sao nhiều thế

\(\Leftrightarrow y\times\left(\dfrac{1}{7}+\dfrac{9}{21}+\dfrac{7}{3}\right)=\dfrac{5}{3}\)

\(\Leftrightarrow y\times\dfrac{61}{21}=\dfrac{5}{3}\)

\(\Leftrightarrow y=\dfrac{5}{3}:\dfrac{61}{21}=\dfrac{35}{61}\)

y x (1/7 +9/21 +7/3)= 5/3

y x 61/21 = 5/3

y = 5/3 : 61/21

y= 35/61

\(Q=10xy^2-\frac{3}{7}xy-8xy^2-\frac{4}{7}xy-y\)

a) \(Q=\left(10xy^2-8xy^2\right)+\left(-\frac{3}{7}xy-\frac{4}{7}xy\right)-y\)

\(Q=2xy^2-xy-y\)

b) Chỗ này sửa thành Q nhá 

Thay x = -7 ; y = -2 vào Q ta được :

\(Q=2\cdot\left(-7\right)\cdot\left(-2\right)^2-\left(-7\right)\cdot\left(-2\right)-\left(-2\right)\)

\(Q=2\cdot\left(-7\right)\cdot4-14+2\)

\(Q=-56-14+2\)

\(Q=-68\)

Vậy giá trị của Q = -68 khi x = -7 ; y = -2

Cộng 2 phương trình ta có 

\(x^3+y^3+\left(7xy+y-x\right)=\left(1+y-x+xy\right)+7\)

\(\Leftrightarrow x^3+y^3+6xy=8\)

\(\Rightarrow x^3+y^3+6xy-8=0\)

\(\Leftrightarrow\left(x+y-2\right)\left(x^2+y^2+4-xy+2y+2x\right)-6xy+6xy=0\)

\(\Leftrightarrow\left(x+y-2\right)\left(x^2+y^2+4-xy+2y+2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+y-2\\x^2+y^2+4-xy+2y+2x=0\end{cases}}\)

nếu \(x+y=2=>x=y=1\)

nếu \(x^2+y^2+4-xy+2y+2x=0=>x=y=-2\left(zô\right)lý\)

zậy x=y=1

\(\left(\frac{5}{7}\cdot x^2\cdot y\right)^3:\left(\frac{1}{7}\cdot x\cdot y\right)^3\)

=\(\left(\frac{5}{7}\cdot x^2\cdot y:\frac{1}{7}:x:y\right)^3\)

\(=\left(\frac{5}{7}\cdot7\cdot x^2:x\cdot y:y\right)^3\)

\(=\left(5\cdot x\right)^3\)

\(=5^3\cdot x^3\)

\(=125\cdot x^3\)

5-/2x+6/-/7-y/

bài 1:

   (-2x^5 y)^8:(-2x^5 y)^6

=(-2x^5 y)^8-6

=(-2x^5y)^2

=4x^25 y^2