3x²-51 = 24

=>3x²=75

=>x²=25

=> x = + - 5

(5x+1)²=36/49

(5x+1)²=\(\left(\frac{6}{7}\right)^2\)

5x+1=6/7

5x=-1/7

x=-1/35

5^x=125

5^x=5³

^x=3

(3x-2)⁵=-243

(3x-2)⁵=(-3)⁵

3x-2=-3

3x=-1

x=-1/3

\(\left(3x^2-51\right)^{2n}=\left(-24\right)^{2n}\)

\(3x^2-51=-24\)

\(3x^2=27\)

\(x^2=9\)

\(x^2=3^2=\left(-3\right)^2\)

TH1: x=3

TH2: x=-3

=.= hok tốt!!

a) x = 1

b) x = -1

a) x²-2x=-1

<=> x²-2x+1=0

<=> (x - 1)² = 0
=> x = 1

a: Ta có: \(x^2-2x=-1\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x-1=0\)

hay x=1

b: Ta có: \(x^3+3x^2=-3x-1\)

\(\Leftrightarrow x^3+3x^2+3x+1=0\)

\(\Leftrightarrow\left(x+1\right)^3=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

\(\left(3x^2-51\right)^{2n}=\left(-24\right)^{2n}\)

\(\Leftrightarrow3x^2-51=-24\)

\(\Leftrightarrow3x^2=27\)

\(\Leftrightarrow x^2=9\)

\(\Leftrightarrow x=\pm3\)

(3x² - 51)²ⁿ = (-24)²ⁿ

=> \(\orbr{\begin{cases}3x^2-51=-24\\3x^2-51=24\end{cases}=>\orbr{\begin{cases}3x^2=27\\3x^2=75\end{cases}}}\) 

=>\(\orbr{\begin{cases}x^2=9\\x^2=25\end{cases}=>}\orbr{\begin{cases}x=3\\x=5\end{cases}}\)

a)  3n = 35 => n = 5

b) 2n = 28 => n = 8

--thodagbun--

( h tớ lm b đt nè :( )

a) \(3^n=243\)

\(=>3^n=3^5\)

\(=>n=5\)

b) \(2^n=256\)

\(=>2^n=2^8\)

\(=>n=8\)

#Nothings -_-

Hài

a) 3ⁿ = 243

=> 3ⁿ = 3⁵

=> n = 5

b) 2ⁿ = 256 

=> 2ⁿ = 2⁸

=> n = 8

(3x² - 51)²ⁿ = 576ⁿ

=> (3x² - 51)²ⁿ = 24²ⁿ = (-24)²ⁿ

\(\Rightarrow\left[\begin{array}{nghiempt}3x^2-51=24\\3x^2-51=-24\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}3x^2=75\\3x^2=27\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x^2=25\\x^2=9\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}\left[\begin{array}{nghiempt}x=5\\x=-5\end{array}\right.\\\left[\begin{array}{nghiempt}x=3\\x=-3\end{array}\right.\end{array}\right.\)

Vậy \(\left[\begin{array}{nghiempt}x=5\\x=-5\\x=3\\x=-3\end{array}\right.\)