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a, Ta có: 3 x = 3 2 nên x = 2
b, Ta có: 5 x = 5 3 nên x = 3
c, Ta có: 3 x + 1 = 3 2 nên x +1 = 2, do đó x = 1
d, Ta có: 6 x - 1 = 6 2 nên x - 1 = 2, đo đó x = 3
e) Ta có: 3 2 x + 1 = 3 3 nên 2x +1 = 3, do đó x = 1
f) Ta có: x 50 = x nên x 50 - x = 0 , do đó x x 49 - 1 = 0 = 0
Vì thế x = 0 hoặc x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x = 0
b) x = -1
c) x = -9
d) x = 24
e) x = 2 hoặc x = -4
f) x = 5 hoặc x = -3
![](https://rs.olm.vn/images/avt/0.png?1311)
32x +3x +3 =759
32x +3x =759-3
32x +3x =756
Ta thấy 32x =3x.3
Tính nốt nhé
Ta có :
3^2x+3^x+3 = 759
3^x . 3^x + 3^x = 759 -3 = 756 ( 3^2x = 3^x . 3^x )
3^x . (3^x +1 ) = 756 = 3^3 . ( 3^3+1)
Suy ra : 3^x = 3^3
x = 3
Vậy x = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
32x + 3x + 3 = 759
3x.3x + 3x + 3 = 759
3x.3x + 3x = 759 - 3
3x.3x + 3x = 756
3x(3x + 1) = 756
3x(3x + 1) = 27.28
=> 3x = 27
3x = 33
=> x = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
3 2 x + 1 + 10 4 x + 2 − 6 6 x + 3 = 12 26 ⇒ 3 2 x + 1 + 10 2 2 x + 1 − 6 3 2 x + 1 = 12 26 ⇒ 3 2 x + 1 + 5 2 x + 1 − 2 2 x + 1 = 12 26 ⇒ 3 + 5 − 2 2 x + 1 = 12 26 ⇒ 6 2 x + 1 = 6 13 ⇒ 2 x + 1 = 13 ⇒ x = 6
![](https://rs.olm.vn/images/avt/0.png?1311)
32x + 3x + 3 = 759
3x.3x + 3x + 3 = 759
3x.3x + 3x = 759 - 3
3x.3x + 3x = 756
3x(3x + 1) = 756
3x(3x + 1) = 27.28
=> 3x = 27
3x = 33
=> x = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2x^3-32x=0\)
\(2x\left(x^2-16\right)=0\)
\(2x\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow2x=0\)hoặc \(\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(x=0\) hoặc \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
b) \(\left(3x-2\right)^2-\left(x+5\right)^2=0\)
\(\left(3x-2-x-5\right)\left(3x-2+x+5\right)=0\)
\(\left(2x-7\right)\left(4x+3\right)=0\)
\(\orbr{\begin{cases}2x-7=0\\4x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-3}{4}\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-3}{4}\end{cases}}\)
c) \(2\left(x+3\right)-x^2-3x=0\)
\(2\left(x+3\right)-\left(x^2+3x\right)=0\)
\(2\left(x+3\right)-x\left(x+3\right)=0\)
\(\left(2-x\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2-x=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
vậy \(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
d) \(4x^2-25-\left(2x+5\right)\left(x+7\right)=0\)
\(\left(4x^2-25\right)-\left(2x+5\right)\left(x+7\right)=0\)
\(\left(2x-5\right)\left(2x+5\right)-\left(2x+5\right)\left(x+7\right)=0\)
\(\left(2x+5\right)\left(2x-5-x-7\right)=0\)
\(\left(2x+5\right)\left(x-12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+5=0\\x-12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{2}\\x=12\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{-5}{2}\\x=12\end{cases}}\)
3^2x+3^X+3=759
(3x)^2+3^x=759-3=756
3^x.3^x+3^x=756
3^x(3^x+1)=756
3^x(3^x+1)=27x28
Vậy 3^x =27 và x = 3
32x+3x+3=759
=>32x+3x=759-3
=>32x+3x=756
=>3x.3x+3x=756
=>3x(3x+1)=756
=>3x(3x+1)=27.28
=>3x=27
=>3x=33
=>x=3