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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
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
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/ \(\dfrac{x+1}{2}=\dfrac{2x+3}{5}\)
\(\Leftrightarrow5\left(x+1\right)=2\left(2x+3\right)\)
\(\Leftrightarrow5x+5=4x+6\)
\(\Leftrightarrow5x-4x=6-5\)
\(\Leftrightarrow x=1\left(tm\right)\)
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
b: 2x-3<0
=>2x<3
hay x<3/2
c: \(\left(2x-4\right)\left(9-3x\right)>0\)
=>(x-2)(x-3)<0
=>2<x<3
d: \(\dfrac{2}{3}x-\dfrac{3}{4}>0\)
=>2/3x>3/4
hay x>9/8
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)
3x : 2,7 = \(\dfrac{1}{3}:2\dfrac{1}{4}=\dfrac{1}{3}:\dfrac{9}{4}=\dfrac{1}{3}.\dfrac{4}{9}=\dfrac{4}{27}\)
=> 3x = \(\dfrac{4}{27}\). 2,7 =\(\dfrac{1028}{270}=4\)
=> x = 4 : 3 = 1,333...
h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây
a) Ta có: |2x-5| \(\ge\)0 với mọi x
mà |2x-5|=-4
=> x\(\in\varnothing\)
b)\(\dfrac{1}{3}-\left|\dfrac{5}{4}-2x\right|=\dfrac{1}{4}\)
=>\(\left|\dfrac{5}{4}-2x\right|=\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{1}{12}\)
=>\(\left[{}\begin{matrix}\dfrac{5}{4}-2x=\dfrac{1}{12}\\\dfrac{5}{4}-2x=-\dfrac{1}{12}\end{matrix}\right.=>\left[{}\begin{matrix}2x=\dfrac{5}{4}-\dfrac{1}{12}=\dfrac{7}{6}\\2x=\dfrac{5}{4}+\dfrac{1}{12}=\dfrac{4}{3}\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\dfrac{7}{12}\\x=\dfrac{2}{3}\end{matrix}\right.\)
phần c và d cũng tương tự bạn tự làm nha
a) |2x - 1| = 3x + 2
<=> \(\left[{}\begin{matrix}2x-1=3x+2\\2x-1=-3x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3=x\\5x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{5}\end{matrix}\right.\)
b) \(2x-\dfrac{4}{3}-1\dfrac{2}{3}=1\)
\(2x-\dfrac{4}{3}-\dfrac{5}{3}=1\)
\(2x-\dfrac{9}{3}=1\)
\(2x-3=1\)
<=> x = 4 : 2 = 2
c) \(\left|3x-\dfrac{1}{4}\right|=x-1\dfrac{3}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{1}{4}=x-1\dfrac{3}{4}\\3x-\dfrac{1}{4}=-x+1\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{7}{4}+\dfrac{1}{4}\\4x=\dfrac{7}{4}+\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{3}{2}\\4x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{1}{2}\end{matrix}\right.\)