A=7 B=1 OR =0 C=2 D=4.5.6

a: \(2^{x}\cdot4=128\)

=>\(2^{x}=\frac{128}{4}=32=2^5\)

=>x=5

b: \(x^{15}=x\)

=>\(x^{15}-x=0\)

=>\(x\left(x^{14}-1\right)=0\)

=>\(\left[\begin{array}{l}x=0\\ x^{14}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^{14}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\end{array}\right.\)

c: \(\left(2x+1\right)^3=125\)

=>\(\left(2x+1\right)^3=5^3\)

=>2x+1=5

=>2x=5-1=4

=>\(x=\frac42=2\)

d: \(\left(x-5\right)^4=\left(x-5\right)^6\)

=>\(\left(x-5\right)^6-\left(x-5\right)^4=0\)

=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)

=>\(\left(x-5\right)^4\cdot\left(x-5-1\right)\left(x-5+1\right)=0\)

=>\(\left(x-5\right)^4\cdot\left(x-6\right)\left(x-4\right)=0\)

=>\(\left[\begin{array}{l}x-5=0\\ x-6=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=6\\ x=4\end{array}\right.\)

a) (5x - 1) : 3 + 1 = 4

=> (5x - 1) : 3 = 3

=> (5x - 1) = 9

=> 5x - 1 = 9

=> 5x = 10

=> x = 2

b) 54 : (16 - x) - 1=5

=> 54:(16-x) = 6

=> 16-x = 9

=> x = 7

các bạn làm ơn giúp mk với mk đang gấp lắm

1, Ta có :

a . 81 = 3⁴ => 3^x= 3⁴ => x = 4 .

b. 125 = 5³ => 5^x+2 = 5³ =>x + 2 = 3 => x = 1

c. 2³ * 2^{x - 1} = 64

=> 2^{3 + ( x - 1 )} = 64 = 2⁶ 

^{=>  3 + ( x - 1 ) = 6} 

^{=>           x - 1   = 6 - 3 = 3}

                      ^{x  = 3 + 1}

                       ^{x  = 4}

\(a,x^2=4\Rightarrow x^2=2^2\Rightarrow x=2\)

\(b,x^2=64\Rightarrow x^2=8^2\Rightarrow x=8\)

\(c,6x^3-8=40\Rightarrow6x^3=48\Rightarrow x^3=8\Rightarrow x^3=2^3\Rightarrow x=2\)

\(d,\left(2x-1\right)^2=49\Rightarrow\left(2x-1\right)^2=7^2\Rightarrow2x-1=7\Rightarrow x=4\)

\(e,2^x:16=2^5\Rightarrow2^x:16=32\Rightarrow2^x=512\Rightarrow2^x=2^9\Rightarrow x=9\)

\(f,4^5:4^x=16\Rightarrow1024:4^x=16\Rightarrow4^x=64\Rightarrow4^x=4^3\Rightarrow x=3\)

a, x^2 = 4

=> x = 2 hoặc x = -2

b, x^2 = 64 

=> x = 8 hoặc x = -8

c, 6x^3 - 8 = 40

=> 6x^3 = 48

=> x^3 = 8

=> x = 2

d, (2x - 1)^2 = 49

=> 2x - 1 = 7 hoặc 2x - 1 = -7

=> 2x = 8 hoặc 2x = -6

=> x = 4 hoặc x = -3

e, 2^x : 16 = 2^5

=> 2^x : 2^4 = 2^5

=> 2^x = 2^9

=> x = 9 

f, 4^5 : 4^x = 16

=> 4^5 - x = 4^2

=> 5 - x = 2

=> x = 3

Bài 1:

a, \(\left(x-2\right)^2=9\)

\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)

b, \(\left(3x-1\right)^3=-8\)

\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)

\(\Rightarrow x=-\dfrac{1}{3}\)

c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)

\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)

d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)

\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)

Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)

e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)

Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)

làm câu nào cũng đc gấp gấp!!!

\(3^x.3^3=81\)

<=> \(3^x=3\)

<=> \(x=1\)