\(\frac{x}{2013}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)

\(\Leftrightarrow\frac{x}{2013}-\left(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\right)=\frac{5}{8}\)

\(\Leftrightarrow\frac{x}{2013}-\left[2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\right]=\frac{5}{8}\)

\(\Leftrightarrow\frac{x}{2013}-\left[2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\right]=\frac{5}{8}\)

\(\Leftrightarrow\frac{x}{2013}-2\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)

\(\Leftrightarrow\frac{x}{2013}-\frac{3}{8}=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2013}=\frac{5}{8}+\frac{3}{8}=1\Rightarrow x=2013\)

Vậy x = 2013

\(S=\left(\frac{1}{7}\right)^2+\left(\frac{2}{7}\right)^2+\left(\frac{3}{7}\right)^2+...+\left(\frac{10}{7}\right)^2\)

\(=\frac{1^2}{7^2}+\frac{2^2}{7^2}+\frac{3^2}{7^2}+...+\frac{10^2}{7^2}\)

\(=\frac{1^2+2^2+3^2+...+10^2}{7^2}\)

\(=\frac{385}{49}=\frac{55}{7}\)

Vậy S = \(\frac{55}{7}\)

\(\frac{x}{y^2}=\frac{x}{y.y}=\frac{x}{y}.\frac{1}{y}=27.\frac{1}{y}=3\)

\(\Rightarrow\frac{1}{y}=\frac{3}{27}=\frac{1}{9}\Rightarrow y=9\)

\(\Rightarrow\frac{x}{9}=27\Rightarrow x=27.9=243\)

Vậy x = 243; y = 9

Nhờ cô Loan,Minh Triều,... giải đi

\(S=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{100}\left(1+2+3+...+100\right)\)

Ta có công thứ \(1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)

Áp dụng vào bài toán ta được :

\(=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+....+\frac{1}{100}\cdot\frac{100.101}{2}\)

\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{101}{2}\)

\(=\frac{2+3+4+...+101}{2}=\frac{\frac{101.102}{2}-1}{2}=2575\)

\(A=\frac{2a+9}{a+3}+\frac{5a+17}{a+3}-\frac{3a}{a+3}=\frac{\left(2a+9\right)+\left(5a+17\right)-3a}{a+3}=\frac{4a+26}{a+3}=\frac{4\left(a+3\right)+14}{a+3}=3+\frac{14}{a+3}\)Để \(A=3-\frac{14}{a+3}\) là số nguyên <=> \(\frac{14}{a+3}\) là số nguyên

=> a + 3 thuộc ước nguyên dương của 14 ( vì a dương => a + 3 dương) => Ư(14) = { 1; 2; 7; 14 }

Ta có : a + 3 = 1 => a = - 2 (loại)

a + 3 = 2 => a = - 1 (loại)

a + 3 = 7 => a = 4 (TM)

a + 3 = 14 => a = 11 (TM)

Vậy a = { 4; 11 }

\(\frac{x}{y+z+t}=\frac{y}{x+z+t}=\frac{z}{x+y+t}=\frac{t}{x+y+z}\)

\(\Leftrightarrow\frac{x}{y+z+t}+1=\frac{y}{x+z+t}+1=\frac{z}{x+y+t}+1=\frac{t}{x+y+z}+1\)

\(\Leftrightarrow\frac{x}{x+y+z+t}=\frac{y}{x+y+z+t}=\frac{z}{x+y+z+t}=\frac{t}{x+y+z+t}\)

\(\Rightarrow x=y=z=t\) Thay vào p ta được

\(p=\frac{x+x}{x+x}+\frac{x+x}{x+x}+\frac{x+x}{x+x}+\frac{x+x}{x+x}=1+1+1+1=4\)

=> p là số nguyên (đpcm)

\(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)

\(\Rightarrow\frac{a}{a+b+d}+1=\frac{b}{c+d+a}+1=\frac{c}{a+b+d}+1=\frac{d}{a+b+c}+1\)

\(=\frac{a}{a+b+c+d}=\frac{b}{a+b+c+d}=\frac{c}{a+b+c+d}=\frac{d}{a+b+c+d}\)

\(\Rightarrow a=b=c=d\) Thay vào A ta được :

\(A=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}=1+1+1+1=4\)

x - y = 9 => x = 9 + y . Thay B ta có :

\(B=\frac{4\left(9+y\right)-9}{3\left(9+y\right)+y}-\frac{4y+9}{3y+9+y}=\frac{36+4y-9}{27+3y+y}-\frac{4y+9}{4y+9}=\frac{27+4y}{27+4y}-\frac{4y+9}{4y+9}=1-1=0\)

Vậy B = 0

X=7

**** cho mình nha