tim x
\(\frac{x-7}{7}\)+\(\frac{x-7}{8}\)=\(\frac{15}{56}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{2}\left(x+1\right):\frac{3}{7}=\frac{32}{135}\)
\(\frac{1}{2}\left(x+1\right)=\frac{32}{135}.\frac{3}{7}\)
\(\frac{1}{2}\left(x+1\right)=\frac{32}{315}\)
\(x+1=\frac{32}{315}:\frac{1}{2}\)
\(x+1=\frac{64}{315}\)
\(x=\frac{64}{315}-1=-\frac{251}{315}\)
\(\frac{1+0,6-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{\frac{3}{3}+\frac{3}{5}-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{3.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}{8.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}=\frac{3.1}{8.1}=\frac{3}{8}\)
\(\frac{\frac{1}{3}+0,25-\frac{1}{5}+0,125}{\frac{7}{6}+\frac{7}{8}-0,7+\frac{7}{16}}=\frac{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}}{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}+\frac{7}{16}}=\frac{1.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}{7.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}=\frac{1.1}{7.1}=\frac{1}{7}\)
=>\(\frac{3}{8}-\frac{1}{7}=\frac{13}{56}\)
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
mk muốn xem bài của mk đúng hay sai thôi !
chứ làm thì mk làm xong rồi !
Theo đầu bài ta có:
\(\frac{x-7}{7}+\frac{x-7}{8}=\frac{15}{56}\)
\(\Rightarrow\left(x-7\right)\cdot\frac{1}{7}+\left(x-7\right)\cdot\frac{1}{8}=\frac{15}{56}\)
\(\Rightarrow\left(x-7\right)\cdot\left(\frac{1}{7}+\frac{1}{8}\right)=\frac{15}{56}\)
\(\Rightarrow\left(x-7\right)\cdot\frac{15}{56}=\frac{15}{56}\)
\(\Rightarrow x-7=1\)
\(\Rightarrow x=8\)