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14 tháng 9 2017
a/ \(\dfrac{x}{-\dfrac{5}{9}}=\dfrac{\dfrac{4}{3}}{-\dfrac{2}{5}}\)
\(\Leftrightarrow x.\left(\dfrac{-2}{5}\right)=\left(\dfrac{-5}{9}\right).\dfrac{4}{3}\)
\(\Leftrightarrow x.\left(-\dfrac{2}{5}\right)=\dfrac{-20}{27}\)
\(\Leftrightarrow x=\)\(\dfrac{-54}{100}\)
b/ tương tự
14 tháng 9 2017
b/ \(\dfrac{7}{\dfrac{5}{3x}}=\dfrac{-\dfrac{4}{9}}{\dfrac{5}{6}}\)
\(\Leftrightarrow\dfrac{5}{3x}.\left(-\dfrac{4}{9}\right)=7.\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{5}{3x}.\left(-\dfrac{4}{9}\right)=\dfrac{35}{6}\)
\(\Leftrightarrow\dfrac{5}{3x}=\dfrac{-315}{24}\)
\(\Leftrightarrow3x\left(-315\right)=5.24\)
\(\Leftrightarrow3x\left(-315\right)=120\)
tự tính tiếp, đang lười
14 tháng 9 2017
a/ \(\dfrac{\dfrac{-5}{12}}{\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|}=\dfrac{\dfrac{-4}{9}}{\dfrac{8}{15}}\)
\(\Leftrightarrow\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|.\left(-\dfrac{4}{9}\right)=\left(-\dfrac{5}{12}\right).\left(\dfrac{8}{15}\right)\)
\(\Leftrightarrow\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|.\left(-\dfrac{4}{9}\right)=\dfrac{-2}{9}\)
\(\Leftrightarrow\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\\dfrac{2}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=0\\\dfrac{2}{3}x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy ....
1 tháng 12 2022
a: \(\Leftrightarrow\dfrac{3}{4}:\dfrac{22}{9}-\left|3x-\dfrac{8}{3}\right|=\dfrac{3}{4}\)
\(\Leftrightarrow\left|3x-\dfrac{8}{3}\right|=\dfrac{3}{4}\cdot\dfrac{9}{22}-\dfrac{3}{4}=\dfrac{3}{4}\cdot\dfrac{-13}{22}=\dfrac{-39}{88}\)(loại)
b: \(\Leftrightarrow\left[{}\begin{matrix}3x-2=x-\dfrac{1}{3}\\3x-2=\dfrac{1}{3}-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{5}{3}\\4x=\dfrac{7}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{7}{12}\end{matrix}\right.\)
22 tháng 5 2022
a: \(\dfrac{31-2x}{x+23}=\dfrac{9}{4}\)
=>121-8x=9x+207
=>-17x=86
hay x=-86/17
b: \(\dfrac{\left|2x-1\right|}{\dfrac{1}{2}}=\dfrac{18}{5}\)
=>|2x-1|=9/5
=>2x-1=9/5 hoặc 2x-1=-9/5
=>2x=14/5 hoặc 2x=-4/5
=>x=7/5 hoặc x=-2/5
LH
21 tháng 9 2017
( 3x - \(\dfrac{1}{2}\))2 = \(\dfrac{4}{9}\)
=> 3x - \(\dfrac{1}{2}\)= \(\dfrac{4}{9}\): 2
=> 3x - \(\dfrac{1}{2}\)= \(\dfrac{2}{9}\)
=> 3x = \(\dfrac{2}{9}\)+ \(\dfrac{1}{2}\)
=> 3x= \(\dfrac{13}{18}\)
=> x = \(\dfrac{13}{54}\)
Chúc bạn học tốt !
HD
21 tháng 9 2017
\(\left(3x-\dfrac{1}{2}\right).2=\dfrac{4}{9}\)
\(\Rightarrow\left(3x-\dfrac{1}{2}\right)=\dfrac{4}{9}:2\)
\(\Rightarrow\left(3x-\dfrac{1}{2}\right)=\dfrac{2}{9}\)
\(\Rightarrow3x=\dfrac{2}{9}+\dfrac{1}{2}\)
\(\Rightarrow3x=\dfrac{13}{18}\)
\(\Rightarrow x=\dfrac{13}{18}:3\)
\(\Rightarrow x=\dfrac{13}{54}\)
Chúc bạn học tốt!
2 tháng 10 2017
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. )
27 tháng 11 2022
b: =>(3x-1)(3x+1)(2x+3)=0
hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)
c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)
=>2x-1/3=19/12 hoặc 2x-1/3=-19/12
=>2x=23/12 hoặc 2x=-15/12=-5/4
=>x=23/24 hoặc x=-5/8
d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)
=>-5/6x=-3/2
=>x=3/2:5/6=3/2*6/5=18/10=9/5
e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4
=>2/5x=5/4 hoặc 2/5x=-1/4
=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8
f: =>14x-21=9x+6
=>5x=27
=>x=27/5
h: =>(2/3)^2x+1=(2/3)^27
=>2x+1=27
=>x=13
i: =>5^3x*(2+5^2)=3375
=>5^3x=125
=>3x=3
=>x=1
DQ
28 tháng 10 2018
a) \(\left[\left(\dfrac{3}{5}\right)^2-\left(\dfrac{2}{5}\right)^2\right]\cdot X=\left(\dfrac{1}{5}\right)^3\)
\(\left(\dfrac{3}{5}-\dfrac{2}{5}\right)\left(\dfrac{3}{5}+\dfrac{2}{5}\right)\cdot X=\dfrac{1}{125}\)
\(\dfrac{1}{5}\cdot1\cdot X=\dfrac{1}{125}\)
\(X=\dfrac{1}{125}:\dfrac{1}{5}=\dfrac{1}{25}\)
b) \(1\dfrac{2}{5}\cdot x+\dfrac{3}{7}=\dfrac{-4}{5}\)
\(1\dfrac{2}{5}\cdot x=\dfrac{-4}{5}-\dfrac{3}{7}\)
\(1\dfrac{2}{5}\cdot x=-\dfrac{43}{35}\)
\(x=-\dfrac{43}{35}:1\dfrac{2}{5}=-\dfrac{43}{49}\)
c) \(\left(3x-2\right)^2=9\)
*Nếu \(9=3^2\) thì:
\(3x-2=3\)
\(3x=5\Rightarrow x=\dfrac{5}{3}\)
*Nếu \(9=\left(-3\right)^2\) thì
\(3x-2=-3\)
\(3x=-1\Rightarrow x=-\dfrac{1}{3}\)
d) \(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Chúc bạn học giỏi.
28 tháng 10 2018
a)\(\dfrac{3^2-2^2}{5^2}.x=\dfrac{1}{5^3}\)
\(\Leftrightarrow\dfrac{5}{5^2}.x=\dfrac{1}{5^3}\)
\(\Leftrightarrow\dfrac{1}{5}.x=\dfrac{1}{5^3}\)
\(\Leftrightarrow x=\dfrac{1}{25}\)
b)\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{7}{5}x=-\dfrac{43}{35}\)
\(\Leftrightarrow x=\dfrac{-43}{49}\)
c)\(9x^2-12x+4=9\)
\(\Leftrightarrow9x^2-12x-5=0\)
\(\Leftrightarrow9x^2-15x+3x-5=0\)
\(\Leftrightarrow3x\left(3x-5\right)+3x-5=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
d)\(\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
\(\left(3x-\dfrac{1}{2}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(3x-\dfrac{1}{2}\right)^2=\left(\dfrac{2}{3}\right)^2\\\left(3x-\dfrac{1}{2}\right)^2=\left(-\dfrac{2}{3}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{1}{2}=\dfrac{2}{3}\\3x-\dfrac{1}{2}=-\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{7}{6}\\3x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{18}\\x=-\dfrac{1}{6}\end{matrix}\right.\)
Vậy ..
thank nha Nguyễn Thanh Hằng tồ