nhiều lắm bn, lm sao tìm đc hết

a) \(x^{12}:\left(-x\right)^6\)

\(=x^{12}:x^6\)

\(=x^6.\)

b) \(\left(-x\right)^7:\left(-x\right)^5\)

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

\(=x^2.\)

c) \(5x^3y^4:10x^2y\)

\(=\frac{1}{2}xy^3.\)

d) \(\frac{3}{4}x^3y^3:\left(-\frac{1}{2}xy^2\right)\)

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

e) \(\left(-xy\right)^{14}:\left(-xy\right)^7\)

\(=\left(-xy\right)^7\)

\(=-x^7y^7.\)

Chúc bạn học tốt!

1) Ta có: \(\dfrac{1}{7}x^2y^3\cdot\left(-\dfrac{14}{3}xy^2\right)\cdot\left(-\dfrac{1}{2}xy\right)\left(x^2y^4\right)\)

\(=\left(-\dfrac{1}{7}\cdot\dfrac{14}{3}\cdot\dfrac{-1}{2}\right)\left(x^2y^3\cdot xy^2\cdot xy\cdot x^2y^4\right)\)

\(=\dfrac{1}{3}x^6y^{10}\)

2) Ta có: \(\left(3xy\right)^2\cdot\left(-\dfrac{1}{2}x^3y^2\right)\)

\(=9xy^2\cdot\dfrac{-1}{2}x^3y^2\)

\(=-\dfrac{9}{2}x^4y^4\)

3) Ta có: \(\left(-\dfrac{1}{4}x^2y\right)^2\cdot\left(\dfrac{2}{3}xy^4\right)^3\)

\(=\dfrac{1}{16}x^4y^2\cdot\dfrac{8}{27}x^3y^{12}\)

\(=\dfrac{1}{54}x^7y^{14}\)

Đặt x+y=a;xy=b

suy ra \(\hept{\begin{cases}4a+b=14\\a^2-b=7\end{cases}}\)

=> a^2+4a-21=0

<=> (a+7)(a-3)=0

Tự giải tiếp

\(-5.\left(2-x\right)+4\left(x-3\right)=10x+15\)

       \(-10+5x+4x-12=10x+15\)

                                \(9x-22=10x+15\)

                              \(10x-9x=-22-15\)

                                              \(x=-37\)

\(7.\left(x-9\right)-5\left(6-x\right)=-6+11x\)

       \(7x-63-30+5x=-6+11x\)

                          \(12x-93=-6+11x\)

                        \(12x-11x=-6+93\)

                                          \(x=87\)

\(xy+14+2y+7x=-10\)

\(\left(xy+2y\right)+\left(14+7x\right)=-10\)

\(y\left(x+2\right)+7\left(2+x\right)=-10\)

\(\left(x+2\right)\left(y+7\right)=-10\)

Câu cuối mk chỉ biết làm đến đó thôi 

bn tự làm nha