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

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

\(\Leftrightarrow\orbr{\begin{cases}3x^2=27\\3x^2=75\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=9\\x^2=25\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\pm3\\x=\pm5\end{cases}}\)

Vậy \(x\in\left\{\pm3;\pm5\right\}\)

Ta có:

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

⇒ \(3x^2-51=-24\)

⇒ \(3x^2=\left(-24\right)+51\)

⇒ \(3x^2=27\)

⇒ \(x^2=27:3\)

⇒ \(x^2=9\)

⇒ \(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Vậy \(x\in\left\{3;-3\right\}.\)

Chúc bạn học tốt!

(3x²-51)²ⁿ=(-24)²ⁿ\(\Rightarrow\)(51-3x²)²ⁿ=(24)²ⁿ\(\Rightarrow\)(51-3x²)=24

\(\Rightarrow\)51-24=3x²\(\Rightarrow\)27\(\Rightarrow\)x²=3²\(\Rightarrow\)x=3 hoặc x=-3

từ đề bài => 3x^2-51=24

=>3x^2=75

=>x^2=75:3=25

=>x=-5 hoặc x=5

2

Tik cho mk nha..................cảm ơn rất nhiều

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

\(\Rightarrow3x^2-51\in\left\{-24;24\right\}\)

+) \(3x^2-51=-24\)

=> 3x² = -24 + 51

=> 3x² = 27

=> x² = 27 : 3

=> x² = 9 = 3² = (-3)²

=> x \(\in\){-3; 3}.

+) \(3x^2-51=24\)

=> 3x² = 24 + 51

=> 3x² = 75

=> x² = 75 : 3

=> x² = 25 = 5² = (-5)²

=> x \(\in\){-5; 5}.

Vậy có 4 giá trị của x thỏa mãn.

4 giá trị                                 

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0.\)

\(1+\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}-4+\frac{x+349}{5}=0\)

\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

\(\Rightarrow x+329=0\)

\(\Rightarrow x=-329\)

Study well 

\(\Rightarrow\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+349}{5}-4=0\)

\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Rightarrow\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Mà \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)

Nên \(x+329=0\Rightarrow x=-329\)

Vậy \(x=-329\)

Chúc bạn học tốt !!!

Bài làm

Không viết lại đề nha !

\(=\frac{\left(x+2\right)}{327}+\frac{\left(x+3\right)}{326}+\frac{\left(x+4\right)}{325}+\frac{\left(x+5\right)}{324}+\frac{\left(x+349\right)}{5}=0\)

\(\Leftrightarrow\frac{\left(x+2\right)}{327}+1+\frac{\left(x+3\right)}{326}+1+\frac{\left(x+4\right)}{325}+1+\frac{\left(x+5\right)}{324}+1+\frac{\left(x+349\right)}{5}\)

\(\Leftrightarrow\frac{\left(x+329\right)}{327}+\frac{\left(x+329\right)}{326}+\frac{\left(x+329\right)}{325}+\frac{\left(x+329\right)}{324}+\frac{\left(x+329\right)}{5}-4=0\)

\(\Leftrightarrow\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)

Nên:x+329=0=>x=-329