kết quả là\(\frac{104}{423}\)à nếu thế thì bài này làm như sau

\(\left(\frac{2}{3}\right)^{x+1}+\left(\frac{2}{3}\right)^{x+3}=\frac{104}{423}\)

\(\left(\frac{2}{3}\right)^x.\left(\frac{2}{3}\right)+\left(\frac{2}{3}\right)^x+\left(\frac{2}{3}\right)^3=\frac{104}{423}\)

\(\left(\frac{2}{3}\right)^x.\left(\frac{2}{3}\right).\left[1+\left(\frac{2}{3}\right)^2\right]=\frac{104}{423}\)

\(\left(\frac{2}{3}\right)^x.\left(\frac{2}{3}\right).\left[1+\frac{4}{9}\right]=\frac{104}{423}\)

\(\left(\frac{2}{3}\right)^x.\left(\frac{2}{3}\right).\frac{13}{9}=\frac{104}{423}\)

\(\left(\frac{2}{3}\right)^x.\left(\frac{2}{3}\right)=\frac{104}{423}:\frac{13}{9}\)

\(\left(\frac{2}{3}\right)^x.\left(\frac{2}{3}\right)=\frac{8}{47}\)

\(\left(\frac{2}{3}\right)^x=\frac{8}{47}:\frac{2}{3}\)

\(\left(\frac{2}{3}\right)^x=\frac{12}{47}\)

Rồi tự suy ra tìm x nha

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\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}...\frac{-99}{100}\)(9 thừa số)

\(=-\frac{3.8.15...99}{4.9.16...100}=\frac{1.3.2.4.3.5...9.11}{2.2.3.3.4.4...10.10}=\frac{\left(1.2.3....9\right).\left(3.4.5...11\right)}{\left(2.3.4...10\right).\left(2.3.4...10\right)}=\frac{11}{10.2}=\frac{11}{20}\)

Đặt \(A=\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).....................\left(\frac{1}{100}-1\right)\)

\(\Rightarrow A=\left(\frac{1}{4}-\frac{4}{4}\right).\left(\frac{1}{9}-\frac{9}{9}\right)..................\left(\frac{1}{100}-\frac{100}{100}\right)\)

\(\Rightarrow A=\frac{-3}{4}.\frac{-8}{9}................\frac{-99}{100}\)

\(\Rightarrow A=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}..............\frac{-9.11}{10.10}\)

\(\Rightarrow A=-\frac{\left(1.2..................9\right).\left(3.4.............11\right)}{\left(2.3...........10\right).\left(2.3...............10\right)}=\frac{1.11}{10.2}=\frac{11}{20}\)

Vậy \(A=\frac{11}{20}\)

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Ta có : 

\(M=a^2+b^2+c^2+2ab+2ac+2bc=\left(a+b+c\right)^2\)* luôn đúng *

Thay a = 1 ; b = 2 ; c = 3 vào biểu thức M ta được : 

\(\left(1+2+3\right)^2=6^2=36\)

\(M=a^2+b^2+c^2+2ab+2ca+2bc\)

\(=\left(a^2+2ab+b^2\right)+\left(2ca+2bc\right)+c^2\)

\(=\left(a+b\right)^2+2c\left(a+b\right)+c^2=\left(a+b+c\right)^2\)

Thay \(a=1\); \(b=2\); \(c=3\)vào biểu thức ta được:

\(M=\left(1+2+3\right)^2=6^2=36\)

\(D=\left(\frac{6}{8}+1\right)\left(\frac{6}{18}+1\right)\left(\frac{6}{30}+1\right)...\left(\frac{6}{10700}+1\right)\)

\(=\frac{14}{8}.\frac{24}{18}...\frac{10706}{10700}=\frac{2.7.8...101.106}{1.8.2.9...100.107}\)

\(=\frac{101.7}{107}=\frac{707}{107}\)

\(D=\left(\frac{6}{8}+1\right).\left(\frac{6}{18}+1\right)..................\left(\frac{6}{10700}+1\right)\)

\(\Rightarrow D=\left(\frac{6}{8}+\frac{8}{8}\right).\left(\frac{6}{18}+\frac{18}{18}\right).................\left(\frac{6}{10700}+\frac{10700}{10700}\right)\)

\(\Rightarrow D=\frac{14}{8}.\frac{24}{18}...........\frac{10706}{10700}\)

\(\Rightarrow D=\frac{2.7}{1.8}.\frac{3.8}{2.9}.............\frac{101.106}{100.107}\)

\(\Rightarrow D=\frac{\left(2.3..............101\right).\left(7.8..........106\right)}{\left(1.2.................100\right).\left(8.9...............107\right)}=\frac{101.7}{1.107}=\frac{707}{107}\)

Vậy \(D=\frac{707}{107}\)

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