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\(\frac{a}{k}=\frac{x}{a}\Rightarrow a^2=xk;\frac{b}{k}=\frac{y}{b}\Rightarrow b^2=ky\)
=>\(\frac{a^2}{b^2}=\frac{xk}{yk}=\frac{x}{y}\)
Có: \(\left\{{}\begin{matrix}\frac{a}{k}=\frac{x}{a}\\\frac{b}{k}=\frac{y}{b}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=kx\\b^2=ky\end{matrix}\right.\\ \Rightarrow\frac{a^2}{b^2}=\frac{kx}{ky}=\frac{x}{y}\)
Ta có:
\(\left\{{}\begin{matrix}\frac{a}{k}=\frac{x}{a}\Rightarrow a^2=kx\\\frac{b}{k}=\frac{y}{b}\Rightarrow b^2=ky\end{matrix}\right.\)
Chia theo vế ta được:
\(a^2:b^2=kx:ky\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{kx}{ky}\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{x}{y}\left(đpcm\right).\)
Chúc bạn học tốt!
Ta có: \(\frac{a}{k}=\frac{x}{a};\frac{b}{k}=\frac{y}{b}\)
=> a2 = x.k; b2 = y.k
=> \(\frac{a^2}{b^2}=\frac{x.k}{y.k}=\frac{x}{y}\left(đpcm\right)\)
a/k = x/a => a2 = kx (1)
b/k = y/b => b2 = ky (2)
chia (1) cho (2) có;
a2/b2 =x/y
Ta có :
\(\begin{cases}\frac{a}{k}=\frac{x}{a}\\\frac{b}{k}=\frac{y}{b}\end{cases}\)
\(\Rightarrow\begin{cases}a^2=kx\\b^2=ky\end{cases}\)
Chia về theo vế
\(\Rightarrow a^2:b^2=\left(kx\right):ky\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{kx}{ky}\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{x}{y}\)
\(x.y=12\Rightarrow y=\frac{12}{x}\) thay vào pt ta có :
\(\frac{x}{3}=\frac{12}{\frac{x}{4}}\)
\(\Leftrightarrow\frac{x}{3}=\frac{3}{x}\) \(\Leftrightarrow x^2=9\) \(\Rightarrow Th1:x=3\Rightarrow y=4\)
\(Th2:x=-3\Rightarrow y=-4\)
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) => \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\left(đpcm\right)\)
\(x^2+y^2=1\)\(\Leftrightarrow\)\(\left(x^2+y^2\right)^2=1\) \(\left(1\right)\)
Thay \(\left(1\right)\) vào \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\) ta được :
\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\)
\(\Leftrightarrow\)\(\frac{x^4b+y^4a}{ab}=\frac{x^4+2x^2y^2+y^4}{a+b}\)
\(\Leftrightarrow\)\(\left(a+b\right)\left(x^4b+y^4a\right)=ab\left(x^4+2x^2y^2+y^4\right)\)
\(\Leftrightarrow\)\(x^4ab+y^4a^2+x^4b^2+y^4ab=x^4ab+2x^2y^2ab+y^4ab\)
\(\Leftrightarrow\)\(x^4b^2+y^4a^2=2x^2y^2ab\)
\(\Leftrightarrow\)\(x^4b^2-2x^2y^2ab+y^4a^2=0\)
\(\Leftrightarrow\)\(\left(x^2b\right)^2-2.x^2b.y^2a+\left(y^2a\right)^2=0\)
\(\Leftrightarrow\)\(\left(x^2b-y^2a\right)=0\)
\(\Leftrightarrow\)\(x^2b-y^2a=0\)
\(\Leftrightarrow\)\(x^2b=y^2a\)
\(\Leftrightarrow\)\(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\) ( thay \(x^2+y^2=1\) )
\(\Leftrightarrow\)\(\left(\frac{x^2}{a}\right)^{1002}=\left(\frac{y^2}{b}\right)^{1002}=\left(\frac{1}{a+b}\right)^{1002}\)
\(\Leftrightarrow\)\(\frac{x^{2004}}{a^{1002}}=\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}\)
Do đó :
\(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}+\frac{1}{\left(a+b\right)^{1002}}=\frac{2}{\left(a+b\right)^{1002}}\) ( đpcm )
Chúc bạn học tốt ~
\(\frac{x^4}{a}=\frac{y^4}{b}=\frac{1}{a+b}=\frac{x^4+y^4}{a+b}\Rightarrow x^4+y^4=1.\)
Mà \(x^2+y^2=1\)=>\(x^4+y^4=x^2+y^2=1.\)
Nếu x =0 => y =1 => a =0 vô lí
Xem lại đề dc ko ( hay mình làm sai?)
ta có
\(\frac{a}{k}=\frac{x}{a}=>a^2=kx\left(1\right)\)
\(\frac{b}{k}=\frac{y}{b}=>b^2=ky\left(2\right)\)
từ (1)và (2) , ta có
\(\frac{a^2}{b^2}=\frac{kx}{ky}=\frac{x}{y}\)
vậy \(\frac{a^2}{b^2}=\frac{x}{y}\)
\(\frac{a^2}{b^2}=\frac{kx}{ky}=\frac{x}{y}\)
Ta có: \(\frac{a}{k}=\frac{x}{a}=>a^2=k\cdot x\)
\(\frac{b}{k}=\frac{y}{b}=>b^2=k\cdot y\)
=> \(\frac{a^2}{b^2}=\frac{kx}{ky}=\frac{x}{y}\)(rút gọn)
=>đpcm
Chúc bạn học tốt!^_^
\(\frac{a}{k}=\frac{x}{a}\)
\(\Rightarrow a^2=kx\)
\(\frac{b}{k}=\frac{y}{b}\)
\(\Rightarrow b^2=ky\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{kx}{ky}=\frac{x}{y}\)