\(\frac{7}{2}+\frac{7}{6}+\frac{7}{12}+\frac{7}{20}+\frac{7}{30}+\frac{7}{42}+\frac{7}{56}+\frac{7}{72}+\frac{7}{90}\)\(\frac{7}{90}\)

=\(\frac{7}{2+6+12+20+30+42+56+72+90}\)

=\(\frac{63}{10}\)

=6.3

Ta có : ''Phần hơn'' của \(\frac{7^{58}+2}{7^{57}+2}\) là :

             \(\frac{7^{58}+2}{^{ }7^{57}+2}\) \(-\) 1 = \(\frac{7^{57}.6}{7^{57}+2}\)

             ''Phần hơn'' của \(\frac{5^{57}+2017}{5^{56}+2017}\) với 1 là :

             \(\frac{7^{57}+2017}{7^{56}+2017}\) \(-\) 1 = \(\frac{7^{56}.6}{7^{56}+2017}\)

           Ta có :\(\frac{7^{56}.6}{7^{56}+2017}\) = \(\frac{7^{56}.7.6}{\left(7^{56}+2017\right)7}\) = \(\frac{7^{57}.6}{7^{57}+14119}\)

         Ta thấy \(\frac{7^{57}.6}{7^{57}+2}\)> \(\frac{7^{57}.6}{7^{57}+14119}\)

         Suy ra \(\frac{7^{57}.6}{7^{57}+2}\) > \(\frac{7^{56}.6}{7^{56}+2017}\)

         Do đó \(\frac{7^{58}+2}{7^{57}+2}\) > \(\frac{7^{57}+2017}{7^{56}+2017}\)

Theo đầu bài ta có:
\(\frac{x-7}{7}+\frac{x-7}{8}=\frac{15}{56}\)
\(\Rightarrow\left(x-7\right)\cdot\frac{1}{7}+\left(x-7\right)\cdot\frac{1}{8}=\frac{15}{56}\)
\(\Rightarrow\left(x-7\right)\cdot\left(\frac{1}{7}+\frac{1}{8}\right)=\frac{15}{56}\)
\(\Rightarrow\left(x-7\right)\cdot\frac{15}{56}=\frac{15}{56}\)
\(\Rightarrow x-7=1\)
\(\Rightarrow x=8\)

\(\frac{3x-7}{5}=\frac{2x-1}{3}\)

\(\Leftrightarrow9x-21=10x-5\)

\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)

\(\frac{4x-7}{12}-x=\frac{3x}{8}\)

\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow-56-64x=36x\)

\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)

\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)

Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0

Vậy x = 2019

\(\frac{5x-8}{3}=\frac{1-3x}{2}\)

\(\Leftrightarrow10x-16=3-9x\)

\(\Leftrightarrow19x=19\Leftrightarrow x=1\)

\(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)

\(\Rightarrow\frac{4x-20-6x+54}{24}=\frac{5x-3+16}{8}\)

\(\Rightarrow\frac{-2x+34}{24}=\frac{5x+13}{8}\)

\(\Rightarrow-16x-272=120x+312\)

\(\Leftrightarrow-136x=584\Leftrightarrow x=\frac{-73}{17}\)

1/5

duyệt đi bạn

a) \(\left(56\times27+56\times35\right)\div62=56\times\left(27+35\right)\div62=56\times62\div62=56\)

b) \(\frac{0,18\times1230+0,9\times4567\times2+3\times5310\times6}{1+4+7+10+....+52+55-514}\)

\(=\frac{0,18\times1230+\left(0,9\times2\right)\times4567+\left(3\times6\right)\times5310}{1+4+5+.....+52+55-514}\)

\(=\frac{0,18\times1230+0,18\times4567+0,18\times5310}{1+4+7+...+52+55-514}\)

\(=\frac{0,18\times\left(1230+4567+5310\right)}{\left(55+1\right)\times55\div2-514}\)

\(=\frac{0,18\times11107}{971}=\frac{1999,26}{971}\)

mình ra cũng giống bạn Forever _ Alone nhé!!!Chẳng qua mình không biết viết phân số

chư học mà

a) \(\frac{62}{7}.x=\frac{29}{9}\)

\(\Rightarrow x=\frac{29}{9}:\frac{62}{7}=\frac{29}{9}.\frac{7}{62}=\frac{203}{468}\)

b)\(\frac{1}{5}:x=\frac{1}{5}-\frac{1}{7}\)

\(\Rightarrow\frac{1}{5}:x=\frac{2}{35}\)

\(\Rightarrow x=\frac{1}{5}:\frac{2}{35}=\frac{1}{5}.\frac{35}{2}=\frac{7}{2}\)

X\(\Leftrightarrow-96x^2+505x+396=0\)ét vế trái : \(\frac{\left(4x+7\right)^2}{7}-\frac{\left(5x-1\right)^2}{7}=\frac{\left(4x+7-5x+1\right)\left(4x+7+5x-1\right)}{7}=\frac{\left(8-x\right)\left(9x+6\right)}{7}\)

pt <=> \(\frac{\left(8-x\right)\left(9x+6\right)}{7}=\frac{\left(8x-3\right)\left(3x+4\right)}{56}\)

\(\Leftrightarrow24\left(8-x\right)\left(3x+2\right)-\left(8x-3\right)\left(3x+4\right)=0\)

\(\Leftrightarrow-96x^2+505x+396=0\)

Giải phương trình trên được : \(x_1=\frac{505}{192}-\frac{\sqrt{407089}}{192}\); \(x_2=\frac{505}{192}+\frac{\sqrt{407089}}{192}\)

Vậy ..............................