\(2x-x^2=0\)
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Bài làm
Phân tích đa thức thành nhân tử
a) x2 + 7x + 12 = x2 + 3x + 4x + 12 = x( x + 3 ) + 4( x + 3 ) = ( x + 3 )( x + 4 )
b) 4x2 + 14x + 6 = 4x2 + 12x + 2x + 6 = 4x( x + 3 ) + 2( x + 3 ) = 2( x + 3 )( 2x + 1 )
4x2 + 2y2 + 2z2 - 4xy - 4xz + 2yz - 6y - 10z + 34 = 0
<=> ( 4x2 - 4xy + y2 - 4xz + 2yz + z2 ) + ( y2 - 6y + 9 ) + ( z2 - 10z + 25 ) = 0
<=> [ ( 4x2 - 4xy + y2 ) - ( 4xz - 2yz ) + z2 ] + ( y - 3 )2 + ( z - 5 )2 = 0
<=> [ ( 2x - y )2 - 2( 2x - y )z + z2 ] + ( y - 3 )2 + ( z - 5 )2 = 0
<=> ( 2x - y - z )2 + ( y - 3 )2 + ( z - 5 )2 = 0
<=> \(\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}}\Rightarrow\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}\)
=> S = ( x - 4 )2012 + ( y - 4 )2017 + ( z - 4 )1282
= ( 4 - 4 )2012 + ( 3 - 4 )2017 + ( 5 - 4 )1282
= 0 - 1 + 1 = 0
Thanks bn nhưng mik vừa làm xog r :)))))))))))))))))))))))))))

\(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\)
\(\Rightarrow\left(2a+13b\right)\left(3c-7d\right)=\left(2c+13d\right)\left(3a-7b\right)\)
\(6ac-14ad+39bc-91bd=6ac-14bc+39ad-91bd\)
\(39bc+14bc=39ad+14ad\)
\(53bc=53ad\)
\(bc=ad\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}\)

\(\frac{3x-7}{x+y}=\frac{3}{4}\)
\(\Rightarrow4\left(3x-y\right)=3\left(x+y\right)\)
\(12x-4y=3x+3y\)
\(9x=7y\)
\(\frac{x}{y}=\frac{7}{9}\)
\(\frac{3x-y}{x+y}=\frac{3}{4}\Leftrightarrow4\left(3x-y\right)=3\left(x+y\right)\)
\(\Leftrightarrow12x-4y=3x+3y\Leftrightarrow9x-7y=0\)
\(\Leftrightarrow9x=7y\Leftrightarrow\frac{x}{y}=\frac{7}{9}\)
Vậy \(\frac{x}{y}=\frac{7}{9}\)
\(2x-x^2=0\)
\(x\left(2-x\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\2-x=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=2\end{cases}}\)
2x-xx=0
x.(2-x)=0
TH1:2x=0
x=0
TH2:2-x=0
x=2