a) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}=\dfrac{15x.2y^2}{7y^3.x^2}=\dfrac{30}{7xy}\)

b) \(\dfrac{4y^2}{11x^4}.\left(-\dfrac{3x^2}{8y}\right)=\dfrac{-4y^2.3x^2}{11x^4.8y}=\dfrac{-3y}{22x^2}\)

c) \(\dfrac{x^3-8}{5x+20}.\dfrac{x^2+4x}{x^2+2x+4}\\ =\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{5\left(x+4\right)}.\dfrac{x\left(x+4\right)}{x^2+2x+4}\\ =\dfrac{x^2-2x}{5}\)

Tớ làm luôn nhé , không chép lại đề đâu

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

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

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

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

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

\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x-4y+5z+3-12-25}{-3\cdot2-4\cdot4+5\cdot6}=\dfrac{16}{8}=2\)

Do đó: x=5; y=5; z=17

\(a,\dfrac{x^3}{8}=\dfrac{y^3}{27}=\dfrac{z^3}{64}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)

Áp dụng t/c dtsbn:

\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{x^2+2y^2-3z^2}{4+18-48}=\dfrac{-650}{-26}=25\\ \Rightarrow\left\{{}\begin{matrix}x^2=100\\y^2=225\\z^2=400\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm10\\y=\pm15\\z=\pm20\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)\) có giá trị là hoán vị của \(\left(\pm10;\pm15;\pm20\right)\)

a: =-4xyz^2

b: =-9x^2y

c: =16x^2y^2

d: =1/6x^2y^3

e: =13/6x^3y^2

f: =7/12x^4y

a) -xyz² - 3xz.yz

= -xyz² - 3xyz²

= -4xyz²

b) -8x²y - x.(xy)

= -8x²y - x²y

= -9x²y

c) 4xy².x - (-12x²y²)

= 4x²y² + 12x²y²

= 16x²y²

d) 1/2 x²y³ - 1/3 x²y.y²

= 1/2 x²y³ - 1/3 x²y³

= 1/6 x²y³

e) 3xy(x²y) - 5/6 x³y²

= 3x³y² - 5/6 x³y²

= 13/6 x³y²

f) 3/4 x⁴y - 1/6 xy.x³

= 3/4 x⁴y - 1/6 x⁴y

= 7/12 x⁴y

bài 1: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{x}{x+2}-\dfrac{x}{x-2}\)

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

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

Bài 2:

1: \(x^2y^2-8-1\)

\(=x^2y^2-9\)

\(=\left(xy-3\right)\left(xy+3\right)\)

2: \(x^3y-2x^2y+xy-xy^3\)

\(=xy\cdot x^2-xy\cdot2x+xy\cdot1-xy\cdot y^2\)

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

\(=xy\left[\left(x-1\right)^2-y^2\right]\)

\(=xy\left(x-1-y\right)\left(x-1+y\right)\)

3: \(x^3-2x^2y+xy^2\)

\(=x\cdot x^2-x\cdot2xy+x\cdot y^2\)

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

4: \(x^2+2x-y^2+1\)

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

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

\(=\left(x+1+y\right)\left(x+1-y\right)\)

5: \(x^2+2x-4y^2+1\)

\(=\left(x^2+2x+1\right)-4y^2\)

\(=\left(x+1\right)^2-4y^2\)

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

6: \(x^2-6x-y^2+9\)

\(=\left(x^2-6x+9\right)-y^2\)

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

\(\left(\dfrac{x}{x+2y}-\dfrac{x+2y}{2y}\right)\left(\dfrac{x}{x-2y}-1+\dfrac{8y^3}{8y^3-x^3}\right)=\dfrac{2xy-\left(x+2y\right)^2}{2y\left(x+2y\right)}\left(\dfrac{2y}{x-2y}+\dfrac{8y^3}{\left(2y-x\right)\left(4y^2+2yx+x^2\right)}\right)=\dfrac{-\left(x^2+2xy+4y^2\right)}{2y\left(x+2y\right)}\cdot\dfrac{2y\left(4y^2+2yx+x^2\right)-8y^3}{\left(x-2y\right)\left(x^2+2xy+4y^2\right)}=\dfrac{-\left(x^2+2xy+4y^2\right)2y\left(4y^2+2xy+x^2-4y^2\right)}{2y\left(x+2y\right)\left(x-2y\right)\left(x^2+2x+4y^2\right)}=\dfrac{-\left(x^2+2xy\right)}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{x}{2y-x}\)

T giải thử thôi nhé :w

a) \(1\frac{1}{4}x^2y\left(\frac{-5}{6}xy\right)^0.\left(-2\frac{1}{3}xy\right)\)

\(=\frac{5}{4}x^2y\left(\frac{-5}{6}xy\right)^0.\left(-\frac{5}{2}xy\right)\)

\(=1.\frac{5}{4}x^2y\left(-\frac{5}{2}xy\right)\)

\(=-\frac{5}{4}x^2y.1.\frac{5}{2}xy\)

\(=-1.\frac{5}{4}.\frac{5}{2}x^3y^2\)

\(=-1.\frac{25x^3y^2}{8}\)

\(=-\frac{25x^3y^2}{8}\)