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b)x2-y2-7x+7y
=(x2-y2)-(7x-7y)
=(x-y)(x+7)-7(x-y)
=(x-y)[(x+y)-7]
=(x-y)(x+y-7)
Thay x=107;y=7 vào biểu thức ta có:
(107-7)(107+7-7)
=100.107
=10700
\(=x^2+8x-x-8=x\left(x+8\right)-\left(x+8\right)=\left(x+8\right)\left(x-1\right)\)
\(=-5x^2+15x+x-3=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)
\(=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
\(=x^2+2\cdot6\cdot x+6^2=\left(x+6\right)^2\\ =\left(94+6\right)^2=100^2=10000\)
\(\Leftrightarrow x^2+3x+2-x^2+6x-9=11\)
=>x=2
\(\left(x+1\right)\left(x+2\right)-\left(x-3\right)^2=11\)
\(\Leftrightarrow x^2+3x+2-x^2+6x-9-11=0\)
\(\Leftrightarrow9x-18=0\)
<=> 9. (x - 2) = 0
<=> x = 2
\(x^2+xy+y^2+1=\left(x^2+xy+\frac{y^2}{4}\right)+\frac{3y^2}{4}+1=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1>0\)
a. \(x\left(3x+12\right)-\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)\\ =\left(3x^2+12x\right)-\left(7x-20\right)+\left(2x^3-3x^2\right)-\left(2x^3+5x\right)\\ =3x^2+12x-7x+20+2x^3-3x^2-2x^3-5x\\ =\left(2x^3-2x^3\right)+\left(3x^2-3x^2\right)+\left(12x-7x-5x\right)+20\\ =0+0+0+20\\ =20\)
\(=\left(x-y\right)\left(x+y\right)-7\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-7\right)\)
\(=100^2=10000\)