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\(\frac{\left(a-b\right)^2}{4}\)- 1 = (\(\frac{a-b}{2}\)- 1)(\(\frac{a-b}{2}\)+ 1)
\(x^2\left(x-3\right)^2-\left(x-3\right)^2-x^2+1=\left(x-3\right)^2\left(x^2-1\right)-\left(x^2-1\right)=\left(x^2-1\right)\left(x-3\right)^2=\left(x-1\right)\left(x+1\right)\left(x-3\right)^2\)
\(3x^2+4x+x^2-4\\ =4x^2+4x-4\\ =4\left(x^2+x-1\right)\)
1/(x+2)2 -(3x-1)2=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x2+12x
2/(x4+x2)(-2x3-2x)=x2(x2+1)-2x(x2+1)=(x2+1)(x2-2x)
\(64x^4+1\)
\(=64x^4+16x^2+1-16x^2\)
\(=\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)\)
\(a,=\left(m-y\right)\left(m+y\right)+a\left(m+y\right)=\left(m+y\right)\left(m-y+a\right)\\ b,=3x\left(y-1\right)+\left(y-1\right)\left(y+1\right)=\left(y-1\right)\left(3x+y+1\right)\)
a: \(=\left(m-y\right)\left(m+y\right)+a\left(m+y\right)\)
\(=\left(m+y\right)\left(m-y+a\right)\)
Ta có :
\(x^4-3x^2+1\)
\(=\left(x^4-2x^2+1\right)-x^2\)
\(=\left(x^2-1\right)^2-x^2\)
\(=\left(x^2-1-x\right)\left(x^2-1+x\right)\)