a: Vì 0,2<1

nên hàm số \(y=\left(0,2\right)^x\) nghịch biến trên R

mà -3<-2

nên \(\left(0,2\right)^{-3}>\left(0,2\right)^{-2}\)

b: Vì \(0< \dfrac{1}{3}< 1\)

nên hàm số \(y=\left(\dfrac{1}{3}\right)^x\) nghịch biến trên R

mà \(2000< 2004\)

nên \(\left(\dfrac{1}{3}\right)^{2000}>\left(\dfrac{1}{3}\right)^{2004}\)

c: Vì 3,2>1

nên hàm số \(y=\left(3,2\right)^x\) đồng biến trên R

mà \(1,5< 1,6\)

nên \(\left(3,2\right)^{1,5}< \left(3,2\right)^{1,6}\)

d: Vì \(0< 0,5< 1\)

nên hàm số \(y=\left(0,5\right)^x\) nghịch biến trên R

mà -2021>-2023

nên \(\left(0,5\right)^{-2021}< \left(0,5\right)^{-2023}\)

a.

\(\left(\dfrac{\sqrt{5}}{5}\right)^{-1,2}=\left(\dfrac{1}{\sqrt{5}}\right)^{-1,2}=\left(5^{-\dfrac{1}{2}}\right)^{-1,2}=5^{\left(-\dfrac{1}{2}\right).\left(-1,2\right)}=5^{0,6}>1\) do \(\left\{{}\begin{matrix}5>1\\0,6>0\end{matrix}\right.\)

b.

\(\left(\dfrac{1}{5}\right)^{\sqrt{2}}=\left(5^{-1}\right)^{\sqrt{2}}=5^{-\sqrt{2}}< 1\) do \(\left\{{}\begin{matrix}5>1\\-\sqrt{2}< 0\end{matrix}\right.\)

a: \(\left(\dfrac{\sqrt{5}}{5}\right)^{-1,2}=\left(\dfrac{1}{\sqrt{5}}\right)^{-\dfrac{6}{5}}=\left(1:\dfrac{1}{\sqrt{5}}\right)^{-\dfrac{5}{6}}=\left(\sqrt{5}\right)^{-\dfrac{5}{6}}\)

\(1=\left(\sqrt{5}\right)^0\)

mà -5/6<0 và \(\sqrt{5}>1\)

nên \(\left(\dfrac{\sqrt{5}}{5}\right)^{-1,2}>1\)

b: \(0< \dfrac{1}{5}< 1\)

=>\(\left(\dfrac{1}{5}\right)^{\sqrt{2}}< \left(\dfrac{1}{5}\right)^0=1\)

a: \(A=3^{\dfrac{2}{5}}\cdot3^{\dfrac{1}{5}}\cdot3^{\dfrac{1}{5}}=3^{\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{1}{5}}=3^{\dfrac{4}{5}}\)

b: \(B=\left(-27\right)^{\dfrac{1}{3}}=\left[\left(-3\right)^3\right]^{\dfrac{1}{3}}=\left(-3\right)^{\dfrac{1}{3}\cdot3}=\left(-3\right)^1=-3\)

c: \(C=\sqrt[3]{-64}\cdot\left(\dfrac{1}{2}\right)^3\)

\(=\sqrt[3]{\left(-4\right)^3}\cdot\dfrac{1}{2^3}=-4\cdot\dfrac{1}{8}=-\dfrac{4}{8}=-\dfrac{1}{2}\)

d: \(D=\left(-27\right)^{\dfrac{1}{3}}\cdot\left(\dfrac{1}{3}\right)^4\)

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

\(=\left(-3\right)^{3\cdot\dfrac{1}{3}}\cdot\dfrac{1}{81}=\dfrac{-3}{81}=\dfrac{-1}{27}\)

e: \(E=\left(\sqrt{3}+1\right)^{106}\cdot\left(\sqrt{3}-1\right)^{106}\)

\(=\left[\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\right]^{106}\)

\(=\left(3-1\right)^{106}=2^{106}\)

f: \(F=360^{\sqrt{5}+1}\cdot20^{3-\sqrt{5}}\cdot18^{3-\sqrt{5}}\)

\(=360^{\sqrt{5}+1}\cdot\left(20\cdot18\right)^{3-\sqrt{5}}\)

\(=360^{\sqrt{5}+1}\cdot360^{3-\sqrt{5}}=360^{\sqrt{5}+1+3-\sqrt{5}}=360^4\)

g: \(G=2023^{3+2\sqrt{2}}\cdot2023^{2\sqrt{2}-3}\)

\(=2023^{3+2\sqrt{2}+2\sqrt{2}-3}\)

\(=2023^{4\sqrt{2}}\)

a: \(A=\dfrac{25^6}{5^3}=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)

b: \(B=32\cdot\left(\dfrac{3}{2}\right)^5=32\cdot\dfrac{3^5}{2^5}=32\cdot\dfrac{243}{32}=243\)

c: \(C=\left(\dfrac{1}{3}\right)^4\cdot3^{-3}=3^{-4}\cdot3^{-3}=3^{-4-3}=3^{-7}\)

d: \(D=4^{-2}\cdot\left(\dfrac{2}{5}\right)^5\cdot5^4\)

\(=\dfrac{1}{4^2}\cdot\dfrac{2^5}{5^5}\cdot5^4\)

\(=\dfrac{1}{16}\cdot\dfrac{32}{5}=\dfrac{2}{5}\)

e: \(E=9^{-5}:\left(\dfrac{5}{3}\right)^4\cdot25^2\)

\(=\dfrac{1}{9^5}:\dfrac{5^4}{3^4}\cdot\left(5^2\right)^2\)

\(=\dfrac{1}{3^{10}}\cdot\dfrac{3^4}{5^4}\cdot5^4=\dfrac{1}{3^6}\)

f: \(F=\left(\dfrac{5}{8}\right)^{-2}:4^2\)

\(=\left(1:\dfrac{5}{8}\right)^2:4^2\)

\(=\left(\dfrac{8}{5}\right)^2\cdot\dfrac{1}{16}=\dfrac{64}{25}\cdot\dfrac{1}{16}=\dfrac{4}{25}\)

g: \(G=\left(\dfrac{5}{3}\right)^3\cdot\left(\dfrac{9}{2}\right)^2:\left(\sqrt{3}\right)^4\)

\(=\dfrac{5^3}{3^3}\cdot\dfrac{9^2}{2^2}:9\)

\(=\dfrac{5^3\cdot3^4}{3^3\cdot2^2}\cdot\dfrac{1}{3^2}\)

\(=\dfrac{125}{2^2\cdot3}=\dfrac{125}{3\cdot4}=\dfrac{125}{12}\)

\(A=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)

\(B=32.\left(\dfrac{3}{2}\right)^5=\dfrac{2^5.3^5}{2^5}=2^5\)

\(C=\left(\dfrac{1}{3}\right)^4.3^{-3}=\dfrac{1}{3^4.3^3}=\dfrac{1}{3^7}\)

\(D=4^{-2}.\left(\dfrac{2}{5}\right)^5.5^4=\dfrac{1}{\left(2^2\right)^2}.\dfrac{2^5}{5^5}.5^4=\dfrac{2}{5}\)

\(E=\dfrac{1}{9^5}.\dfrac{3^4}{5^4}.\left(5^2\right)^2=\dfrac{1}{3^{10}}.\dfrac{3^4}{5^4}.5^4=\dfrac{1}{3^6}\)

\(F=\dfrac{8^2}{5^2}:\left(2^2\right)^2=\dfrac{\left(2^3\right)^2}{5^2.2^4}=\dfrac{2^6}{5^2.2^4}=\dfrac{2^2}{5^2}\)

\(G=\dfrac{5^3}{3^3}.\dfrac{\left(3^2\right)^2}{2^2}:3^2=\dfrac{5^3}{3^3}.\dfrac{3^4}{2^2}.\dfrac{1}{3^2}=\dfrac{5^3}{3.2^2}\)

a: \(A=\dfrac{9^4}{3^2}=\dfrac{\left(3^2\right)^4}{3^2}=\dfrac{3^8}{3^2}=3^6\)=729

b: \(B=81\left(\dfrac{5}{3}\right)^4=81\cdot\dfrac{5^4}{3^4}=\dfrac{81}{3^4}\cdot5^4=5^4=625\)

c: \(C=\left(\dfrac{4}{7}\right)^{-4}\cdot\left(\dfrac{2}{7}\right)^3\)

\(=\left(\dfrac{7}{4}\right)^4\cdot\left(\dfrac{2}{7}\right)^3\)

\(=\dfrac{7^4}{4^4}\cdot\dfrac{2^3}{7^3}\)

\(=\dfrac{2^3}{4^4}\cdot7\)

\(=\dfrac{2^3}{2^8}\cdot7=\dfrac{7}{2^5}=\dfrac{7}{32}\)

d: \(D=7^{-6}\cdot\left(\dfrac{2}{3}\right)^0\left(\dfrac{7}{5}\right)^6\)

\(=7^{-6}\left(\dfrac{7}{5}\right)^6\)

\(=\dfrac{1}{7^6}\cdot\dfrac{7^6}{5^6}=\dfrac{1}{5^6}=\dfrac{1}{15625}\)

e: \(E=8^3:\left(\dfrac{2}{3}\right)^5\cdot\left(\dfrac{1}{3}\right)^2\)

\(=2^6:\dfrac{2^5}{3^5}\cdot\dfrac{1}{3^2}\)

\(=2^6\cdot\dfrac{3^5}{2^5}\cdot\dfrac{1}{3^2}\)

\(=\dfrac{2^6}{2^5}\cdot\dfrac{3^5}{3^2}=3^3\cdot2=54\)

f: \(F=\left(\dfrac{7}{9}\right)^{-2}\cdot\left(\dfrac{1}{\sqrt{3}}\right)^8\)

\(=\left(\dfrac{9}{7}\right)^2\cdot\left(\dfrac{1}{3}\right)^4\)

\(=\dfrac{9^2}{7^2}\cdot\dfrac{1}{3^4}=\dfrac{9^2}{3^4}\cdot\dfrac{1}{7^2}=\dfrac{81}{81}\cdot\dfrac{1}{49}=\dfrac{1}{49}\)

g: \(G=\left(-\dfrac{4}{5}\right)^{-2}\cdot\left(\dfrac{2}{5}\right)^2\cdot\left(\sqrt{2}\right)^3\)

\(=\left(-\dfrac{5}{4}\right)^2\cdot\left(\dfrac{2}{5}\right)^2\cdot2\sqrt{2}\)

\(=\dfrac{25}{16}\cdot\dfrac{4}{25}\cdot2\sqrt{2}=\dfrac{4}{16}\cdot2\sqrt{2}=\dfrac{8\sqrt{2}}{16}=\dfrac{\sqrt{2}}{2}\)

ĐKXĐ:

a.

\(2x-4>0\Rightarrow x>2\Rightarrow D=\left(2;+\infty\right)\)

b.

\(2x+8>0\Rightarrow x>-4\Rightarrow D=\left(-4;+\infty\right)\)

c.

\(4-x>0\Rightarrow x< 4\Rightarrow D=\left(-\infty;4\right)\)

d.

\(\dfrac{1}{x+4}>0\Rightarrow x>-4\Rightarrow D=\left(-4;+\infty\right)\)

e. 

\(\left(x-3\right)\left(x+9\right)>0\Rightarrow\left[{}\begin{matrix}x>3\\x< -9\end{matrix}\right.\) \(\Rightarrow D=\left(-\infty;-9\right)\cup\left(3;+\infty\right)\)

a: ĐKXĐ: 2x-4>0

=>2x>4

=>x>2

b: ĐKXĐ: 2x+8>0

=>2x>-8

=>x>-4

c: ĐKXĐ: 4-x>0

=>-x>-4

=>x<4

d: ĐKXĐ: \(\dfrac{1}{x+4}>0\)

=>x+4>0

=>x>-4

e: ĐKXĐ: \(\left(x-3\right)\left(x+9\right)>0\)

=>\(\left[{}\begin{matrix}x-3>0\\x+9< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -9\end{matrix}\right.\)

Bài 2: 

a: \(=\dfrac{7}{9}\left(\dfrac{7}{6}-\dfrac{19}{20}-\dfrac{1}{15}\right)+\dfrac{22}{5}\cdot\dfrac{1}{24}\)

\(=\dfrac{7}{9}\cdot\dfrac{3}{20}+\dfrac{22}{120}=\dfrac{7}{60}+\dfrac{11}{60}=\dfrac{18}{60}=\dfrac{3}{10}\)

b: \(=\left(\dfrac{35-32}{60}\right)^2+\dfrac{4}{5}\cdot\dfrac{70-45}{80}\)

\(=\dfrac{1}{400}+\dfrac{4\cdot25}{400}=\dfrac{101}{400}\)

a. Dãy là tổng cấp số nhân lùi vô hạn với \(\left\{{}\begin{matrix}u_1=1\\q=\dfrac{1}{10}\end{matrix}\right.\)

Do đó: \(S=\dfrac{u_1}{1-q}=\dfrac{1}{1-\dfrac{1}{10}}=\dfrac{10}{9}\)

b. Tương tự, tổng cấp số nhân lùi vô hạn với \(\left\{{}\begin{matrix}u_1=1\\q=-\dfrac{1}{3}\end{matrix}\right.\) bạn tự ráp công thức

c. \(S=2+S_1\) với \(S_1\) là cấp số nhân lùi vô hạn \(\left\{{}\begin{matrix}u_1=\dfrac{3}{10}\\q=\dfrac{3}{10}\end{matrix}\right.\)