tìm a, b

 \(\frac{a}{b}=\frac{3}{4}\) \(< =>\frac{a}{3}=\frac{b}{4}\)

Đặt \(\frac{a}{3}=\frac{b}{4}=k=>a=3k;b=4k\)

Ta có: \(A=\frac{a^2+b^2}{a^2-b^2}=\frac{\left(3k\right)^2+\left(4k\right)^2}{\left(3k\right)^2-\left(4k\right)^2}=\frac{9k^2+16k^2}{9k^2-16k^2}=\frac{\left(9+16\right).k^2}{\left(9-16\right).k^2}=\frac{28k^2}{-7k^2}=\frac{28}{-7}=-4\)

Vậy A=-4
 

câu hỏi là gì vậy

Co ai giup mk zoi

\(\left(a+b+c\right)^2=3\left(a^2+b^2+c^2\right)\)

\(< =>a^2+b^2+c^2+2ab+2bc+2ac=3a^2+3b^2+3c^2\)

\(< =>a^2+b^2+c^2+2ab+2bc+2ac-3a^2-3b^2-3c^2=0\)

\(< =>-2a^2-2b^2-2c^2+2ab+2bc+2ac=0\)

\(< =>-\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)=0\)

\(< =>2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(< =>a^2+a^2+b^2+b^2+c^2+c^2-2ab-2bc-2ac=0\)

\(< =>\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)

\(< =>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Mà \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) với mọi a;b;c

Để thỏa mãn \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\) thì \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Leftrightarrow a=b=c\left(đpcm\right)}\)

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