Bài 1:
a)Ta có:
\(\frac{4}{5}\left(\frac{7}{2}+\frac{1}{4}\right)^2=\frac{4}{5}\left(\frac{15}{4}\right)^2=\frac{4}{5}.\frac{15}{4}.\frac{15}{4}=\frac{45}{4}\)
b)Ta có:
\(\frac{5^4.20^4}{25^5.4^5}=\frac{\left(5.20\right)^4}{\left(25.4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)
Bài 2:
Ta có:
\(\frac{x}{2}=\frac{y}{-5}=\frac{x-y}{2-\left(-5\right)}=\frac{10}{7}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{20}{7}\\y=\frac{-50}{7}\end{matrix}\right.\)
Bài 3:
Ta có:
\(\left\{{}\begin{matrix}\left(x-\frac{1}{5}\right)^{2004}\ge0\\\left(y+0,4\right)^{100}\ge0\\\left(z-3\right)^{678}\ge0\\\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-\frac{1}{5}\right)^{2004}=0\\\left(y+0,4\right)^{100}=0\\\left(z-3\right)^{678}=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-\frac{1}{5}\right)=0\\\left(y+0,4\right)=0\\\left(z-3\right)=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{5}\\y=-0,4\\z=3\\\end{matrix}\right.\)
Bài 1 :
a,b\(\frac{5^4.20^4}{25^5.4^5}=\frac{1^4.5^4}{5^5.1^5}=\frac{1.1^4}{1^5.1}=\frac{1}{1}=1\)\(\frac{4}{5}.\left(\frac{7}{2}+\frac{1}{4}\right)^2=\frac{4}{5}.\left(\frac{14+1}{4}\right)^2=\frac{4}{5}.\left(\frac{15}{4}\right)^2=\left(\frac{4}{5}.\frac{15}{4}\right)^2=\left(\frac{1.3}{1.1}\right)^2=(\frac{3}{1})^2=3^2=9\)
Bài 2 :\(x\div2=y\div\left(-5\right)\) và x - y = 10
Có \(x\div2=y\div\left(-5\right)\Rightarrow\frac{x}{2}=\frac{y}{-5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{2}=\frac{y}{-5}=\frac{x-y}{2-\left(-5\right)}=\frac{10}{7}\)
Do đó :
\(\frac{x}{2}=\frac{10}{7}\Rightarrow x=\frac{2.10}{7}=\frac{20}{7}\)
\(\frac{y}{-5}=\frac{10}{7}\Rightarrow y=\frac{-5.10}{7}=\frac{-50}{7}\)
Vậy x = \(\frac{20}{7},y=\frac{-50}{7}\)
Chúc bạn học tốt
