\(\dfrac{-2^3\cdot3^3\cdot5^3\cdot7\cdot8}{3\cdot5^3\cdot2^4\cdot42}\)

\(=\dfrac{-2^6\cdot3^3\cdot5^3\cdot7}{3\cdot5^3\cdot2^4\cdot2\cdot3\cdot7}\)

\(=\dfrac{-2^6\cdot3^3\cdot5^3}{2^5\cdot3^2\cdot5^3}=-2\cdot3=-6\)

\(\frac{\left(-2\right)^3.3^3.5^3.7.8}{3.2^4.5^3.14}=\frac{2^5.3^3.5^3.7.\left(-2\right)}{3.2^5.5^3.7}=\frac{3^2.\left(-2\right)}{1}=-18\)

Ta có; \(\frac{\left(2^3.3^4\right)}{\left(2^2.3^3.5\right)}\Leftrightarrow\frac{\left(2^2.2^1.3^3.3^1\right)}{2^2.3^3.5}\Leftrightarrow\frac{\left(2.3\right)}{5}=\frac{6}{5}\)

Đs: \(\frac{6}{5}\)

k bt lm tự mà làm

dễ lắm bạn

\(\frac{4^7}{4^{10}}\)=\(\frac{1}{4^3}\) vì 4^7 : 4^7=1;4^10:4^7=4^3

\(\frac{5^{11}}{5^8}\)=\(5^3\)vì 5^11:5^8=5^3;5^8:5^8=1

chỉ tke thui.mấy cái gạch kia bỏ đi vì mk chưa sử dụng phần mềm rành lắm

4⁷/4¹⁰ = 1/64

5¹¹/5⁸ = 125

Ủng hộ nhaaa

\(\frac{4^7}{4^{10}}=\frac{1}{4^3}\)

\(\frac{5^{11}}{5^8}=\frac{5^3}{1}=5^3\)

dap an la : 1/2 nhe tic minh ne 

\(\frac{2^3}{2^4}=\frac{2.2.2}{2.2.2.2}=\frac{1}{2}\)

kết quả là 1 phần 2

Bạn viết rõ hơn nhé : 

\(\frac{x^4-xy^3}{2xy+y^2}:\frac{x^3+x^2y+xy^2}{2x+y}\)

= \(\frac{x^4-xy^3}{2xy+y^2}.\frac{2x+y}{x^3+x^2y+xy^2}\)

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

= \(\frac{x-y}{y}\)

Chúc bạn học tốt !!!

\(A=\frac{1}{2}-\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3-\left(\frac{1}{2}\right)^4+...-\left(\frac{1}{2}\right)^{20}\)

\(2A=1-\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3+...-\left(\frac{1}{2}\right)^{19}\)

\(2A-A=\)\(\left(1-\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3+...-\left(\frac{1}{2}\right)^{19}\right)-\)\(\left(\frac{1}{2}-\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3-\left(\frac{1}{2}\right)^4+...-\left(\frac{1}{2}\right)^{20}\right)\)

\(A=1-\left(\frac{1}{2}\right)^{20}\)