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

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

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

=0

b: \(\dfrac{69}{157}-\left(2+\left(3+4+5^{-1}\right)^{-1}\right)^{-1}\)

\(=\dfrac{69}{157}-\left(2+\left(3+4+\dfrac{1}{5}\right)^{-1}\right)^{-1}\)

\(=\dfrac{69}{157}-\left(2+1:\dfrac{36}{5}\right)^{-1}\)

\(=\dfrac{69}{157}-\left(2+\dfrac{5}{36}\right)^{-1}\)

\(=\dfrac{69}{157}-\left(\dfrac{77}{36}\right)^{-1}\)

\(=\dfrac{69}{157}-\dfrac{36}{77}=\dfrac{-339}{12089}\)

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

Bài 2: 

x=13 nên x+1=14

\(f\left(x\right)=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+14\)

\(=x^{14}-x^{14}-x^{13}+x^{13}-...+x^3+x^2-x^2-x+14\)

=14-x=1

x=13 nên x+1=14

f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14

=x14−x14−x13+x13−...+x3+x2−x2−x+14=x14−x14−x13+x13−...+x3+x2−x2−x+14

=14-x=1

\(f\left(x\right)=4x^2+3x+1\)

\(g\left(x\right)=3x^2-2x+1.\)

a) \(h\left(x\right)=f\left(x\right)-g\left(x\right)\)

\(\Rightarrow h\left(x\right)=\left(4x^2+3x+1\right)-\left(3x^2-2x+1\right)\)

\(\Rightarrow h\left(x\right)=4x^2+3x+1-3x^2+2x-1\)

\(\Rightarrow h\left(x\right)=\left(4x^2-3x^2\right)+\left(3x+2x\right)+\left(1-1\right)\)

\(\Rightarrow h\left(x\right)=x^2+5x.\)

b) Ta có \(h\left(x\right)=x^2+5x.\)

Đặt \(x^2+5x=0\)

\(\Rightarrow x.\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=0-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy \(x=0\) và \(x=-5\) là các nghiệm của đa thức \(h\left(x\right).\)

Chúc bạn học tốt!

Đặt \(A=\left(1+\dfrac{7}{9}\right)\left(1+\dfrac{7}{20}\right)\left(1+\dfrac{7}{33}\right)....\left(1+\dfrac{7}{2900}\right)\)

\(B=\left(81-\dfrac{3}{4}\right)\left(81-\dfrac{3^2}{5}\right)\left(81-\dfrac{3^3}{6}\right)....\left(81-\dfrac{3^{2014}}{2017}\right)\)

Ta có:

\(A=\left(1+\dfrac{7}{9}\right)\left(1+\dfrac{7}{20}\right)\left(1+\dfrac{7}{33}\right).....\left(1+\dfrac{7}{2900}\right)\)

\(A=\dfrac{16}{9}.\dfrac{27}{20}.\dfrac{40}{33}.....\dfrac{2907}{2900}\)

\(A=\dfrac{2.8}{1.9}.\dfrac{3.9}{2.10}.\dfrac{4.10}{3.11}.....\dfrac{51.57}{50.58}\)

\(A=\dfrac{2.3.4.5.6....56.57}{1.2.3.4.5.....57.58}=\dfrac{1}{58}\)

\(B=\left(81-\dfrac{3}{4}\right)\left(81-\dfrac{3^2}{5}\right).....\left(81-\dfrac{3^{2014}}{2017}\right)\)

Vì trong dãy số trên có một thừa số là \(\left(81-\dfrac{3^6}{9}\right)=\left(81-81\right)=0\)

\(\Rightarrow B=0\)

Vì \(a=A+B\Rightarrow a=\dfrac{1}{58}+0=\dfrac{1}{58}\)(1)

Thay (1) vào đa thức \(f\left(x\right)=5x-29a\) ta được:

\(f\left(x\right)=5x-29.\dfrac{1}{58}=5x-\dfrac{1}{2}\)

Ta lại có:

\(f\left(x\right)=0\Leftrightarrow5x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{10}\)

Vậy nghiệm của đa thức trên là \(\dfrac{1}{10}\)

Chúc bạn học tốt!!!

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

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

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