9: \(=1-\dfrac{1}{99}+1-\dfrac{1}{100}+\dfrac{100}{101}\cdot\dfrac{1-4+3}{12}=2-\dfrac{199}{9900}=\dfrac{19601}{9900}\)

10: \(=\left(\dfrac{78}{79}+\dfrac{79}{80}+\dfrac{80}{81}\right)\cdot\dfrac{6+5+9-20}{30}=0\)

Ta có bất đẳng thức giá trị tuyệt đối: 

\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)

Dấu \(=\)khi \(AB\ge0\).

d) \(\left|x+1\right|+\left|x+2\right|+\left|2x-3\right|\)

\(\ge\left|x+1+x+2\right|+\left|2x-3\right|\)

\(=\left|2x+3\right|+\left|3-2x\right|\)

\(\ge\left|2x+3+3-2x\right|=6\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x+1\right)\left(x+2\right)\ge0\\\left(2x+3\right)\left(3-2x\right)\ge0\end{cases}}\Leftrightarrow-1\le x\le\frac{3}{2}\).

e) \(\left|x+1\right|+\left|x+2\right|+\left|x-3\right|+\left|x-5\right|\)

\(=\left(\left|x+1\right|+\left|3-x\right|\right)+\left(\left|x+2\right|+\left|5-x\right|\right)\)

\(\ge\left|x+1+3-x\right|+\left|x+2+5-x\right|\)

\(=4+7=11\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x+1\right)\left(3-x\right)\ge0\\\left(x+2\right)\left(5-x\right)\ge0\end{cases}}\Leftrightarrow-1\le x\le3\).

Do đó phương trình đã cho vô nghiệm. 

Trong tích có thừa số 9 - 3^2 = 9 - 9 = 0
Vậy (7 - 3^0).(8 - 3^1).(9 - 3^2)...(107 - 3^100) = 0

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

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

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

Đặt \(A=\left(\dfrac{3}{4}-81\right)\left(\dfrac{3^2}{5}-81\right)\left(\dfrac{3^3}{6}-81\right)\cdot...\cdot\left(\dfrac{3^{2000}}{2003}-81\right)\)

\(=\left(\dfrac{3^6}{9}-81\right)\cdot\left(\dfrac{3}{4}-81\right)\left(\dfrac{3^2}{5}-81\right)\cdot...\cdot\left(\dfrac{3^{2000}}{2003}-81\right)\)

\(=\left(81-81\right)\cdot\left(\dfrac{3}{4}-81\right)\left(\dfrac{3^2}{5}-81\right)\cdot...\cdot\left(\dfrac{3^{2000}}{2003}-81\right)\)

=0