Ta có:\(\frac{x}{y^2}=\frac{x}{y.y}=\frac{x}{y}.\frac{1}{y}=16.\frac{1}{y}=2\)

=>\(\frac{1}{y}\)=2:16=\(\frac{1}{8}\)

=>y=8

=\(\frac{x}{8}=16\)

=>x=16.8=128

Có: \(B=x^3-y^3\)

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

Lại có: \(x-y=2\)

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

\(\Rightarrow x^2-2xy+y^2=4\)

Mà: \(xy=35\)

\(\Rightarrow x^2-70+y^2=4\)

\(\Rightarrow x^2+y^2=74\)

\(\Rightarrow x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=2.\left(35+74\right)=2.109=218\)

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}\) => \(\dfrac{2x}{6}=\dfrac{y}{5}=\) \(\dfrac{4z}{8}\) 

Áp dụng tính chất dãy tỉ số bằng nhau  ta có 

\(\dfrac{2x}{6}\) = \(\dfrac{y}{5}\) = \(\dfrac{4z}{8}\)= \(\dfrac{2x-y+4z}{6-5+8}\) = \(\dfrac{27}{9}\) = 3

\(x\) = 3.6:2= 9

y=3.5 = 15

z =  3.8:4 = 6

a) x:2=10:5

=>x:2=2

=>x=4

b) \(\frac{x}{2}=\frac{y}{2}\)và x+y=10

áp dụng t/c dãy tỉ số = nhau

\(\frac{x}{2}=\frac{y}{2}=\frac{x+y}{2+2}=\frac{10}{4}=2,5\)

\(\frac{x}{2}=2,5\Rightarrow x=5\)

\(\frac{y}{2}=2,5\Rightarrow y=5\)

c) 3x=2y

=> \(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{1}{5}.\frac{x}{2}=\frac{y}{3}.\frac{1}{5}\Rightarrow\frac{x}{10}=\frac{y}{15}\)(1)

7y=5z

=> \(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{1}{3}.\frac{y}{5}=\frac{1}{3}.\frac{z}{7}\Rightarrow\frac{y}{15}=\frac{z}{21}\)(2)

từ (1) và (2)

=> \(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)

áp dụng t/c dãy t/s = nhau

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{32}{-16}=-2\)

\(\frac{x}{10}=-2\Rightarrow x=-20\)

\(\frac{y}{15}=-2\Rightarrow y=-30\)

\(\frac{z}{21}=-2\Rightarrow z=-42\)

d) \(2019-\left|x-2019\right|=x\Rightarrow\left|x-2019\right|=2019-x\)

\(\Rightarrow\orbr{\begin{cases}x-2019=-2019+x\\x-2019=2019-x\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=4038\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2019\end{cases}}\)

Để  2 đường thẳng trên song song \(\Leftrightarrow\hept{\begin{cases}m^2+1=5\\m\ne2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}m=\pm2\\m\ne2\end{cases}}\)

\(\Leftrightarrow m=-2\)

Vậy ..

\(\left(-\frac{1}{4}x^2y^2\right).\left(\frac{3}{8}x^5y^3\right)\)

=\(\left(-\frac{1}{4}.\frac{3}{8}\right).\left(x^2.x^5\right).\left(y^2.y^3\right)\)

=\(-\frac{3}{32}x^7y^5\)

Bậc của đơn thức \(-\frac{3}{32}x^7y^5\) có bậc 12

Nhớ tick cho mình nha!

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

\(\Leftrightarrow x^2-2xy+y^2=4\)

\(\Leftrightarrow x^2+y^2=4+2\cdot15=34\)

\(C=x^3-y^3+5x^2+5y^2\)

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

\(C=2\cdot\left(34+15\right)+5\cdot34\)

\(C=268\)