\(5^{2x+1}\times133+13\times5^{2x-1}=4750\)
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Bài 2 :
a, \(2^x+2^{x+4}=272\)
\(2^x+2^x.2^4=272\)
\(2^x.\left(1+2^4\right)=272\)
\(2^x.17=272\)
\(2^x=272:17\)
\(2^x=16=2^4\)
\(\Rightarrow x=4\)
1: Bài này hơi khó đó
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{1}{x\times\left(x+1\right)\div2}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6\times\left(6+1\right)\div2}+\frac{1}{7\times\left(7+1\right)\div2}+...+\frac{1}{x\times\left(x+1\right)\div2}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6\times7\div2}+\frac{1}{7\times8\div2}+...+\frac{1}{x\times\left(x+1\right)\div2}\)
\(\Rightarrow\frac{2}{6\times7}+\frac{2}{7\times8}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow2\times\left(\frac{1}{6}+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow2\times\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\div2\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)
=> x = 18 - 1
=> x = 17
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{96}\)
\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{96}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{5}{16}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{5}{16}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{5}{16}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{5}{16}\)
\(\frac{1}{2x+3}=\frac{1}{48}\)
=> 2x + 3 = 48
=> 2x = 48 - 3
=> 2x = 45
=> x = 45/2
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)
\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{93}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(\Rightarrow2x=90\)
\(\Rightarrow x=45\)
a) Ta có: \(9\cdot5^x=6\cdot5^6+3\cdot5^6\)
\(\Leftrightarrow9\cdot5^x=9\cdot5^6\)
\(\Leftrightarrow5^x=5^6\)
hay x=6
b) Ta có: \(2^{2x+1}+4^{x+3}=264\)
\(\Leftrightarrow4^x\cdot2+4^x\cdot64=264\)
\(\Leftrightarrow4^x=4\)
hay x=1
Ta có : \(\left(2.x-1\right)^2=3^2.5^2\)
\(\Leftrightarrow\left(2.x-1\right)^2=\left(3.5\right)^2\)
\(\Leftrightarrow\left(2.x-1\right)^2=15^2\)
\(\Leftrightarrow2.x-1=15\)
\(\Leftrightarrow2.x=15+1\)
\(\Leftrightarrow2.x=16\)
\(\Leftrightarrow x=16:2\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\)
\(\left(2x-1\right)^2=3^2.5^2\)
\(\left(2x-1\right)^2=225\)
\(\left(2x-1\right)^2=\left(\pm15\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x-1=15\\2x-1=-15\end{cases}\Rightarrow\orbr{\begin{cases}2x=16\\2x=-14\end{cases}\Rightarrow}\orbr{\begin{cases}x=8\\x=-7\end{cases}}}\)
\(\Rightarrow x\in\left\{-7;8\right\}\)
nhanh hộ mình với mình sắp đi học rồi.Ai trả lời đc mình cho 1 k nhanh nhanh lên nha, nguy cấp rồi đó
\(720:\left[41-\left(2x-5\right)\right]=2^3\times5\)
\(720:\left[41-\left(2x-5\right)\right]=40\)
\(\left[41-\left(2x-5\right)\right]=720:40\)
\(2x=23+5\)
\(x=28:2\)
\(x=14\)
5\(^{2\text{x}+1}\) . 133 + 13 . 5\(^{2\text{x}-1}\) = 4750
5\(^{2\text{x}-1}\) . 5\(^2\) . 133 + 13 . 5\(^{2\text{x}-1}\) = 4750
5\(^{2\text{x}-1}\) ( 5\(^2\) . 133 + 13 ) = 4750
5\(^{2\text{x}-1}\) . 3338 = 4750
5\(^{x-1}\) = \(\dfrac{2375}{1669}\)
=> Không có giá trị x