1. Tìm tỉ số \(\dfrac{x}{y}\) biết \(\dfrac{2x-y}{x+y}\)=\(\dfrac{2}{3}\)
2. Tìm x:
a) 3,8 : (2x) = \(\dfrac{1}{4}:2\dfrac{2}{3}\)
b) \(1\dfrac{1}{3}\) : 0,8 = \(\dfrac{2}{3}\) : (0,1x)
c) \(\dfrac{x}{-15}\) = \(\dfrac{-60}{x}\)
d) \(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\)
e) 0,2 : \(1\dfrac{1}{5}\)= \(\dfrac{2}{3}\) : (6x+7)
3. Tính A = \(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}\)
bài 1) ta có : \(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow2\left(x+y\right)=3\left(2x-y\right)\)
\(\Leftrightarrow2x+2y=6x-3y\Leftrightarrow4x=5y\Leftrightarrow\dfrac{x}{y}=\dfrac{5}{4}\)
vậy \(\dfrac{x}{y}=\dfrac{5}{4}\)
bài 1
\(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow\dfrac{2.\dfrac{x}{y}-1}{\dfrac{x}{y}+1}=\dfrac{2.\dfrac{x}{y}+2-3}{\dfrac{x}{y}+1}=2-\dfrac{3}{\dfrac{x}{y}+1}=\dfrac{2}{3}\)
\(2-\dfrac{2}{3}=\dfrac{4}{3}=\dfrac{3}{\dfrac{x}{y}+1}\)
\(\left(\dfrac{x}{y}+1\right)=\dfrac{9}{4}\Rightarrow\dfrac{x}{y}=\dfrac{9}{4}-\dfrac{4}{4}=\dfrac{5}{4}\)
3) \(A=\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}\)
\(=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=\dfrac{2^{20}}{2^{12}}\)
\(=2^{20-12}=2^8\)
Vậy \(A=2^8\)