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ai chơi minecraft hay blockman go thì hãy sud kênh
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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!
![](https://rs.olm.vn/images/avt/0.png?1311)
(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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(3x^2-51\right)^{2n}=\left(-24\right)^{2n}\)
\(\Rightarrow3x^2-51\in\left\{-24;24\right\}\)
+) \(3x^2-51=-24\)
=> 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.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0.\)
\(1+\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}-4+\frac{x+349}{5}=0\)
\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
\(\Rightarrow x+329=0\)
\(\Rightarrow x=-329\)
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 !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{20}+19+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}\right)\)
\(S=19\)
Nếu ko hiểu kb vs mền rồi mền giải thích cho
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 \(x\in\left\{\pm3;\pm5\right\}\)