Sai đề ?

sr nha này : \(\frac{3}{a+\sqrt{3}}-\frac{2}{a-b\sqrt{3}}=7-20\sqrt{3}\)tìm a,b hữu tỉ

LEVER

\(\frac{3}{a+b\sqrt{3}}-\frac{2}{a-b\sqrt{3}}=7-20\sqrt{3}\)\(\Leftrightarrow\frac{3\left(a-b\sqrt{3}\right)-2\left(a+b\sqrt{3}\right)}{a^2-3b^2}=7-20\sqrt{3}\)

\(\Leftrightarrow\frac{a-5\sqrt{3}b}{a^2-3b^2}=7-20\sqrt{3}\)\(\Leftrightarrow\frac{a-5\sqrt{3}b}{a^2-3b^2}=\frac{7-20\sqrt{3}}{49-48}\Leftrightarrow\frac{a-5\sqrt{3}b}{a^2-3b^2}=\frac{7-20\sqrt{3}}{7^2-3.4^2}\Leftrightarrow\hept{\begin{cases}a=7\\b=4\end{cases}}\)

Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow\)\(x^2+y^2+z^2=4\)

\(P=\frac{x^3}{x+3y}+\frac{y^3}{y+3z}+\frac{z^3}{z+3x}=\frac{x^4}{x^2+3xy}+\frac{y^4}{y^2+3yz}+\frac{z^4}{z^2+3zx}\)

\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(x^2+y^2+z^2\right)}=\frac{4^2}{4+3.4}=1\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{2}{\sqrt{3}}\)

à nhầm, \(a=b=c=\frac{4}{3}\) nhé 

có a giải đc bài này giúp em vs ạ

\(\frac{2}{a+b\sqrt{5}}-\frac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\)

\(\Leftrightarrow\frac{2a-2b\sqrt{5}-3a-3b\sqrt{5}}{a^2-5b^2}=-9-20\sqrt{5}\)

\(\Leftrightarrow\frac{a+5b\sqrt{5}}{a^2-5b^2}=9+20\sqrt{5}\)

\(\Leftrightarrow\sqrt{5}\left(100b^2+5b-20a^2\right)=9a^2-a-45b^2\)

Ta nhận thây VT là sô vô tỷ còn VP là sô hữu tỷ.

\(\Rightarrow\hept{\begin{cases}100b^2+5b-20a^2=0\\9a^2-a-45b^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=0\\b=0\end{cases}\left(loai\right)}\)hoặc \(\hept{\begin{cases}a=9\\b=4\end{cases}\left(nhan\right)}\)