Bài 1: Tìm x,y
a) Ta có: \(x^{10}:x^7=\frac{1}{27}\)
\(\Leftrightarrow x^3=\left(\frac{1}{3}\right)^3\)
hay \(x=\frac{1}{3}\)
Vậy: \(x=\frac{1}{3}\)
b) Ta có: \(\frac{1}{8}x-1=0.25\)
\(\Leftrightarrow\frac{1}{8}x=\frac{1}{4}+1=\frac{5}{4}\)
\(\Leftrightarrow x=\frac{5}{4}:\frac{1}{8}=\frac{5}{4}\cdot8=\frac{40}{4}=10\)
Vậy: x=10
c) Ta có: \(\left|2\frac{1}{2}-x\right|=4\)
\(\Leftrightarrow\left|\frac{5}{2}-x\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{5}{2}-x=4\\\frac{5}{2}-x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=4-\frac{5}{2}=\frac{3}{2}\\-x=-4-\frac{5}{2}=-\frac{13}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-3}{2}\\x=\frac{13}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-3}{2};\frac{13}{2}\right\}\)
d) Ta có: \(\frac{x}{6}=\frac{y}{7}\) và x+y=-39
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{x}{6}=\frac{y}{7}=\frac{x+y}{6+7}=\frac{-39}{13}=-3\)
Do đó:
\(\left\{{}\begin{matrix}\frac{x}{6}=-3\\\frac{y}{7}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-18\\y=-21\end{matrix}\right.\)
Vậy: (x,y)=(-18;-21)
