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\(\left(4+2x\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}4+2x=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=-4\\x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)
vậy ta chọn : B
A = ( x - 2 )2 - ( 2x + 1 )2
A = x2 - 4x + 4 - 4x2 + 4x + 1
A = - 3x2 + 5
B = ( x - 2y )2 - ( x - 2y ) . ( 2y + x )
B = x2 - 4xy + 4y2 - ( 2xy + x2 - 4y2 - 2xy )
B = x2 - 4xy + 4y2 - 2xy - x2 + 4y2 + 2xy
B = 8y2 - 4xy
a) \(\left(x^2-2x+1\right)\left(x-1\right)=\left(x-1\right)^2\left(x-1\right)=\left(x-1\right)^3=x^3-3x^2+27x-1\)
\(a,\)\(\left(x^2-2x+1\right)\left(x-1\right)\)
\(=\left(x-1\right)^2\left(x-1\right)\)
\(=\left(x-1\right)^3\)
\(=x^3-3x^2+3x-1\)
\(b,\)\(\left(x^3-2x^2+x-1\right)\left(5-x\right)\)
\(=5x^3-x^4-10x^2+2x^3+5x-x^2-5+x\)
\(-x^4+7x^3-11x^2+6x-5\)
1.X2-2X-4y2-4y
=x2-2x+1-(4y2+4y+1)
=(x+1)2-(2y+1)2
=>(x+1-2y-1)(x+1+2y+1)
=(x-2y)(x+2y+2)
2.x4+2x3-4x-4
=(x2)2-22+2x3-4x
=(x2-2)(x2+2)+2x(x2-2)
=(x2-2)(x2+2+2x)
a, \(x^3-2x^2+3x-6=x\left(x^2+3\right)-2\left(x^2+3\right)=\left(x-2\right)\left(x^2+3\right)\)
b, \(x^2+2x+1-4y^2=\left(x+1\right)^2-\left(2y\right)^2=\left(x+1-2y\right)\left(x+1+2y\right)\)