Ta có : \(\left(2,7x-1\frac{1}{2}x\right):\frac{2}{7}=\frac{-21}{4}\)

\(\Rightarrow\)    \(\left(\frac{27}{10}x-\frac{3}{2}x\right):\frac{2}{7}=\frac{-21}{4}\)

\(\Rightarrow\)    \(x.\left(\frac{27}{10}-\frac{3}{2}\right)=\frac{-21}{4}.\frac{2}{7}\)

\(\Rightarrow\)    \(x.\frac{6}{5}=\frac{-3}{2}\)

\(\Rightarrow\)    \(x=\frac{-3}{2}:\frac{6}{5}\)

\(\Rightarrow\)    \(x=\frac{-5}{4}\)

Vậy \(x=\frac{-5}{4}\)

~ Chúc các bn thi cuối năm đạt điểm cao ~

x=\(-\frac{5}{4}\)

x+7/2010+x+6/2011=x+5/2012+x+4/2013

((x+7/2010)-1)+((x+6/2011)-1)=(x+5/2012)-1)+(x+4/2013)-1)

x+2017/2010+x+2017/2011-x+2017/2012-x+2017/2013=0

x+2017(1/2010+1/2011-1/2012-1/2013)=0

x+2017=0(vì 1/2010+1/2011-1/2012-1/2013<0)

x=-2017

vậy.......

tk mk nha bn

1.
a) \(\frac{11}{2}-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3\)
               \(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3-\frac{11}{2}\)
               \(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=-\frac{5}{2}\)
                          \(\left|2x+-\frac{3}{2}\right|=-\frac{2}{3}:\left(-\frac{5}{2}\right)\)
                          \(\left|2x+-\frac{3}{2}\right|=\frac{4}{15}\)
\(\Rightarrow\left|2x+-\frac{3}{2}\right|\in\text{{}\frac{4}{15};-\frac{4}{15}\)}
Nếu, \(2x+\left(-\frac{3}{2}\right)=\frac{4}{15}\)
                               \(2x=\frac{53}{30}\)
                                  \(x=\frac{53}{60}\)
Nếu, \(2x+\left(-\frac{3}{2}\right)=-\frac{4}{15}\)
                               \(2x=\frac{37}{30}\)
                                  \(x=\frac{37}{60}\)
Vậy \(x\in\text{{}\frac{53}{60};\frac{37}{60}\)}
b) \(\left|\frac{2}{7}x-\frac{1}{5}\right|-\left|-x+\frac{4}{9}\right|=0\)
    \(\left|\frac{2}{7}x-\frac{1}{5}\right|=\left|-x+\frac{4}{9}\right|\)
\(\Rightarrow\left|\frac{2}{7}x-\frac{1}{5}\right|\in\text{{}-x+\frac{4}{9};-\left(x+\frac{4}{9}\right)\)}
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-x+\frac{4}{9}\)
                          \(x=\frac{203}{405}\)
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-\left(-x+\frac{4}{9}\right)\)
         \(\frac{2}{7}x-\frac{1}{5}=x-\frac{4}{9}\)
            \(\frac{2}{7}x-x=\frac{1}{5}-\frac{4}{9}\)
                 \(-\frac{5}{7}x=-\frac{11}{45}\)
                           \(x=\frac{77}{225}\)
Vậy \(x\in\text{{}\frac{203}{405};\frac{77}{225}\)}

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{8}\)

\(=\frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{3.7}+\frac{1}{7.8}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)

\(=1-\frac{1}{8}+0+0+...+0\)

\(=\frac{7}{8}\)

= \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{6.7}+\frac{1}{7.8}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
= \(1-\frac{1}{8}=\frac{7}{8}\)

Áp ụng t/c của dãy tỉ số = nhau ta được :

 \(\frac{x}{7}=\frac{y}{13}=\frac{x+y}{7+13}=\frac{40}{20}=2\)

\(\Rightarrow\frac{x}{7}=2\Rightarrow x=14\)

\(\Rightarrow\frac{y}{13}=2\Rightarrow y=26\)

Để A nguyên thì x-3 chia hết cho 7.

Khi đó:x-3=7a(với a thuộc Z)

<=>x=7a+3

Vậy để A=\(\frac{x-3}{7}\) thuộc Z thì x=7a+3 với a thuộc Z

\(\frac{x^2+y^2}{10}=\frac{x^2-2y^2}{7}\Rightarrow10\left(x^2-2y^2\right)=7.\left(x^2+y^2\right)\)

\(10x^2-20y^2=7x^2+7y^2\)

\(10x^2-7x^2=20y^2+7y^2\)

\(3x^2=27y^2\)

\(x^2=9y^2\)

\(\Rightarrow x^4=81y^4\)

\(\text{Thay }x^4=81y^4\text{ vào }x^4y^4=81\text{ ta được:}\)

\(81y^4.y^4=81\)

\(y^8=1\)

\(\Rightarrow y=1\text{Hoặc }y=-1\)

\(\text{Với }y=1\text{ thì }x^4=81.1=81\Rightarrow x=3\text{ hoặc }x=-3\)

\(\text{Với }y=-1\text{ thì }x^4=81.1=81\Rightarrow x=3\text{ hoặc }x=-3\)

\(\text{Vậy }x=\left\{3;-3\right\};y=\left\{1;-1\right\}\)

1

x + 7/12= 15/18

x= 15/18-7/12

x= 1/4

k mk nhé

3

4/10= 2/5

\(\frac{x}{15}\)=\(\frac{2}{5}\)

x= 15:5x2= 6

x= 6

k mk nhé

\(A=3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}\)

\(A=3-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)

\(A=3-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)

\(A=3-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)

\(A=3-\left(1-\frac{1}{8}\right)\)

\(A=3-\frac{5}{8}\)

\(A=\frac{19}{8}\)