tìm x biết 2+4+6+...=544
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
tổng trên có số số hạng là: \(\frac{\left(x-2\right)}{2}+1=\frac{x}{2}-1+1=\frac{x}{2}\)
tổng của dãy là: \(\frac{\left(x+2\right)x}{2}=544\Leftrightarrow\frac{x^2}{2}+x-544=0\Leftrightarrow x^2+2x-1088=0\Leftrightarrow x^2-32x+34x-1008=0\Leftrightarrow\left(x-32\right)\left(x+34\right)=0\)
=> x=32(t/m) hoặc x=-34(loại)
=> x=32
\(1,\sqrt{3}x-3=\sqrt{27}\)
\(\Leftrightarrow\sqrt{3}x-3=3\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}\left(x-\sqrt{3}\right)=3\sqrt{3}\)
\(\Leftrightarrow x-\sqrt{3}=3\)
\(\Leftrightarrow x=3+\sqrt{3}\)
\(2,\sqrt{2}x-\sqrt{28}=\sqrt{32}\)
\(\Leftrightarrow\sqrt{2}x-2\sqrt{7}=4\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}x=4\sqrt{2}+2\sqrt{7}\)
\(\Leftrightarrow x=\dfrac{\sqrt{2^2}\left(2\sqrt{2}+\sqrt{7}\right)}{\sqrt{2}}\)
\(\Leftrightarrow x=\sqrt{2}\left(2\sqrt{2}+\sqrt{7}\right)\)
\(\Leftrightarrow x=4+\sqrt{14}\)
\(3,\sqrt{6}x-2\sqrt{6}=\sqrt{54}\)
\(\Leftrightarrow\sqrt{6}\left(x-2\right)=3\sqrt{6}\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=5\)
\(4,\sqrt{3}x-\sqrt{2}x=\sqrt{3}+\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)x=\sqrt{3}+\sqrt{2}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
\(a,\Leftrightarrow2^x\left(1+2^4\right)=544\\ \Leftrightarrow2^x=\dfrac{544}{17}=32=2^5\\ \Leftrightarrow x=5\\ b,\Leftrightarrow\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\3x-\dfrac{2}{5}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Sửa lại đề
\(2^x+2^x+4=516\)
\(\Rightarrow2^x+2^x=512\)
\(\Rightarrow2^x+2^x=2^8+2^8\)
\(\Rightarrow x=8\)
\(2^x+2^{x+4}=544\\ \Leftrightarrow2^x.\left(1+2^4\right)=544\\ \Leftrightarrow2^x.17=544\\ \Leftrightarrow2^x=\dfrac{544}{17}=32=2^5\\ Vậy:x=5\)
\(...2^x\left(1+16\right)=544\Rightarrow2^x=544:17=32=2^5\)
\(\Rightarrow x=5\)
\(2^x+2^{x+4}=544\)
\(\Rightarrow2^x.1+2^x.2^4=544\)
\(\Rightarrow2^x\left(1+2^4\right)=544\)
\(\Rightarrow2^x.17=544\)
\(\Rightarrow2^x=544\div17\)
\(\Rightarrow2x=32=2^5\)
\(\Rightarrow x=5\)