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1) \(a^2-2ab+b^2-9=\left(a-b\right)^2-9=\left(a-b-3\right)\left(a-b+3\right)\)
2) \(x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
3) \(25-x^2-2xy-y^2=25-\left(x+y\right)^2=\left(5-x-y\right)\left(5+x+y\right)\)
\(1,=\left(a-b\right)^2-9=\left(a-b-3\right)\left(a-b+3\right)\\ 2,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 3,=25-\left(x+y\right)^2=\left(5-x-y\right)\left(5+x+y\right)\)
Lời giải:
a.
$2x(3x^2-4x+2)=2x.3x^2-2x.4x+2x.2$
$=6x^3-8x^2+4x$
b.
$2x(3x+5)-3(2x^2-2x+3)=2x.3x+2x.5-(3.2x^2-3.2x+3.3)$
$=6x^2+10x-6x^2+6x-9=16x-9$
a) \(\left(x+3\right)^2=x^2+6x+9\)
b) \(\left(2x+1\right)^2-\left(2x+3\right)\left(2x-3\right)=4x^2+4x+1-4x^2+9\)
\(=4x+10\)
\(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)=\left(x+1\right)\left(x-2\right)\)
\(2x\left(x-2\right)-\left(x-2\right)^2=\left(x-2\right)\left[2x-\left(x-2\right)\right]=\left(x-2\right)\left(2x-x+2\right)=\left(x-2\right)\left(x+2\right)\)
\(4x^2-20xy+25y^2=\left(2x\right)^2-2.2x.5y+\left(5y\right)^2=\left(2x-5y\right)^2\)
\(x^2+3x-x-3=x\left(x+3\right)-\left(x+3\right)=\left(x-1\right)\left(x+3\right)\)
\(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)
\(2y\left(x+2\right)-3x-6=2y\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(2y-3\right)\)
Bài 1:
\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)
Bài 2:
\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)
\(6x^3-4x^2+4x+3x^2-2x+2=6x^3-x^2+2x+2\)
\(=6x^3-4x^2+4x+3x^2-2x+2=6x^3-x^2+2x+2\)