Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, Ta co : \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\)\(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)(1)
Xet :\(\frac{a}{a+b}=\frac{c}{c+d}\Rightarrow\frac{a}{c}=\frac{a+b}{c+d}\)(2)
Tu (1) va (2) \(\Rightarrow\frac{a}{a+b}=\frac{c}{c+d}\)
b
a) \(C=\frac{\left(\frac{2}{3}\right)^3\times\left(-\frac{3}{4}\right)^2\times\left(-1\right)^5}{\left(\frac{2}{5}\right)^2\times\left(-\frac{5}{12}\right)^2}\)
\(C=\frac{\frac{2^3}{3^3}.\frac{\left(-3\right)^2}{4^2}.\left(-1\right)^5}{\frac{2^2}{5^2}.\frac{\left(-5\right)^2}{12^2}}\)
\(C=\frac{\frac{-\left(2^3.3^2\right)}{3^3.2^4}}{\frac{2^2.5^2}{5^2.2^4.3^2}}\)
\(C=\frac{\frac{-1}{3.2}}{\frac{1}{2^2.3^2}}\)
\(C=\frac{\frac{-1}{6}}{\frac{1}{36}}\)
\(C=-6\)
b) \(D=\frac{6^6+6^3\times3^3+3^6}{-73}\)
\(D=\frac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}\)
\(D=\frac{2^6.3^6+2^3.3^6+3^6}{-73}\)
\(D=\frac{3^6\left(2^6+2^3+1\right)}{-73}\)
\(D=\frac{3^6.73}{\left(-1\right).73}\)
\(D=-3^6=-729\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)\(=>\hept{\begin{cases}a=b.k\\c=d.k\end{cases}}\)
\(\left(\frac{a-b}{c-d}\right)^2=\left(\frac{b.k-b}{d.k-d}\right)^2=\left(\frac{b.\left(k-1\right)}{d.\left(k-1\right)}\right)^2\)\(=\frac{\left(b^2.\left(k-1\right)^2\right)}{\left(d^2.\left(k-1\right)^2\right)}=\frac{b^2.\left(k-1\right)^2}{d^2.\left(k-1\right)^2}=\frac{b^2}{d^2}\)\(\left(1\right)\)
\(\frac{ab}{cd}=\frac{b.k.b}{d.k.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\left(2\right)\)
Từ (1) và (2) => \(\left(\frac{a-b}{c-d}\right)^2=\frac{ab}{cd}\)
Đặt \(\frac{a}{b}\)= \(\frac{c}{d}\)= k => a= bk ; c = dk
\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) = \(\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}\)= \(\frac{b^2.\left(k-1\right)^2}{d^2.\left(k-1\right)^2}\)= \(\frac{b^2}{d^2}\) (1)
\(\frac{ab}{cd}\)= \(\frac{bk.b}{dk.d}\)= \(\frac{b^2}{d^2}\) (2)
Từ (1) và (2) ->> \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) = \(\frac{ab}{cd}\)