Tìm x:
a. \(\frac{x-1}{x+5}\)= \(\frac{6}{7}\)
b. (2x - \(\frac{1}{2}\))4 + \(\frac{11}{16}\)= \(\frac{3}{4}\)
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Đặt \(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(\Rightarrow2A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{99}}\)
\(\Rightarrow A=1-\frac{1}{2^{99}}=\frac{2^{99}-1}{2^{99}}\)
Có:\(\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)=0\)Mà \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}>0\)
=>x+2020=0<=>x=-2020
\(\left(x-\frac{1}{2}\right)^3=\frac{8}{27}\Leftrightarrow\sqrt[3]{\left(x-\frac{1}{2}\right)^3}=\sqrt[3]{\frac{8}{27}}\Leftrightarrow x-\frac{1}{2}=\frac{2}{3}\Leftrightarrow x=\frac{1}{6}\)
chúc bạn học tốt!!!! :)))
\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)và 2x2 + 2y2 - 3z2 = -100
\(\Rightarrow\frac{2x^2}{2.3^2}=\frac{2y^2}{2.4^2}=\frac{3z^2}{3.5^2}=\frac{2x^2+2y^2-3z^2}{2.9+2.16-3.25}\frac{-100}{-25}=\frac{100}{25}=4\)
=> x2 = 4 . 3 = 12 =>\(x=\sqrt{12}\)
y2 = 4 . 4 = 16 => y = 4
z2 = 4 . 5 = 20 => z = \(\sqrt{20}\)
Bài giải
Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\text{ }\Rightarrow\hept{\begin{cases}x=3k\\y=4k\\z=5k\end{cases}}\)
\(\Rightarrow\text{ }2x^2+2y^2-3z^2=2\cdot9k^2+2\cdot16k^2-3\cdot25k^2=18k^2+32k^2-75k^2\)
\(=k^2\left(18+32-75\right)=-25k^2=-100\)
\(\Rightarrow\text{ }k^2=4\text{ }\Rightarrow\text{ }k=\pm2\)
\(\Rightarrow\hept{\begin{cases}x=-6\\y=-8\\z=-10\end{cases}}\text{ hoặc }\hept{\begin{cases}x=6\\y=8\\z=10\end{cases}}\)
a, \(\frac{x-1}{x+5}=\frac{6}{7}\)
\(\Rightarrow\left(x-1\right).7=\left(x+5\right).6\)
\(\Rightarrow7x-7=6x+30\)
\(\Rightarrow7x-6x=7+30\)
\(\Rightarrow x=37\)
Vậy x=37
b, \(\left(2x-\frac{1}{2}\right)^4+\frac{11}{16}=\frac{3}{4}\)
\(\Rightarrow\left(2x-\frac{1}{2}\right)^4=\frac{1}{16}\)
\(\Rightarrow\left(2x-\frac{1}{2}\right)^4=\left(\frac{1}{2}\right)^4\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=\frac{1}{2}\\2x-\frac{1}{2}=-\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=1\\2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=0\end{cases}}\)
Vậy \(x=\frac{1}{2}\) hoặc \(x=0\)