Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu b) tạm thời ko bít làm =.=
Bài 1 :
\(d)\) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2x\)
\(\Leftrightarrow\)\(\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=2x\)
\(\Leftrightarrow\)\(\frac{4^6}{3^6}.\frac{6^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{2^6.3^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{3^6}{1}=2x\)
\(\Leftrightarrow\)\(2^{12}=2x\)
\(\Leftrightarrow\)\(x=\frac{2^{12}}{2}\)
\(\Leftrightarrow\)\(x=2^{11}\)
\(\Leftrightarrow\)\(x=2048\)
Vậy \(x=2048\)
Chúc bạn học tốt ~
Bài 1 :
\(a)\) Ta có :
\(4+\frac{x}{7+y}=\frac{4}{7}\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{4}{7}-4\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{-24}{7}\)
\(\Leftrightarrow\)\(\frac{x}{-24}=\frac{7+y}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{-24}=\frac{7+y}{7}=\frac{x+7+y}{-24+7}=\frac{22+7}{-17}=\frac{29}{-17}=\frac{-29}{17}\)
Do đó :
\(\frac{x}{-24}=\frac{-29}{17}\)\(\Rightarrow\)\(x=\frac{-29}{17}.\left(-24\right)=\frac{696}{17}\)
\(\frac{7+y}{7}=\frac{-29}{17}\)\(\Rightarrow\)\(y=\frac{-29}{17}.7-7=\frac{-322}{17}\)
Vậy \(x=\frac{696}{17}\) và \(y=\frac{-322}{17}\)
Chúc bạn học tốt ~
\(A=\frac{7\times9+14\times27+21\times36}{21\times27+42\times81+63\times108}=\frac{7\times9+7\times2\times9\times3+7\times3\times9\times4}{21\times27+21\times2\times27\times3+21\times3\times27\times4}=\frac{7\times9\times\left(1+2\times3+3\times4\right)}{21\times27\times\left(1+2\times3\times3\times4\right)}=\frac{7\times3\times3}{7\times3\times3\times9}=\frac{1}{9}\)
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
1. \(-3x+\frac{5}{6}=-\frac{1}{2}\)
\(\Leftrightarrow-3x=-\frac{4}{3}\)
\(\Leftrightarrow x=-\frac{4}{3}:\left(-3\right)\)
\(\Leftrightarrow x=\frac{4}{9}\)
2) \(\frac{5}{6}.\left(0,875-\frac{7}{5}\right)+\left(\frac{1}{8}+0,4\right):\frac{6}{5}\)
= \(\frac{5}{6}.\left(\frac{7}{8}-\frac{7}{5}\right)+\left(\frac{1}{8}+\frac{2}{5}\right):\frac{6}{5}\)
= \(\frac{5}{6}.\frac{-21}{40}+\frac{21}{40}:\frac{6}{5}\)
= \(\frac{-7}{16}+\frac{7}{16}\)
\(=0\)
1/ Tìm x: \(-3x+\frac{5}{6}=\frac{-1}{2}\)
=> \(-3x=\frac{-1}{2}-\frac{5}{6}\)
=> \(-3x=\frac{-3-5}{6}\)
=> \(-3x=\frac{-8}{6}\)
=> \(-3x=\frac{-4}{3}\)
=> \(3x=\frac{4}{3}\)
=> \(x=\frac{\frac{4}{3}}{3}=\frac{4}{9}\)
2/ Tính: \(\frac{5}{6}.\left(0,875-\frac{7}{5}\right)+\frac{\left(\frac{1}{8}+0,4\right)}{\frac{6}{5}}\)
= \(\frac{5}{6}\left(0,875-\frac{7}{5}\right)+\left(\frac{1}{8}+0,4\right).\frac{5}{6}\)
= \(\frac{5}{6}\left(0,875-\frac{7}{5}+\frac{1}{8}+0,4\right)\)
= \(\frac{5}{6}\left(0,875+0,4+\frac{1}{8}-\frac{7}{5}\right)\)
= \(\frac{5}{6}\left(\frac{51}{40}+\frac{1}{8}-\frac{7}{5}\right)\)
= \(\frac{5}{6}\left(\frac{51+5-56}{40}\right)\)
= \(\frac{5}{6}.0=0\)