Tìm x thỏa mãn: |3x-4|10 = |3x-4|20
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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left|3x+4\right|+\left|3x-1\right|=\left|3x+4\right|+\left|1-3x\right|\)
Theo bất đẳng thức: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\), ta có:
\(\left|3x+4\right|+\left|1-3x\right|\ge\left|3x+4+1-3x\right|=5\Rightarrow\left|3x+4\right|+\left|3x-1\right|\ge5\) (*)
Mặt khác:
Với mọi x ta có:
\(3\left(x+1\right)^2\ge0\Rightarrow3\left(x+1\right)^2+4\ge4\Rightarrow\dfrac{20}{3\left(x+1\right)^2+4}\le\dfrac{20}{4}\Rightarrow\dfrac{20}{3\left(x+1\right)^2+4}\le5\) (**)
Từ (*)(**) \(\Rightarrow\dfrac{20}{3\left(x+1\right)^2+4}=5\)
\(\Rightarrow3\left(x+1\right)^2+4=4\)
\(\Rightarrow3\left(x+1\right)^2=0\)
\(\Rightarrow\left(x+1\right)^2=0\)
\(\Rightarrow x=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>-2x=90/91
hay x=-45/91
b: =>2x=-7
hay x=-7/2
c: ->-3x=-12
hay x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
(3x - 1)10 = (3x - 1)20
(3x - 1)20 - (3x - 1)10 = 0
(3x - 1)10 . (3x - 1)10 - (3x - 1)10 . 1 = 0
(3x - 1)10 . [(3x - 1)10 - 1] = 0
\(\Rightarrow\orbr{\begin{cases}\left(3x-1\right)^{10}=0\\\left(3x-1\right)^{10}-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x-1=0\\\left(3x-1\right)^{10}=1\end{cases}}\)
3x - 1 = 0 (3x - 1)10 = 1
=> 3x = 1 3x - 1 = 1 hoặc 3x - 1 = -1
=> x = \(\frac{1}{3}\) 3x = 2 hoặc 3x = 0
x = \(\frac{2}{3}\) hoặc x = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: =>2x-9=10/91
=>2x=829/91
hay x=829/182
b: =>2x=-7
hay x=-7/2
c: =>-3x=-12
hay x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(3x-2\right)-3x^2=\dfrac{3}{4}\)
\(\Leftrightarrow3x^2-2x-3x^2=\dfrac{3}{4}\)
\(\Leftrightarrow-2x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{3}{4}:\left(-2\right)\)
\(\Leftrightarrow x=-\dfrac{3}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{3x+2}{3}=\dfrac{2y-6}{9}=\dfrac{\left(3x+2\right)+\left(2y-6\right)}{3+9}=\dfrac{3x+2y-4}{12}=\dfrac{3x+2y-4}{6x}\)
Suy ra 6x = 12 <=> x = 12 : 6 = 2
Khi đó \(\dfrac{3x+2}{3}=\dfrac{3\cdot2+2}{3}=\dfrac{8}{3}\)
Suy ra \(\dfrac{2y-6}{9}=\dfrac{8}{3}\Leftrightarrow2y-6=\dfrac{8\cdot9}{3}=24\)
\(\Leftrightarrow2y=24+6=30\Leftrightarrow y=30:2=15\)
Vậy x = 2; y = 15
![](https://rs.olm.vn/images/avt/0.png?1311)
TH1:3x-1<0 <=>x<1/3
Ta có:1-3x+2-x=4
<=>-4x=1
<=>x=-1/4(TMĐK)
TH2: x>=1/3 và x=<2 <=>1/3=<x=<2
Ta có:3x-1+2-x=4
<=>2x=3
<=>x=3/2(TMĐK)
TH3:x-2>0<=>x>2
Ta có:3x-1+x-2=4
<=>4x=7
<=>x=7/4(loại)
l3x-4l10=l3x-4l10
khi 3x-4=0 hoặc 3x-4=1
3x=4 hoặc 3x=5
x=4/3 hoặc x=5/3
Vậy x=4/3 hoặc x=5/3