\(VT=\dfrac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}=\dfrac{\left(x+y\right)\left(x+2y\right)}{\left(x+2y\right)\left(x-y\right)\left(x+y\right)}=\dfrac{1}{x-y}\)

\(\dfrac{1}{2}x.\dfrac{1}{4}x^2.\dfrac{x^3}{8}.2y.4y-8y^3=x.x^2.x^3.y.y.\dfrac{2.4}{2.4.8}-8y^3\\ =x^6.y^2.\dfrac{1}{8}-8y^3\)

\(=\dfrac{1}{2}\cdot\dfrac{1}{4}\cdot\dfrac{1}{8}\cdot x^3\cdot x^3\cdot8y^2-8y^3\)

\(=\dfrac{1}{8}x^6y^2-8y^3\)

Tớ làm luôn nhé , không chép lại đề đâu

\(\dfrac{x+2y+x-2y}{\left(x-2y\right)\left(x+2y\right)}+\dfrac{8y^2}{x\left(4y^2-x^2\right)}\)

\(=\dfrac{2x}{\left(x-2y\right)\left(x+2y\right)}-\dfrac{8y^2}{x\left(x-2y\right)\left(x+2y\right)}\)

=\(\dfrac{2x^2-8y^2}{x\left(x+2y\right)\left(x-2y\right)}=\dfrac{2\left(x^2-4y^2\right)}{x\left(x-2y\right)\left(x+2y\right)}\)

\(=\dfrac{2\left(x-2y\right)\left(x+2y\right)}{x\left(x-2y\right)\left(x+2y\right)}=\dfrac{2}{x}\)

Ta có: \(\dfrac{x-1}{6}=\dfrac{-2y+3}{30}\)

\(\Leftrightarrow\dfrac{3x-3}{18}=\dfrac{-8y+12}{120}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{3x-3}{18}=\dfrac{-8y+12}{120}=\dfrac{3x-3+8y-12}{18-120}=\dfrac{2-15}{-102}=\dfrac{13}{102}\)

Do đó: 

\(\left\{{}\begin{matrix}\dfrac{x-1}{6}=\dfrac{13}{102}\\\dfrac{3-2y}{30}=\dfrac{13}{102}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=\dfrac{13}{17}\\-2y+3=\dfrac{65}{17}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{30}{17}\\-2y=\dfrac{14}{17}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{30}{17}\\y=\dfrac{-7}{17}\end{matrix}\right.\)

Ta có: 5x - 5 = 3 - 2y

=> 5x+2y = 8

=> 20x + 8y = 32

Mà 3x +8y = 2

=> 17x = 30

=> x = \(\dfrac{30}{7}\)

=> y = ... giải tiếp nha bạn.

Xin 1 like nha bạn. Thx bạn

\(A=\dfrac{1}{5}x^2y^3+\dfrac{2}{3}x^2y^3-\dfrac{3}{4}x^2y^3+x^2y^3=\left(\dfrac{1}{5}+\dfrac{2}{3}-\dfrac{3}{4}+1\right)x^2y^3=\dfrac{67}{60}x^2y^3\\ B=\left(x^2y\right)^3\left(\dfrac{1}{2}xy^2z\right)^2=x^6y^3.\dfrac{1}{4}x^2y^4z^2=\dfrac{1}{4}x^8y^7z^2\)

Lời giải:
1.

\(\frac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}=\frac{a^2(a-4)-(a-4)}{(a^3-8)-(7a^2-14a)}=\frac{(a-4)(a^2-1)}{(a-2)(a^2+2a+4)-7a(a-2)}\)

\(=\frac{(a-4)(a-1)(a+1)}{(a-2)(a^2-5a+4)}=\frac{(a-4)(a-1)(a+1)}{(a-2)(a-1)(a-4)}=\frac{a+1}{a-2}\)

2.

\(\frac{x^2y^2+1+(x^2-y)(1-y)}{x^2y^2+1+(x^2+y)(1+y)}=\frac{x^2y^2+1+x^2-x^2y-y+y^2}{x^2y^2+1+x^2+x^2y+y+y^2}\)

\(=\frac{(x^2y^2-x^2y+x^2)+(y^2-y+1)}{(x^2y^2+x^2y+x^2)+(y^2+y+1)}\)

\(=\frac{x^2(y^2-y+1)+(y^2-y+1)}{x^2(y^2+y+1)+(y^2+y+1)}=\frac{(x^2+1)(y^2-y+1)}{(x^2+1)(y^2+y+1)}=\frac{y^2-y+1}{y^2+y+1}\)