Trả lời:

a, \(\left(x^2-2y\right)\left(x^4+2x^2y+4y^2\right)-x^3\left(x-y\right)\left(x^2+xy+y^2\right)+8y^3\)

\(=\left(x^2\right)^3-\left(2y\right)^3-x^3\left(x^3-y^3\right)+8y^3\)

\(=x^6-8y^3-x^6+x^3y^3+8y^3\)

\(=x^3y^3\)

b, \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3+7\)

\(=x^3-8-\left(x^3-3x^2+3x-1\right)+7\)

\(=x^3-8-x^3+3x^2-3x+1+7\)

\(=3x^2-3x\)

c, \(x\left(x+2\right)\left(2-x\right)+\left(x+3\right)\left(x^2-3x+9\right)\)

\(=x\left(4-x^2\right)+x^3+27\)

\(=4x-x^3+x^3+27\)

\(=4x+27\)

\(^{ }\)

a. \(x^2+x-2=0\)

\(\Leftrightarrow x^2+2x-x-2=0\)

\(\Leftrightarrow\left(x-1\right).\left(x+2\right)=0\)

tới đây chắc bạn giải tiếp đc r 

b.\(x^2+2x-3=0\)

\(\Leftrightarrow x^2+3x-x-3=0\)

\(\Leftrightarrow\left(x-1\right).\left(x+3\right)=0\)

rồi bạn giải tiếp nha

có chỗ nào ko hiểu cứ hỏi

Bài 1 : 

a, \(-2x^2y.3xy^2\left(-\frac{1}{2}xy\right)=3x^4y^4\)

b, \(4xy^2\left(\frac{1}{2}x^2y\right)^2\left(-12yz\right)=4xy^2.\frac{1}{4}x^4y^2\left(-12yz\right)=-12x^5y^5z\)

Trả lời:

d, \(-2x^2y.3xy^2.\left(-\frac{1}{2}xy\right)=-6x^3y^3.\left(-\frac{1}{2}xy\right)=3x^4y^4\)

e, \(4xy^2.\left(\frac{1}{2}x^2y\right)^2\left(-12yz\right)=4xy^2.\frac{1}{4}x^4y^2.\left(-12yz\right)=x^5y^4.\left(-12yz\right)=-12x^5y^5z\)

a) \(-x-y^2+x^2-y=-x-y+x^2-y^2=-\left(x+y\right)+\left(x-y\right)\left(x+y\right)\)\(=\left(x+y\right)\left(x-y-1\right)\)

b) \(x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)

c) \(x^2-5x+5y-y^2=x^2-y^2+5y-5x=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)\(=\left(x-y\right)\left(x+y-5\right)\)

d) \(5x^3-5x^2y-10x^2+10xy=5x^2\left(x-y\right)-10x\left(x-y\right)\)\(=\left(x-y\right)\left(5x^2-10x\right)=\left(x-y\right)5x\left(x-2\right)\)

e) \(27x^3-8y^3=\left(3x\right)^3-\left(2y\right)^3=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)

f) \(x^6-y^6=\left(x^2\right)^3-\left(y^2\right)^3=\left(x-y\right)\left(x+y\right)\left(x^4+x^2y^2+y^4\right)\)

\(x\left(a-b\right)-\left(a-b\right)=x\left(a-b\right)-1\left(a-b\right)=\left(x-1\right)\left(a-b\right)\)