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3x+1 + 2x.3x - 18x - 27 = 0
=> 3x.3 + 2x.3x - (18x + 27) = 0
=> 3x.(3 + 2x) - 9.(2x + 3) = 0
=> (2x + 3).(3x - 9) = 0
\(\Rightarrow\orbr{\begin{cases}2x+3=0\\3^x-9=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}2x=-3\\3^x=9\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=2\end{cases}}\)
Vậy ...
<=> 3x(2x+3)-9(2x+3)=0
<=> (2x+3)(3x-9)=0
<=> 2x+3=0 => x=-3/2
Và: 3x-9=0 => 3x=9=32 => x=2
Đs: x=-3/2 và x=2
\(3^{x+1}+2x+3^x-18x-27=0\)
<=> \(3^x\left(3+2x\right)-9\left(2x+3\right)=0\)
<=> \(\left(2x+3\right)\left(3^x-9\right)=0\)
<=>\(\orbr{\begin{cases}x=-\frac{3}{2}\\x=2\end{cases}}\)
vậy.......
3x + 1 + 2x .3x - 18x - 27 = 0
<=> 3x ( 3 + 2x ) - 9 ( 2x + 3 ) = 0
<=> ( 3x - 9 ) ( 2x + 3 ) = 0
<=> \(\orbr{\begin{cases}3^x-9=0\\2x+3=0\end{cases}}\)<=>\(\orbr{\begin{cases}3^x=9\\2x=-3\end{cases}}\)
<=>\(\orbr{\begin{cases}3^x=3^2\\x=-\frac{3}{2}\end{cases}}\)<=>\(\orbr{\begin{cases}x=2\\x=-\frac{3}{2}\end{cases}}\)
a: \(\Leftrightarrow3^x\cdot3+2x\cdot3^x-18x-27=0\)
\(\Leftrightarrow3^x\left(2x+3\right)-9\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3^x-9\right)=0\)
=>x=2 hoặc x=-3/2
b: \(\Leftrightarrow\left|2x+5\right|\cdot\dfrac{1}{2}-\dfrac{5}{4}\cdot2\cdot\left|2x+5\right|+\dfrac{7}{3}\cdot4\cdot\left|2x+5\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left|2x+5\right|=\dfrac{1}{44}\)
=>2x+5=1/44 hoặc 2x+1=-1/44
=>x=-219/88 hoặc x=-221/88
\(\left|x\right|=\frac{4}{7}\)
\(\Rightarrow x=\frac{4}{7}\)
b,\(\left(2x-3\right)^2=64\)
\(\Leftrightarrow\left(2x-3\right)^2=\left(\pm8\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=8\\2x-3=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=-\frac{5}{2}\end{cases}}}\)
c,\(\left(\frac{1}{2}\right)^x=\frac{1}{16}\)
\(\Rightarrow\left(\frac{1}{2}\right)^x=\left(\pm\frac{1}{2}\right)^4\)
\(\Rightarrow x=\pm\frac{1}{2}\)
d,\(3^{x+1}=27\)
\(\Leftrightarrow3^{x+1}=3^3\)
\(\Leftrightarrow x+1=3\)
\(\Leftrightarrow x=2\)
a) x3 = -27
<=> -33 = -27
=> x = -3
b) (2x - 1)3 = 8
<=> 8x3 - 12x2 + 6x - 1 = 8
<=> 8x3 - 12x2 + 6x - 1 - 8 = 0
<=> (2x - 3)(4x2 + 3) = 0
<=> 2x - 3 = 0 hoặc 4x2 + 3 = 0
2x = 0 + 3
2x = 3
x = 3/2
=> x = 3/2
c) x3 = x5
<=> x3 - x5 = 0
<=> x3(1 - x2) = 0
<=> x = 0; 1; -1
=> x = 0; 1; -1
d) (x - 2)2 = 16
<=> (x - 2)2 = 42
<=> x - 2 = 4 hoặc x - 2 = -4
x = 4 + 2 x = -4 + 2
x = 6 x = -2
=> x = 6; -2
g) (2x - 3)2 = 9
<=> (2x - 3)2 = 32
<=> 2x - 3 = 3 hoặc 2x - 3 = -3
2x = 3 + 3 2x = -3 + 3
2x = 6 2x = 0
x = 3 x = 0
=> x = 3; 0
y) 3x3 - 4x = 0
<=> x(3x - 4) = 0
<=> x = 0 hoặc 3x - 4 = 0
3x = 0 + 4
3x = 4
x = 4/3
3x+1+2X.3x-18x+27=0
<=> 3x(3+2x)-9(3+2x)=0
<=> (3+2x)(3x-9)=0
\(\Leftrightarrow\orbr{\begin{cases}3+2x=0\\3^x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1,5\\x=2\end{cases}}\)
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