\(\frac{x}{12}=\frac{5}{6}\)
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WH
12 tháng 3 2018
\(-\frac{5}{9}-\left(-\frac{5}{12}\right)\)
\(=-\frac{5}{9}+\frac{5}{12}\)
\(=-\frac{5}{36}\)
*Tìm x
\(-\frac{5}{6}-x=\frac{7}{12}+\left(-\frac{1}{3}\right)\)
\(\Rightarrow-\frac{5}{6}-x=\frac{1}{4}\)
\(\Rightarrow x=-\frac{5}{6}-\frac{1}{4}\)
\(\Rightarrow x=-\frac{13}{12}\)
HN
12 tháng 3 2018
\(\frac{-5}{9}-\frac{-5}{12}\)
= \(\frac{-20}{36}-\frac{-15}{36}\)
= \(\frac{-5}{36}\)
Tìm x:
\(\frac{-5}{6}-x=\frac{7}{12}+\frac{-1}{3}\)
\(\frac{-5}{6}-x=\frac{7}{12}+\frac{-4}{12}\)
\(\frac{-5}{6}-x=\frac{1}{4}\)
\(x=\frac{-5}{6}-\frac{1}{4}\)
\(x=\frac{-10}{12}-\frac{3}{12}\)
\(x=\frac{-13}{12}\)
21 tháng 1 2019
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot....\cdot\frac{30}{62}\cdot\frac{31}{64}=2^x\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.....\cdot\frac{30}{31}\cdot\frac{31}{32}\right)=2^x\)
\(\Leftrightarrow\frac{1}{32}=2^{x+1}\)
Làm nốt.
ko làm được câu này hay câu b ib với tớ nha.khẳng định tối giải.
29 tháng 3 2018
\(x-\frac{1}{9}-\frac{3}{5}=\frac{3}{6}\)
\(x=\frac{3}{6}+\left(\frac{1}{9}-\frac{3}{5}\right)\)
\(x=\frac{3}{6}+\left(-\frac{22}{45}\right)\)
\(x=\frac{1}{90}\)
\(\frac{-12}{25}.\frac{3}{4}-x+\frac{6}{-11}-\frac{5}{6}=0\)
\(\frac{-9}{25}-x+\left(\frac{-91}{66}\right)=0\)
\(\frac{-9}{25}-x=0+\left(\frac{-91}{66}\right)\)
\(\frac{-9}{25}-x=\frac{-91}{66}\)
\(x=\left(\frac{-9}{25}\right)-\left(\frac{-91}{66}\right)\)
\(x=\frac{1681}{1650}\)
26 tháng 3 2020
\(B=81\cdot\left(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}\right)\cdot\frac{158158158}{711711711}\)
\(B=81\cdot\left(\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)+1}\right)\cdot\frac{2}{9}\)
\(B=81\cdot\left(\frac{12}{4}:\frac{6470}{7653}\right)\cdot\frac{2}{9}\)
Xem lại đề bài bẹn owii -.-
CP
10
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 8 2020
d/
\(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}cos\left(\frac{x}{5}-\frac{\pi}{12}\right)-\frac{\sqrt{3}}{2}sin\left(\frac{x}{5}-\frac{\pi}{12}\right)\right)=sin\left(\frac{x}{5}+\frac{2\pi}{3}\right)-sin\left(\frac{3x}{5}+\frac{\pi}{6}\right)\)
\(\Leftrightarrow\sqrt{2}cos\left(\frac{x}{5}-\frac{\pi}{12}+\frac{\pi}{3}\right)=2cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)sin\left(\frac{\pi}{4}-\frac{x}{5}\right)\)
\(\Leftrightarrow cos\left(\frac{x}{5}-\frac{\pi}{4}\right)=\sqrt{2}cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)cos\left(\frac{x}{5}-\frac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\frac{x}{5}-\frac{\pi}{4}\right)=0\\cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{5}-\frac{\pi}{4}=\frac{\pi}{2}+k\pi\\\frac{2x}{5}+\frac{5\pi}{12}=\frac{\pi}{4}+k2\pi\\\frac{2x}{5}+\frac{5\pi}{12}=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{15\pi}{4}+k5\pi\\x=-\frac{5\pi}{12}+k5\pi\\x=-\frac{5\pi}{3}+k5\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 8 2020
c/
\(\Leftrightarrow\sqrt{3}sin\left(x-\frac{\pi}{3}\right)+cos\left(\frac{\pi}{3}-x\right)=2sin1972x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin\left(x-\frac{\pi}{3}\right)+\frac{1}{2}cos\left(x-\frac{\pi}{3}\right)=sin1972x\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}+\frac{\pi}{6}\right)=sin1972x\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=sin1972x\)
\(\Leftrightarrow\left[{}\begin{matrix}1972x=x-\frac{\pi}{6}+k2\pi\\1972x=\frac{7\pi}{6}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{11826}+\frac{k2\pi}{1971}\\x=\frac{7\pi}{11838}+\frac{k2\pi}{1973}\end{matrix}\right.\)
\(\frac{x}{12}=\frac{5}{6}\)
=> 6x = 12.5
=> 6x = 60
=> x = 10
Vậy x = 10
\(\frac{x}{12}=\frac{5}{6}=\frac{10}{12}\)\(\Rightarrow x=10\)
Vậy \(x=10\)