Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
câu E
\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )
1: =>1/3:x=3/5-2/3=9/15-10/15=-1/15
=>x=-1/3:1/15=5
2: \(\Leftrightarrow x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)
=>x*2/3=-1
=>x=-3/2
3: \(\Leftrightarrow\dfrac{8}{3}:x=\dfrac{25}{12}:\dfrac{-3}{50}=\dfrac{25}{12}\cdot\dfrac{-50}{3}\)
hay x=-48/625
9: =>x=-2*3/1,5=-4
8: =>2/3:x=5/2:-3/10=5/2*(-10)/3=-50/6=-25/3
=>x=-2/3:25/3=-2/3*3/25=-2/25
a) \(\dfrac{1}{5}+x=-\dfrac{1}{2}\)
\(x=-\dfrac{1}{2}-\dfrac{1}{5}=-\dfrac{7}{10}\)
b) \(1\dfrac{3}{4}x+1\dfrac{1}{2}=-\dfrac{4}{5}\)
\(\dfrac{7}{4}x+\dfrac{3}{2}=-\dfrac{4}{5}\)
\(\dfrac{7}{4}x=-\dfrac{4}{5}-\dfrac{3}{2}=-\dfrac{23}{10}\)
\(x=\dfrac{-47}{35}\)
c) \(\left|x+\dfrac{3}{4}\right|-5=-2\)
\(\left|x+\dfrac{3}{4}\right|=-2+5=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{15}{4}\end{matrix}\right.\)
d) \(-\dfrac{5}{8}+x=\dfrac{4}{9}\)
\(x=\dfrac{4}{9}+\dfrac{5}{8}\)
\(x=\dfrac{77}{72}\)
\(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|-3.2+\dfrac{2}{5}\right|\)
\(\Rightarrow\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|\dfrac{-28}{5}\right|\)
\(\Rightarrow\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\dfrac{28}{5}\)
\(\Rightarrow\left|x-\dfrac{1}{3}\right|=\dfrac{28}{5}-\dfrac{4}{5}=\dfrac{24}{5}\)
\(\Rightarrow x-\dfrac{1}{3}=\left(\pm\dfrac{24}{5}\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{24}{5}\\x-\dfrac{1}{3}=\dfrac{-24}{5}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{24}{5}+\dfrac{1}{3}\\x=\dfrac{-24}{5}+\dfrac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{77}{15}\\x=\dfrac{-67}{15}\end{matrix}\right.\)
Vậy...
a) \(2\left(4x-30\right)-3\left(x+5\right)+4\left(x-10\right)=5\left(x+2\right)\)
\(\Leftrightarrow8x-60-3x+15+4x-40=5x+10\)
\(\Leftrightarrow9x-35=5x+10\)
\(\Leftrightarrow9x-5x=10+35\)
\(\Leftrightarrow4x=45\)
\(\Leftrightarrow x=\dfrac{45}{4}=11,25\)
b) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\left(6x+1\right)\)
\(\Leftrightarrow\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=4x+\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{31}{60}+x=4x+\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{31}{60}-\dfrac{2}{3}=4x-x\)
\(\Leftrightarrow3x=\dfrac{1}{60}\)
\(\Leftrightarrow x=\dfrac{1}{180}\)
c) \(\dfrac{7}{3}-\left(2x-\dfrac{1}{3}\right)=\left(-2\dfrac{1}{6}+1\dfrac{1}{2}\right):0,25\)
\(\Leftrightarrow\dfrac{7}{3}-2x+\dfrac{1}{3}=-1\dfrac{2}{3}:\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{8}{3}-2x=\dfrac{-5}{3}.4\)
\(\Leftrightarrow\dfrac{8}{3}-2x=\dfrac{-20}{3}\)
\(\Leftrightarrow2x=\dfrac{8}{3}+\dfrac{20}{3}\)
\(\Leftrightarrow2x=\dfrac{28}{3}\)
\(\Leftrightarrow x=4\dfrac{2}{3}\)
d) \(0,75+\dfrac{5}{9}:x=5\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{3}{4}+\dfrac{5}{9}:x=\dfrac{11}{2}\)
\(\Leftrightarrow\dfrac{5}{9}:x=\dfrac{11}{2}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{9}:x=\dfrac{19}{4}\)
\(\Leftrightarrow x=\dfrac{5}{9}:\dfrac{19}{4}\)
\(\Leftrightarrow x=\dfrac{20}{171}\)
1,
\(\left(2x+1\right)^3=-0,001\\ \left(2x+1\right)^3=\left(-0.1\right)^3\\ \Leftrightarrow2x+1=-0.1\\ 2x=-1.1\\ x=-\dfrac{11}{10}:2\\ x=-\dfrac{11}{20}\\ Vậy...\)
2,
\(\left(2x-3\right)^4=\left(2x-3\right)^6\\ \Leftrightarrow\left(2x-3\right)^6-\left(2x-3\right)^4=0\\ \Leftrightarrow\left(2x-3\right)^4\cdot\left[\left(2x-3\right)^2-1\right]=0\\ \Rightarrow\left\{{}\begin{matrix}\left(2x-3\right)^4=0\\\left(2x-3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\\left(2x-3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x=3\\2x-3=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\\ Vậyx\in\left\{\dfrac{3}{2};2\right\}\)
3, Làm tương tự câu 2
5,
\(9^x:3^x=3\\ \left(9:3\right)^x=3\\ 3^x=3\\ \Rightarrow x=1\\ Vậy...\)
6,
\(3^x+3^{x+3}=756\\ 3^x+3^x\cdot3^3\\ 3^x\cdot\left(1+27\right)=756\\ 3^x\cdot28=756\\ \Leftrightarrow3^x=27\\ 3^x=3^3\\ \Rightarrow x=3\\ vậy...\)
7,
\(5^{x+1}+6\cdot5^{x+1}=875\\ 5^{x+1}\cdot\left(1+6\right)=875\\ 5^{x+1}\cdot7=875\\ \Leftrightarrow5^{x+1}=125\\ \Leftrightarrow5^{x+1}=5^3\Leftrightarrow x+1=3\\ \Rightarrow x=2\\ Vậy...\)
9,
1) \(\dfrac{1}{4}x-\dfrac{1}{3}=\dfrac{-5}{9}\)
\(\Rightarrow\dfrac{1}{4}x=-\dfrac{2}{9}\Rightarrow x=-\dfrac{8}{9}\)
2) \(2^{x-3}-3.2^x=-92\)
\(\Rightarrow2^x\left(2^{-3}-3\right)=-92\)
\(\Rightarrow2^x.\dfrac{-23}{9}=-92\)
\(\Rightarrow2^x=32\Rightarrow x=5\)