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\(x\)(\(x\) - 2) = 16
\(x^2\) - 2\(x\) - 16 = 0
(\(x^2\) - \(x\)) - (\(x\) - 1) - 17 = 0
\(x\)(\(x\) - 1) - (\(x-1\)) = 17
(\(x\) - 1)(\(x\) - 1) = 17
(\(x-1\))2 = 17
\(\left[{}\begin{matrix}x-1=\sqrt{17}\\x-1=-\sqrt{17}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=1+\sqrt{17}\\x=1-\sqrt{17}\end{matrix}\right.\)
Vậy: \(\in\) {1 - \(\sqrt{17}\); 1 + \(\sqrt{17}\)}
Bài 2:
2\(x\) + 3.2\(x\) = 144
2\(x\).(1 + 3) = 144
2\(x\).4 = 144
2\(x\) = 144 : 4
2\(x\) = 36
2\(^x\) = 36
Nếu \(x\) = 6 ⇒ 2\(^x\) = 64 > 36 (loại)
Nếu \(x\) ≤ 5 ⇒2\(^x\) ≤ 25 = 32 < 36 (loại)
Vậy \(x\in\) \(\varnothing\)
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\(2^{x-3}-3.2^x=-92\)
\(\Rightarrow2^x\left(2^{-3}-3\right)=-92\)
\(\Rightarrow2^x.\dfrac{-23}{8}=-92\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\Rightarrow x=5\)
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ngoctram sai
vì 2^x ko pải 2.x nên k đc lấy 36:2
đúng nha
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\(2^{x-2}-3.2^x=-88\)
\(\Rightarrow2^x:2^2-3.2^x=-88\)
\(\Rightarrow2^x.\left(\dfrac{1}{4}-3\right)=-88\)
\(\Rightarrow\)\(2^x.\left(-2\dfrac{3}{4}\right)=-88\)
\(\Rightarrow\)\(2^x=\)32
=> x=5
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\(2^{x-2}-3.2^x=-88\)
\(\Leftrightarrow2^x.\frac{1}{4}-3.2^x=-88\)
\(\Leftrightarrow2^x\left(\frac{1}{4}-3\right)=-88\)
\(\Leftrightarrow2^x.\frac{-11}{4}=-88\)
\(\Leftrightarrow2^x=-88.\frac{-4}{11}\)
\(\Leftrightarrow2^x=-8.\left(-4\right)\)
\(\Leftrightarrow2^x=32=2^5\)
\(\Leftrightarrow x=5\)
Vậy x = 5
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\(\left|x-3,2\right|+\left|\dfrac{2x-1}{5}\right|=x+3\) (1)
TH1: \(\left\{{}\begin{matrix}x>3,2\Rightarrow\left|x-3,2\right|=x-3,2\\x>\dfrac{1}{2}\Rightarrow\left|\dfrac{2x-1}{5}\right|=\dfrac{2x-1}{5}\end{matrix}\right.\)
\(\left(1\right)\Rightarrow x-3,2+\dfrac{2x-1}{5}=x+3\)
\(\Rightarrow5x-16+2x-1=5x+15\Rightarrow2x=32\Leftrightarrow x=16\left(tm\right)\)
TH2: \(\left\{{}\begin{matrix}x>3,2\Rightarrow\left|x-3,2\right|=x-3,2\\x< \dfrac{1}{2}\Rightarrow\left|\dfrac{2x-1}{5}\right|=\dfrac{1-2x}{5}\end{matrix}\right.\)
\((1)\)\(\Rightarrow x-3,2+\dfrac{1-2x}{5}=x+3\Rightarrow5x-16+1-2x=5x+15\)
\(\Rightarrow-2x=0\Rightarrow x=0\left(l\right)\)
TH3: \(\left\{{}\begin{matrix}x< 3,2\Rightarrow\left|x-3,2\right|=3,2-x\\x>\dfrac{1}{2}\Rightarrow\left|\dfrac{2x-1}{5}\right|=\dfrac{2x-1}{5}\end{matrix}\right.\)
\(\left(1\right)\Rightarrow3,2-x+\dfrac{2x-1}{5}=x+3\)
\(\Rightarrow16-5x+2x-1=5x+15\Rightarrow8x=0\Leftrightarrow x=0\left(l\right)\)
TH4: \(\left\{{}\begin{matrix}x< 3,2\Rightarrow\left|x-3,2\right|=3,2-x\\x< \dfrac{1}{2}\Rightarrow\left|\dfrac{2x-1}{5}\right|=\dfrac{1-2x}{5}\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow3,2-x+\dfrac{1-2x}{5}=x+3\)
\(\Rightarrow16-5x+1-2x=5x+15\Rightarrow12x=2\Rightarrow c=\dfrac{1}{6}\left(tm\right)\)
Vậy \(x=\left\{16;\dfrac{1}{6}\right\}\)
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\(6.8^{x-1}+8^{x+1}=6.8^{19}+8^{21}\)
\(\Rightarrow\hept{\begin{cases}x-1=19\\x+1=21\end{cases}\Rightarrow\hept{\begin{cases}x=20\\x=20\end{cases}}}\)
\(5.2^x+3.2^{x+2}=5.2^5+3.2^7\)
\(\Rightarrow\hept{\begin{cases}x=5\\x+2=7\end{cases}\Rightarrow\hept{\begin{cases}x=5\\x=5\end{cases}}}\)
P/s:Kết quả thì chắc chắn đúng nhưng cách trình bày bài giải có thể sai,mong bn thông cảm =.=
2^x x3 +2 ^x =128
2^x x4=128
2^x=32
mà 2^5 =32
=>x=5
chúc bạn học giỏi nha