giup minh voi huhu

\(\frac{2x+3}{3y-2}=1\Rightarrow2x+3=3y-2\Rightarrow2x-3y=-5\)\(\left(1\right)\)

\(3\left(3y+2\right)-4\left(x-2y\right)=0\)

\(\Rightarrow9y+6-4x+2y=0\)

\(\Rightarrow5x+2y=-6\)\(\left(2\right)\)

Từ ( 1 ) và ( 2 ) ta có :

\(\hept{\begin{cases}2x-3y=-5\\5x+2y=-6\end{cases}\Rightarrow\hept{\begin{cases}10x-15y=-25\\10x+4y=-12\end{cases}}}\)

Trừ xuống ta có : \(-19y=13\Rightarrow y=-\frac{13}{19}\)

\(\Rightarrow x=...\)

\(\hept{\begin{cases}2x^2+3y=1\\3x^2-2y=2\end{cases}}\Rightarrow\hept{\begin{cases}x^2=\frac{1-3y}{2}\text{ (*)}\\3x^2-2y=2\text{ (**)}\end{cases}}\)

Thay \(\text{ (*)}\)vào \(\text{ (**)}\),ta được:

\(\text{ (**)}=\frac{3-9y}{2}-2y=2\)

\(\Leftrightarrow3-9y-4y=4\)

\(\Leftrightarrow3-13y=4\)

\(\Leftrightarrow y=\frac{3-4}{13}=\frac{-1}{13}\text{ (***)}\)

Thay \(\text{ (***) vào (*), ta được:}\)

\(x^2=\frac{1+\frac{3}{13}}{2}=\frac{\frac{16}{13}}{2}=\frac{16}{13.2}=\frac{8}{13}\)

\(\Rightarrow x=\pm\sqrt{\frac{8}{13}}\)

\(\text{ Vậy cặp (x;y) thỏa mãn là }\left(\pm\sqrt{\frac{8}{13}};\frac{-1}{13}\right)\)

~Chúc bạn _HỌ_C T_ỐT^^~

\(\sqrt{\left(3-\sqrt{5}\right)}.\left(\sqrt{10}-\sqrt{2}\right).\left(3+\sqrt{5}\right)\)

\(=\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3+\sqrt{5}}\)

\(=\sqrt{9-5}.\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3+\sqrt{5}}\)

\(=2\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3+\sqrt{5}}\)

\(=\sqrt{2}\left(\sqrt{10}-\sqrt{2}\right).\sqrt{2}.\sqrt{3+\sqrt{5}}\)

\(=\left(\sqrt{20}-\sqrt{4}\right).\sqrt{6+2\sqrt{5}}\)

\(=\left(2\sqrt{5}-2\right)\sqrt{5+2\sqrt{5}+1}\)

\(=\left(2\sqrt{5}-2\right)\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\)

\(=2.\left(5-1\right)=2.4=8\)

\(\sqrt{x^3}-1=\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right).\)