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\(\left(-\dfrac{3}{4}\right)^{1000}\cdot\left(-\dfrac{4}{3}\right)^{1000}\)

\(=\left[\left(-\dfrac{3}{4}\right)\cdot\left(-\dfrac{4}{3}\right)\right]^{1000}\)

\(=1^{1000}\)

\(=1\)

\(\left(-\dfrac{3}{4}\right)^{1000}.\left(-\dfrac{4}{3}\right)^{1000}\\ =\left[\left(-\dfrac{3}{4}\right).\left(-\dfrac{4}{3}\right)\right]^{1000}\\ =1^{1000}\\ =1\)

Quy ước:\(a^1=a\)

Lời giải:

1.

$(\frac{5}{6})^{10}.(\frac{3}{10})^{10}=(\frac{5}{6}.\frac{3}{10})^{10}=(\frac{1}{4})^{10}$

$=\frac{1}{4^{10}}$

2.

$(\frac{4}{7})^{19}: (\frac{-12}{35})^{19}=(\frac{4}{7}: \frac{-12}{35})^{19}=(\frac{-5}{3})^{19}$

3.

$(\frac{-3}{7})^7:\frac{-3}{5}=\frac{(-3)^7}{7^7}.\frac{5}{-3}=\frac{5.(-3)^6}{7^7}=\frac{5.3^6}{7^7}$

giúp mình với ạ

3\(^{x+1}\) - 3\(^x\) = 1428

3\(^x\).( 3 - 1) = 1428

3\(^x\).2        = 1428

3\(^x\)           = 1428: 2

3\(^x\)           = 714 

3\(^{x+1}\) là số lẻ \(\forall\) \(x\) ⇒ 3\(^x\) \(\ne\) 714 ∀ \(x\) ⇒ \(x\) \(\in\) \(\varnothing\)

\(\dfrac{1}{2}x+2\dfrac{1}{2}=3\dfrac{1}{2}x.\left(-\dfrac{1}{3}\right)\\ \Rightarrow\dfrac{1}{2}x+\dfrac{5}{2}=\dfrac{7}{2}x.\left(-\dfrac{1}{3}\right)\\ \Rightarrow\dfrac{1}{2}x+\dfrac{5}{2}+\dfrac{7}{2}x=-\dfrac{1}{3}\\ \Rightarrow\left(\dfrac{1}{2}+\dfrac{7}{2}\right)x+\dfrac{5}{2}=-\dfrac{1}{3}\\ \Rightarrow4x=-\dfrac{17}{6}\\ \Rightarrow x=-\dfrac{17}{24}.\)

\(\dfrac{1}{2}x+2\dfrac{1}{2}=3\dfrac{1}{2}x-\dfrac{1}{3}\\ \Rightarrow\dfrac{1}{2}x-3\dfrac{1}{2}x=-\dfrac{1}{3}-2\dfrac{1}{2}\\ \Rightarrow\left(\dfrac{1}{2}-\dfrac{7}{2}\right)x=-\dfrac{1}{3}-\dfrac{5}{2}\\ \Rightarrow\dfrac{-6}{2}x=-\dfrac{17}{6}\\ \Rightarrow-3x=-\dfrac{17}{6}\\ \Rightarrow x=\left(-\dfrac{17}{6}\right):\left(-3\right)\\ \Rightarrow x=\dfrac{17}{18}\)

\(A=\dfrac{27^{10}+9^5}{9^{13}+27^2}\\ =\dfrac{\left(3^3\right)^{10}+\left(3^2\right)^5}{\left(3^2\right)^{13}+\left(3^3\right)^2}\\ =\dfrac{3^{30}+3^{10}}{3^{26}+3^6}\\ =\dfrac{3^{10}\left(3^{20}+1\right)}{3^6\left(3^{20}+1\right)}\\ =\dfrac{3^{10}}{3^6}\\=3^4\\ =81\)

\(A=\dfrac{27^{10}+9^5}{9^{13}+27^2}\)

\(A=\dfrac{\left(3^3\right)^{10}+\left(3^2\right)^5}{\left(3^2\right)^{13}+\left(3^3\right)^2}\)

\(A=\dfrac{3^{30}+3^{10}}{3^{26}+3^6}\)

\(A=\dfrac{3^{10}+\left(3^{20}+1\right)}{3^6.\left(3^{20}+1\right)}\)

\(A=\dfrac{3^{10}}{3^6}\)

\(A=3^4\)

\(A=81\)