)
\(\Leftrightarrow x^2=2,6\)
\(\Leftrightarrow1,61\approx1,6\)
\(\Rightarrow x=1,6\)
\(16-5x^2-3=0\)
\(\Leftrightarrow16-5x^2=3\)
\(\Leftrightarrow5x^2=16-3\)
\(\Leftrightarrow5x^2=13\Leftrightarrow x^2=\frac{13}{5}\)
\(\Rightarrow\orbr{\begin{cases}x=\sqrt{\frac{13}{5}}\\x=-\sqrt{\frac{13}{5}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{65}}{5}\\x=-\frac{\sqrt{65}}{5}\end{cases}}\)
2.(x+5) - x2 - 5x = 0
2(x+5) - x(x+5) = 0
(x+5)(2-x) = 0
=> x+5=0 hoặc 2-x=0
=> x=-5 hoặc x=2
\(\frac{x^5-1}{x^3-1}=\frac{ }{x+1}\)