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\(10x-25-x^2=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
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= (5x-25) + (5x - x2)
= 5(x-5) + x(5-x)
= 5(x-5) - x(x-5)
= (5 - x)(x - 5)
\(a,x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right).\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right).\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(b,9x^2+y^2+6xy=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)
\(c,6x-9-x^2=-\left(x^2-6x+9\right)=-\left(x^2-2.x.3+3^2\right)=-\left(x-3\right)^2\)
= (3x + 1 - x - 1)(3x + 1 + x + 1)
= 2x(4x + 2)
Em áp dụng hđt số 3 trong sgk nhé.
\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2=\left(x+2y\right)^2\)
\(a,8x^3+12x^2y+6xy^2+y^3=\left(2x\right)^3+3.\left(2x\right)^2.y+3.2x.y^2+y^3=\left(2x+y\right)^3\)
\(b,x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
\(c,4x^2-25=\left(2x\right)^2-5^2=\left(2x-5\right)\left(2x+5\right)\)
\(\left(x-2\right)^3-1=\left(x-2\right)\left[\left(x-3\right)^2+x-2\right]=\left(x-2\right)\left(x^2+5x+7\right)\)
\(\left(x+3y\right)^2-9y^2=x\left(x+6y\right)\)
\(\left(x+3\right)^2-\left(x-1\right)^2=4\left(2x+4\right)=8\left(x+2\right)\)
a) \(\left(x-2\right)^3-1=\left(x-2\right)^3-1^3=\left(x-2-1\right)\left[\left(x-2\right)^2+\left(x-2\right)\cdot1+1^2\right]\)\(=\left(x-3\right)\left(x^2-4x+4+x-2+1\right)\)
\(=\left(x-3\right)\left(x^2-3x+3\right)\)
b) \(\left(x+3y\right)^2-9y^2\)
\(=\left(x+3y\right)^2-\left(3y\right)^2\)
\(=\left(x+3y+3y\right)\left(x+3y-3y\right)\)
\(=x\left(x+6y\right)\)
c) \(\left(x+3\right)^2-\left(x-1\right)^2\)
\(=\left(x+3-x+1\right)\left(x+3+x-1\right)\)
\(=4\left(2x+2\right)\)
\(=8\left(x+1\right)\)
\(\left(x^2-2xz+z^2\right)-\left(y^2-2yt+t^2\right)=\left(x-z\right)^2-\left(y-t\right)^2=\left(x-z+y-1\right)\left(x-z-y+t\right)\)
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T I C K nha
\(x^2-x=x\left(x-1\right)\)