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\(2^x+2^{x+4}=544\\ \Leftrightarrow2^x.\left(1+2^4\right)=544\\ \Leftrightarrow2^x.17=544\\ \Leftrightarrow2^x=\dfrac{544}{17}=32=2^5\\ Vậy:x=5\)
\(...2^x\left(1+16\right)=544\Rightarrow2^x=544:17=32=2^5\)
\(\Rightarrow x=5\)
1) \(5^{x+1}-5^x=20\Leftrightarrow5^x\left(5-1\right)=20\Leftrightarrow5^x=5\Leftrightarrow x=1\)
2) \(2^x+2^{x+4}=544\Leftrightarrow2^x\left(1+2^4\right)=544\Leftrightarrow2^x=32\Leftrightarrow x=5\)
3) \(4^{2x+1}+4^{2x}=80\Leftrightarrow4^{2x}\left(4+1\right)=80\Leftrightarrow16^x=16\Leftrightarrow x=1\)
4) \(3^{2x+2}+3^{2x+1}=108\Leftrightarrow3^{2x}\left(3^2+3\right)=108\Leftrightarrow9^x=9\Leftrightarrow x=1\)
5) \(7^{x+3}-7^{x+1}=16464\Leftrightarrow7^x\left(7^3-7\right)=16464\Leftrightarrow7^x=49\Leftrightarrow x=2\)
\(2x\left(4x+3\right)+5=x\left(8x+4\right)+1\\ \Leftrightarrow8x^2+6x+5=8x^2+4x+1\\ \Leftrightarrow8x^2-8x^2+6x-4x=1-5\\ \Leftrightarrow2x=-4\\ \Leftrightarrow x=-4:2\\ \Leftrightarrow x=-2\)
Vậy \(x=-2\)
\(a,\Leftrightarrow2^x\left(1+2^4\right)=544\\ \Leftrightarrow2^x=\dfrac{544}{17}=32=2^5\\ \Leftrightarrow x=5\\ b,\Leftrightarrow\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\3x-\dfrac{2}{5}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Sửa lại đề
\(2^x+2^x+4=516\)
\(\Rightarrow2^x+2^x=512\)
\(\Rightarrow2^x+2^x=2^8+2^8\)
\(\Rightarrow x=8\)
\(2^x+2^{x+4}=544\)
\(\Rightarrow2^x.1+2^x.2^4=544\)
\(\Rightarrow2^x\left(1+2^4\right)=544\)
\(\Rightarrow2^x.17=544\)
\(\Rightarrow2^x=544\div17\)
\(\Rightarrow2x=32=2^5\)
\(\Rightarrow x=5\)
2x + 2x+4 = 544
\(\Leftrightarrow\)2x . 1 + 2x . 24 = 544
\(\Leftrightarrow\)2x . 17 = 544
\(\Leftrightarrow\)2x = 32
\(\Leftrightarrow\)x = 5
\(2^x+2^{x+4}=544\Leftrightarrow2^x\left(1+2^4\right)=544\)
\(\Leftrightarrow2^x=\frac{544}{17}=32=2^5\Rightarrow x=5\)
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