Question

tìm x,y 

\(\frac{x}{3}\)= \(\frac{y}{4}\)và x.y = 12

2.cho \(\frac{a}{b}\)= \(\frac{c}{d}\).CMR

\(\frac{a^2+b^2}{c^2+d^2}\)= \(\frac{ab}{cd}\)

3.Cho \(\frac{a+b}{a-b}\)= \(\frac{c+d}{c-d}\) . CMR

\(^{a^2}\)= b.c

 

Answer 1

\(x.y=12\Rightarrow y=\frac{12}{x}\) thay vào pt ta có : 

\(\frac{x}{3}=\frac{12}{\frac{x}{4}}\)

\(\Leftrightarrow\frac{x}{3}=\frac{3}{x}\) \(\Leftrightarrow x^2=9\) \(\Rightarrow Th1:x=3\Rightarrow y=4\)

\(Th2:x=-3\Rightarrow y=-4\)

Answer 2

đặt \(\frac{x}{3}=\frac{y}{4}=k\Rightarrow x=3k,y=4k\)

ta có:

\(x.y=3k.4k=12.k^2=12\Rightarrow k^2=1\Rightarrow\orbr{\begin{cases}k=1\\k=-1\end{cases}}\)

\(k=1\Rightarrow x=3.1=3,y=4.1=4\)

\(k=\left(-1\right)\Rightarrow x=3.\left(-1\right)=-3,y=4.\left(-1\right)=-4\)

vậy x=3,y=4 hay x=-3, y=-4

2.\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(1\right)\)

\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}\left(2\right)\)

từ (1) và (2) => \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\left(đpcm\right)\)