\(\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}\)

\(=\frac{6}{1x4}+\frac{6}{4x7}+\frac{6}{7x10}+\frac{6}{10x13}\)

\(=2\left(\frac{3}{1x4}+\frac{3}{4x7}+\frac{3}{7x10}+\frac{3}{10x13}\right)\)

\(=2\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}\right)\)

\(=2\left(1-\frac{1}{13}\right)\)

\(=2x\frac{12}{13}\)

\(=\frac{24}{13}\)

\(\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}\)

\(=\frac{6}{1.4}+\frac{6}{4.7}+\frac{6}{7.10}+\frac{6}{10.13}\)

\(=2\left(\frac{3}{1.4}+\frac{3}{1.7}+\frac{3}{7.10}+\frac{3}{10.13}\right)\)

\(=2\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}\right)\)

\(\Leftrightarrow2=\left(1-\frac{1}{13}\right)\)

\(=2.\frac{12}{13}\)

\(=\frac{24}{13}\)

A=\(\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}+\frac{6}{208}\)

A=\(\frac{6}{1\cdot4}+\frac{6}{4\cdot7}+\frac{6}{7\cdot10}+\frac{6}{10\cdot13}+\frac{6}{13\cdot16}\)

A:2=\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\)

A:2=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\)\(\frac{1}{16}\)

A:2=\(1-\frac{1}{16}\)

A:2=\(\frac{15}{16}\)

A=\(\frac{15}{8}\)

vậy ...

Kết quả là

A = 15/8

   Đs...

\(A=\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}+\frac{1}{208}\)

\(\Rightarrow A=\frac{6}{1\cdot4}+\frac{6}{4\cdot7}+\frac{6}{7\cdot10}+\frac{6}{10\cdot13}+\frac{6}{13\cdot16}\)

\(\Rightarrow\)\(\frac{3}{6}A=\frac{3}{6}\cdot\left(\frac{6}{1\cdot4}+\frac{6}{4\cdot7}+\frac{6}{7\cdot10}+\frac{6}{10\cdot13}+\frac{1}{13\cdot16}\right)\)

\(\Rightarrow\)\(\frac{3}{6}A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\)

\(\Rightarrow\)\(\frac{3}{6}A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)

\(\Rightarrow\)\(\frac{3}{6}A=1-\frac{1}{16}\)

\(\Rightarrow\)\(\frac{3}{6}A=\frac{15}{16}\)

\(\Rightarrow A=\frac{15}{16}:\frac{3}{6}\Rightarrow A=\frac{15}{8}\)

1/3xD=1/(2x4)+1/(4x6)+...+1/(98x100)
2/3xD=2/(2x4)+2/(4x6)+...+1/(98x100)
2/3xD= 1/2-1/4+1/4-1/6+...+1/98-1/100
2/3xD=1/2-1/100
2/3xD=49/100
D=147/200

 

A=1/2+1/3+1/6+1/12
A=(1/2+1/3+1/6)+1/12
A=1+1/12
A=13/12

B=20/3+61/8+10/3+37/8
B=30/3+98/8
B=10+49/4
B=89/4

\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)

\(\Rightarrow\frac{7}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{9}\)

\(\Rightarrow\frac{21}{18}< \left|x-\frac{12}{18}\right|< \frac{52}{18}\)

còn lại cậu tự tính nha

\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)

\(\frac{7}{6}< x-\frac{2}{3}< \frac{26}{9}\)

\(\frac{11}{6}< x< \frac{32}{9}\)

Coi cái giá trị tuyệt đối là ẩn đi

\(\frac{4}{5}-|x-\frac{1}{6}|=\frac{2}{3}\)

\(\Rightarrow|x-\frac{1}{6}|=\frac{2}{15}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{6}=\frac{2}{15}\\x-\frac{1}{6}=-\frac{2}{15}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{1}{30}\end{cases}}\)

Vậy.....

b) \(\frac{26+x}{39-x}=\frac{6}{7}\)

=> 7( 26+ x) = 6(39-x) 

=>182 +7x  = 234 - 6x

=> 7x+6x = 234-182

=> 13x= 52 

=> x=4

a) \(\frac{26+x}{39+x}=\frac{6}{7}\)

=> 7(26+x) = 6(39+x) 

=> 182 + 7 x = 234 + 6x 

=> 7x - 6x = 234 - 182 

=> x = 52 

Thế nha ^^

1)

a)

\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)

\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)

\(\frac{-20}{5}< x< \frac{-3}{10}\)

\(\frac{-40}{10}< x< \frac{-3}{10}\)

\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)

\(\left(\frac{-5}{3}\right)^3< x< \frac{-24}{35}.\frac{-5}{6}\)

\(\frac{25}{3}< x< \frac{-4}{7}.\frac{1}{1}\)

\(\frac{-25}{3}< x< \frac{-4}{7}\)

\(\frac{-175}{21}< x< \frac{-12}{21}\)

\(\Rightarrow Z\in\left\{-13;-14;-15;-16;...;-174\right\}\)

2)

a)

\(\left(\frac{9}{10}-\frac{15}{16}\right).\left(\frac{5}{12}-\frac{11}{15}-\frac{7}{20}\right)\)

\(=\left(\frac{72}{80}-\frac{75}{80}\right).\left(\frac{25}{60}-\frac{44}{60}-\frac{21}{60}\right)\)

\(=\frac{-3}{80}.\frac{-40}{60}\)

\(=\frac{-1}{-2}.\frac{-1}{-20}\)

\(=\frac{1}{40}\)

\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)

\(\Leftrightarrow-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)

=> x \(\in\) {-1;0;1;2;3;4;5;6}

\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)

\(\Leftrightarrow\)\(\frac{9-10}{12}\le\frac{x}{12}< 1-\left(\frac{8-3}{12}\right)\)

\(\Leftrightarrow\)\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)

\(\Leftrightarrow-1\le x< 7\)

Mà x nguyên

=>x={-1;0;1;2;3;4;5;6}