Tim x biet :
3x+1-3x=162
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3x + 17 = 3x + 1 + 16 .: 3x + 1 => 16 .: 3x + 1 => 3x + 1 = 1 ; 2 ; 4 ; 8 ; 16 => 3x = 0 ; 1 ; 3 ; 7 ; 15 => x = 0 ; 1 ; 5
ta co (2x-1)(3x+1)+(3x+4)(3-2x)=5
(=)6x2-3x+2x-1+6x-6x2+12-8x=5
(=)-4x+11=5
(=)-4x=-6
(=)x=3/2
(2x-1)(3x+1)+(3x-4)(3-2x)=5
<=> 6x2+2x-3x-1+9x-6x2-12+8x=5
<=> 16x-13=5
<=> 16x = 18
<=> x=9/8
3/4x + 5-(2/3x-4)-(1/6+1)=(1/3x+4)-(1/3x-3)
=3/4x+5-2/3x-4-1/6x+1=1/3x+4-1/3x-3
=-1/12=7
x=84
Đ/S...
\(\left(\dfrac{3x}{4}+5\right)-\left(\dfrac{2x}{3}-4\right)-\left(\dfrac{x}{6}+1\right)=\left(\dfrac{1}{3}+4\right)-\left(\dfrac{1}{3}x-3\right)\)
\(\Leftrightarrow\dfrac{3x}{4}-\dfrac{2x}{3}-\dfrac{x}{6}+5+4-1=\dfrac{13}{3}-\dfrac{1}{3}x+9\)
\(\Leftrightarrow\dfrac{9x-8x-2x}{12}+8=\dfrac{13-x}{3}+\dfrac{27}{3}\)
\(\Leftrightarrow\dfrac{-x}{12}+\dfrac{96}{12}=\dfrac{40-x}{3}\Leftrightarrow\dfrac{96-x}{12}=\dfrac{160-4x}{12}\)
\(\Rightarrow96-160=-4x+x\Leftrightarrow-64=-3x\Leftrightarrow x=\dfrac{64}{3}\)
|2-3x|=x-1
2-3x=x-1 hoặc 2-3x=x-1
TH1:2-3x=x-1
3x-x=2-(-1)
2x=2+1
2x=3
x=3/2
TH2:2-3x=-(x-1)
2-3x=-x+1
3x+x=2-1
4x=1
x=1/4
Ta có : |2 - 3x| = x - 1
\(\Leftrightarrow\orbr{\begin{cases}2-3x=x-1\\2-3x=-x+1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-3x-x=-1-2\\-3x+x=1-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-4x=-3\\-2x=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{2}\end{cases}}\)
3x + 1 - 3x = 162
3x . 3 - 3x . 1 = 162
3x . ( 3 - 1 ) = 162
3x . 2 = 162
3x = 162 : 2
3x = 81
3x = 34
=> x = 4 =
\(3^{x+1}-3^x=162\Leftrightarrow3^x\left(3-1\right)=162\Leftrightarrow3^x.2=162\Leftrightarrow3^x=81\Leftrightarrow x=4\)