2+4+6+...+x=544. Tìm x
tìm x biết 2+4+6+...+x=544
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
tìm x biết 2+4+6+...=544
Tìm x, biết:
1/ √3 × x - 3 = √27
2/ √2 × x - √28 = √32
3/ √6 × x - 2√6 = √54
4/ √3 × x - √2 × x = √3 + √2
\(1,\sqrt{3}x-3=\sqrt{27}\)
\(\Leftrightarrow\sqrt{3}x-3=3\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}\left(x-\sqrt{3}\right)=3\sqrt{3}\)
\(\Leftrightarrow x-\sqrt{3}=3\)
\(\Leftrightarrow x=3+\sqrt{3}\)
\(2,\sqrt{2}x-\sqrt{28}=\sqrt{32}\)
\(\Leftrightarrow\sqrt{2}x-2\sqrt{7}=4\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}x=4\sqrt{2}+2\sqrt{7}\)
\(\Leftrightarrow x=\dfrac{\sqrt{2^2}\left(2\sqrt{2}+\sqrt{7}\right)}{\sqrt{2}}\)
\(\Leftrightarrow x=\sqrt{2}\left(2\sqrt{2}+\sqrt{7}\right)\)
\(\Leftrightarrow x=4+\sqrt{14}\)
\(3,\sqrt{6}x-2\sqrt{6}=\sqrt{54}\)
\(\Leftrightarrow\sqrt{6}\left(x-2\right)=3\sqrt{6}\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=5\)
\(4,\sqrt{3}x-\sqrt{2}x=\sqrt{3}+\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)x=\sqrt{3}+\sqrt{2}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
tìm X
2x+2x+4=544
\(2^x+2^{x+4}=544\Leftrightarrow2^x\left(1+2^4\right)=544\)
\(\Leftrightarrow2^x=\frac{544}{17}=32=2^5\Rightarrow x=5\)
~ Học tốt ~
2^x+2^(x+4)=544
2 mũ x +2 mũ (x+4)=544
Ta có : 2x + 2x + 4 = 544
=> 2x(1 + 24) = 544
=> 2x.17 = 544
=> 2x = 32
=> 2x = 25
=> x = 5
Vậy x = 5
Tìm x biết:
a) \(2^x+2^{x+4}=544\)
b) \(\left(\dfrac{2}{5}-3x\right)^2-\dfrac{1}{5}=\dfrac{4}{25}\)
\(a,\Leftrightarrow2^x\left(1+2^4\right)=544\\ \Leftrightarrow2^x=\dfrac{544}{17}=32=2^5\\ \Leftrightarrow x=5\\ b,\Leftrightarrow\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\3x-\dfrac{2}{5}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Tìm x
2^x+2^x+4=544
cảm ơn đã trả lời kb nhé
Sửa lại đề
\(2^x+2^x+4=516\)
\(\Rightarrow2^x+2^x=512\)
\(\Rightarrow2^x+2^x=2^8+2^8\)
\(\Rightarrow x=8\)
hnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
\(2^x+2^{x+4}=544\)
\(2^x\left(1+2^4\right)=544\)
\(2^x.17=544\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
Tìm x:
\(2^x+2^{x+4}=544\)
\(2^x+2^{x+4}=544\)
\(\Rightarrow2^x.1+2^x.2^4=544\)
\(\Rightarrow2^x\left(1+2^4\right)=544\)
\(\Rightarrow2^x.17=544\)
\(\Rightarrow2^x=544\div17\)
\(\Rightarrow2x=32=2^5\)
\(\Rightarrow x=5\)
Tìm x:
2x+2x+4=544
\(2^x+2^{x+4}=544\\ \Leftrightarrow2^x.\left(1+2^4\right)=544\\ \Leftrightarrow2^x.17=544\\ \Leftrightarrow2^x=\dfrac{544}{17}=32=2^5\\ Vậy:x=5\)
\(...2^x\left(1+16\right)=544\Rightarrow2^x=544:17=32=2^5\)
\(\Rightarrow x=5\)