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1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
a, \(\dfrac{5}{6}-\left|2-x\right|=\dfrac{1}{3}\Rightarrow\dfrac{5}{6}-\dfrac{1}{3}=\left|2-x\right|\)
<=> \(\dfrac{1}{2}=\left|2-x\right|\) \(\Leftrightarrow\left[{}\begin{matrix}2-x=\dfrac{1}{2}\\2-x=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
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Mấy câu sau tương tự thôi
a)\(\dfrac{3}{2}hay\dfrac{-3}{2}\)
b)\(\dfrac{13}{20}hay\dfrac{-13}{20}\)
c)\(\dfrac{11}{6}hay\dfrac{-11}{6}\)
d)\(\dfrac{4}{3}hay\dfrac{-4}{3}\)
e)\(\dfrac{1}{5}hay\dfrac{-1}{5}\)
Đây là câu trả lời của mình
Hay có nghĩa là hoặc
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
1: \(\Leftrightarrow\left(x+1\right)^2=4\)
=>x+1=2 hoặc x+1=-2
=>x=1 hoặc x=-3
2: \(\Leftrightarrow7x-21=5x+25\)
=>2x=46
=>x=23
3: \(\Leftrightarrow x^2+4x+3=x^2+0.5x+4x+2\)
=>4,5x+2=4x+3
=>x=1
a/ \(\dfrac{1}{3}-\dfrac{2}{5}+3x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1}{15}+3x=\dfrac{3}{4}\)
\(\Leftrightarrow3x=\dfrac{49}{60}\)
\(\Leftrightarrow x=\dfrac{49}{180}\)
Vậy....
b/ \(\dfrac{3}{2}-1+4x=\dfrac{2}{3}-7x\)
\(\Leftrightarrow\dfrac{1}{2}+4x=\dfrac{2}{3}-7x\)
\(\Leftrightarrow4x+7x=\dfrac{2}{3}-\dfrac{1}{2}\)
\(\Leftrightarrow11x=\dfrac{1}{6}\)
\(\Leftrightarrow x=\dfrac{1}{66}\)
Vậy....
c/ \(2\left(\dfrac{3}{4}-5x\right)=\dfrac{4}{5}-3x\)
\(\Leftrightarrow\dfrac{3}{2}-10x=\dfrac{4}{5}-3x\)
\(\Leftrightarrow-10x+3x=\dfrac{4}{5}-\dfrac{3}{2}\)
\(\Leftrightarrow-7x=-\dfrac{7}{10}\)
\(\Leftrightarrow x=-\dfrac{1}{10}\)
Vậy .....
d/ \(4\left(\dfrac{1}{2}-x\right)-5\left(x-\dfrac{3}{10}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow2-4x-5x-\dfrac{3}{2}=\dfrac{7}{4}\)
\(\Leftrightarrow2+\left(-4x\right)+\left(-5x\right)+\left(\dfrac{-3}{2}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow-9x+\dfrac{1}{2}=\dfrac{7}{4}\)
\(\Leftrightarrow-9x=\dfrac{5}{4}\)
\(\Leftrightarrow x=-\dfrac{5}{36}\)
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. )
a/ \(\dfrac{x+1}{2}=\dfrac{2x+3}{5}\)
\(\Leftrightarrow5\left(x+1\right)=2\left(2x+3\right)\)
\(\Leftrightarrow5x+5=4x+6\)
\(\Leftrightarrow5x-4x=6-5\)
\(\Leftrightarrow x=1\left(tm\right)\)
Vậy ...
b/ \(\left|x-1\right|+3\left|y+1\right|+\left|z+2\right|=0\)
Mà với \(\forall x;y;z\) ta có :
\(\left\{{}\begin{matrix}\left|x-1\right|\ge0\\3\left|y+1\right|\ge0\\\left|z+2\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-1\right|=0\\3\left|y+1\right|=0\\\left|z+2\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+1=0\\z+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\\z=-2\end{matrix}\right.\)
Vậy ...
c/ \(\dfrac{x-2}{4}=\dfrac{5-3x}{4}\)
\(\Leftrightarrow x-2=5-3x\)
\(\Rightarrow x+3x=5+2\)
\(\Leftrightarrow4x=7\)
\(\Leftrightarrow x=\dfrac{7}{4}\)
Vậy ......
d/ \(\dfrac{x+2}{4}=\dfrac{4}{x+2}\)
\(\Leftrightarrow\left(x+2\right)\left(x+2\right)=16\)
\(\Leftrightarrow\left(x+2\right)^2=4^2=\left(-4\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Vậy ...
e/ \(\dfrac{x-1}{5}=\dfrac{-20}{x-1}\)
\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=-100\)
\(\Leftrightarrow\left(x-1\right)^2=-100\)
Lại có : \(\left(x-1\right)^2\ge0\)
\(\Leftrightarrow\) k tồn tại x
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
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
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
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.
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)\(\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
`#3107.101107`
\(\dfrac{3x}{5}=\dfrac{x-1}{3}\)
\(\Rightarrow3x\cdot3=5\left(x-1\right)\)
\(\Rightarrow5\left(x-1\right)=9x\)
\(\Rightarrow5x-5=9x\)
\(\Rightarrow5x-9x=5\)
\(\Rightarrow-4x=5\)
\(\Rightarrow x=-\dfrac{5}{4}\)
Vậy, \(x=-\dfrac{5}{4}.\)