Ta có x+y+1=0=>xây =-1

A = x³+x².y- x.y²-y3 + x² - y² +2.x+2.y +3

A = x² .(x+y)- y² .(x+y) + x² - y² +2.(x+y)+3

A= x².(-1)-y².(-1)+ x²-y²+ (-2)+3

A= x².0-y².0+1=1

\(A=\left(\dfrac{-3}{7}.x^3.y^2\right).\left(\dfrac{-7}{9}.y.z^2\right).\left(6.x.y\right)\)

\(A=\left(\dfrac{-3}{7}x^3y^2\right).\left(\dfrac{-7}{9}yz^2\right).6xy\)

\(A=\left(\dfrac{-3}{7}.\dfrac{-7}{9}.6\right).\left(x^3.x\right)\left(y^2.y.y\right).z^2\)

\(A=2x^4y^4z^2\)

\(B=-4.x.y^3\left(-x^2.y\right)^3.\left(-2.x.y.z^3\right)^2\)

\(B=\left[\left(-4\right).\left(-2\right)\right].\left(x.x^6.x^2\right)\left(y^3.y^3.y^2\right)\left(z^6\right)\)

\(B=8x^7y^{y^8}z^6\)

\(^{2^{25}}\) là \(2^{25}\) mé các bạn, mình sợ mọi người nhầm

Đợi tí nha bạn Phạm Mai Linh

Câu 1 :                                               Bài giải

Theo đề bài : \(x\text{ : }y\text{ : }z=5\text{ : }4\text{ : }3\text{ }\Rightarrow\text{ }\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=\frac{x+y-z}{5+4-3}=\frac{x+y-z}{6}=\frac{x-y+z}{5-4+3}=\frac{x-y+z}{4}\)

( Áp dụng t/c dãy tỉ số bằng nhau )

\(\Rightarrow\text{ }x+y-z=x-y+z\)

\(\Rightarrow\text{ }y=x-y+z+z-x=2z+y\)

\(A=\frac{x+2\cdot y-3\cdot z}{x-2\cdot y+3\cdot z}=\frac{\left(x+y-z\right)+\left(y-2z\right)}{\left(x-y+z\right)+\left(2z-y\right)}=\frac{\left(x+y-z\right)+\left(2z+y-2z\right)}{\left(x-y+z\right)+\left(2z-2z-y\right)}=\frac{\left(x+y-z\right)+y}{\left(x-y+z\right)+\left(-y\right)}\)

Đến đây chịu ! Nhưng giải gần xong rồi !

a. Ta có:

\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)=a^2\left(b-c\right)-b^2\left(b-c+a-b\right)+c^2\left(a-b\right)=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)

\(=\left(a-b\right)\left(c-a\right)\left(c-b\right)\)

và \(ab^2-ac^2-b^3+bc^2=a\left(b^2-c^2\right)-b\left(b^2-c^2\right)=\left(a-b\right)\left(b-c\right)\left(b+c\right)\)

Vậy, \(A=\frac{\left(a-b\right)\left(c-a\right)\left(c-b\right)}{\left(a-b\right)\left(b-c\right)\left(b+c\right)}=\frac{c-a}{-c-b}=\frac{a-c}{c+b}\)