Tìm x: (Giải chi tiết giúp mình nha)
\(2^x+4^x=20.\)
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\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
x2+4x+4=0
(x+2)2=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)2+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0 Th2: x+1=0
x=-3 x=-1
vậy ...
\(20^{3x}=20^{13}:20^4\\ \Leftrightarrow20^{3x}=20^9\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\)
\(20^{3x}=20^{13}:\left(4.5\right)^4\\ \Leftrightarrow20^{3x}=20^{13}:20^4\\ \Leftrightarrow20^{3x}=20^9\\ \Rightarrow3x=9\\ \Rightarrow x=9:3\\ \Rightarrow x=3\)
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\)- \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\) - \(\dfrac{1}{10.11}\) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\) =\(x\)-\(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{11}\) = \(x\) - \(\dfrac{5}{13}\)
\(x-\dfrac{5}{13}=\dfrac{1}{11}\)
\(x\) = \(\dfrac{1}{11}\) + \(\dfrac{5}{13}\)
\(x\) = \(\dfrac{68}{143}\)
<=> 2x^2 +x-4x-2-5x-15=2x^2-6x+4+8x-2-2x
2x^2-8x-17-2x^2-2=0
-8x-19=0
x=-19/8
\(\left(x^2-5\right)\left(x+2\right)+5x=2x^2+17\)
\(\Rightarrow\left(x^3+2x^2-5x-10\right)+5x=2x^2+17\)
\(\Rightarrow x^3+2x^2-5x-10+5x=2x^2+17\)
\(\Rightarrow x^3+2x^2-10=2x^2+17\)
\(\Rightarrow x^3-10=17\)
\(\Rightarrow x^3=17+10=27\)
\(\Rightarrow x^3=3^3\)
\(\Rightarrow x=3\)
(x2−5)(x+2)+5x=2x2+17
⇒(x3+2x2−5x−10)+5x=2x2+17
⇒x3+2x2−5x−10+5x=2x2+17
⇒x3+2x2−10=2x2+17
⇒x3−10=17
⇒x3=17+10=27
⇒x3=33
⇒x=3
bn viết đề lại rõ hơn đc k? bạn viết như thế thì có thể hiểu thành 2 nghĩa:
\(\orbr{\begin{cases}3^x+2-3^x=24\\3^{x+2}-3^x=24\end{cases}}\)
Đề trên chắc k đúng nên mik giải cái đề dưới nhé
\(3^{x+2}-3^x=24\)
<=> \(3^x.3^2-3^x=24\)
<=> \(3^x.\left(3^2-1\right)=24\)
<=> \(3^x.8=24\Rightarrow3^x=24:8=3\Rightarrow x=1\)
2x + 4x = 20
2x ( 1 + 22 ) = 20
2x . 5 = 20
2x = 4
2x = 22
=> x = 2
Vậy x = 2
2x+4x=20
=)2x+4x=22+42=)x= 2 ngắn gọn nha ! (^-^)