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\(3C=3+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2013}}.\)

\(\Rightarrow3C-C=\frac{1}{3^{2013}}-3\)

\(\Rightarrow C=\frac{\frac{1}{3^{2013}}-3}{2}\le\frac{3}{2}\)

Study well 

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Đúng đầu bài ko v bn ?

\(\frac{x}{2}=\frac{y}{1}=\frac{z}{3}\)

=> \(\frac{x^2}{4}=\frac{y^2}{1}=\frac{z^2}{9}\)

=> \(\frac{y^2}{1}.\frac{y^2}{1}=\frac{x^2}{4}.\frac{y^2}{1}=\frac{162}{4}=\frac{81}{2}\)

=> \(y^4=\frac{81}{2}\)=> y=> x,z

\(8\left(x+1\right)^2+y^2=35\)(1)

Dễ suy ra được \(y^2\)lẻ\(\Leftrightarrow\)y lẻ

Từ (1) suy ra \(y^2\le35\Leftrightarrow-6< y< 6\)

Từ đó suy ra \(y\in\left\{\pm5;\pm3;\pm1\right\}\)

*Nếu \(y=\pm1\)\(\Rightarrow8\left(x+1\right)^2=34\left(L\right)\)

*Nếu \(y=\pm3\Rightarrow8\left(x+1\right)^2=26\left(L\right)\)

*Nếu \(y=\pm5\Rightarrow8\left(x+1\right)^2=10\left(L\right)\)

Vậy không có x,y cần tìm

D=(2(x-1)/(x-1))-(1/x-1)          (đk  x-1 khác 0 => x khác 1)

để D đạt gtri nguyên thì x-1 phải là Ư(1)

=>x-1=1;x-1=-1

=>x=2;x=0 

Để D coa giá trị là một số nguyên:

\(\Rightarrow2x-3⋮x-1\)

\(\Rightarrow\left(2x-3\right)-2\left(x-1\right)⋮x-1\)

\(\Rightarrow2x-3-2x-2⋮x-1\)

\(\Rightarrow1⋮x-1\)

\(\Rightarrow x-1\in\left\{\pm1\right\}\)

Vậy \(x\in\left\{0;2\right\}\)

\(\)

\(\frac{2}{x-1}=\frac{3}{y-2}=\frac{4}{z.3}\)

=> \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{3.z}{4}\)

=> \(\frac{2\left(x-1\right)}{2.2}=\frac{3\left(y-2\right)}{3.3}=\frac{3z:3}{4:3}\)

=> \(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z}{\frac{4}{3}}=\frac{2x-2+3y-6-z}{4+9-\frac{4}{3}}=\frac{\left(2x+3y-z\right)-8}{\frac{35}{3}}=\frac{95-8}{\frac{35}{3}}=\frac{261}{35}\)

=> \(\hept{\begin{cases}\frac{2x-2}{4}=\frac{261}{35}\\\frac{3y-6}{9}=\frac{261}{35}\\\frac{z}{\frac{4}{3}}=\frac{261}{35}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{557}{35}\\y=\frac{853}{35}\\z=\frac{348}{35}\end{cases}}}\)

Câu hỏi của Nguyễn Ngọc Thảo Linh - Toán lớp 7 - Học toán với OnlineMath

Em tham khảo nhé! 

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có - 11/3^7.7^3 =-77/3^7.7^4 rồi so sánh với -78/3^7.7^4 nhé :))