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

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^^~

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

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

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

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

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

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

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

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

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

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

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

Áp dụng bđt Cô si cho 3 số ta đc

\(\frac{x}{y^2+2}+\frac{y}{z^2+2}+\frac{z}{x^2+2}\ge3\sqrt[3]{\frac{xyz}{\left(y^2+2\right)\left(z^2+2\right)\left(x^2+2\right)}}\)

\(VT\ge3\sqrt[3]{\frac{1}{27}=}1\)

Dấu " = " xảy ra <=> x = y = z = 1

p/s : quên cách làm khúc giữa

Áp dụng bất đẳng thức Cô si cho 3 số thực ko âm ta đc :

\(\frac{x}{y^2+2}+\frac{y}{z^2+2}+\frac{z}{x^2+2}\ge3\sqrt[3]{\frac{xyz}{\left(y^2+2\right)\left(z^2+2\right)\left(x^2+2\right)}}\)

\(\Rightarrow VT\ge3\sqrt[3]{\frac{1}{1+2y^2x^2+2z^2x^2+2z^2y^2+4x^2+4z^2+4y^2+8}}\)( phân tích đa thức thành nhân tử )

\(\Rightarrow VT\ge3\sqrt[3]{\frac{1}{9+\frac{2}{z^2}+\frac{2}{y^2}+\frac{2}{x^2}+4x^2+4z^2+4y^2}}\)( vì \(xyz=1\Rightarrow x^2y^2z^2=1\Rightarrow x^2y^2=\frac{1}{z^2}\)các phân số khác chứng minh tương tự )

\(\Rightarrow VT\ge3\sqrt[3]{\frac{1}{9+\frac{2+4z^4}{z^2}+\frac{2+4y^4}{y^2}+\frac{2+4x^4}{x^2}}}\)( quy đồng mẫu số  ) ( A )

Áp dụng bất đẳng thức Cô si cho 3 số thực ko âm ta được :

\(\frac{2+4z^4}{z^2}+\frac{2+4y^4}{y^2}+\frac{2+4x^4}{x^2}\ge3\sqrt[3]{\frac{\left(2+4z^4\right)\left(2+4y^4\right)\left(2+4x^4\right)}{x^2y^2z^2}}\) ( 1 )

Ta có :

\(2+4x^4\ge2+4.1^4=6\) ( 2 ) 

\(2+4y^4\ge2+4.1^4=6\) ( vì x^4 , y^4 , z^4 đều là các lũy thừa số mũ chẵn ) ( 3 )

\(2+4z^4\ge2+4.1^4=6\)( 4 ) 

x^2 . y^2 . z^2 = ( xyz )^2 = 1^2 = 1 ( 5 )

Từ ( 1 ) , ( 2 ) , ( 3 ) , ( 4 ) , ( 5 )  suy ra :

\(\frac{2+4z^4}{z^2}+\frac{2+4y^4}{y^2}+\frac{2+4x^4}{x^2}\ge3\sqrt[3]{\frac{6^3}{1}}=18\) ( B )

Thay B vào A ta đc  :

\(\Rightarrow VT\ge3\sqrt[3]{\frac{1}{9+\frac{2+4z^4}{z^2}+\frac{2+4y^4}{y^2}+\frac{2+4x^4}{x^2}}}\ge3\sqrt[3]{\frac{1}{9+18}}=1\)

a.ĐK:\(x\ge0;x\ne1\)

\(M=\frac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\)

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

\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)

b.\(\frac{3\sqrt{x}}{x-\sqrt{x}+1}< 1\Leftrightarrow3\sqrt{x}< x+\sqrt{x}+1\Leftrightarrow\left(\sqrt{x}-1\right)^2>0\)

(Luôn đúng với mọi \(x\ge0;x\ne1\)) . Ta có đpcm

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\)\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2+2}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\sqrt{2}+1\)

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)

\(\sqrt{4x+1}-\sqrt{3x+1}=1\)

\(\Rightarrow\left(\sqrt{4x+1}-\sqrt{3x+1}\right)^2=1\)

\(\Rightarrow|4x+1|-2\sqrt{\left(4x+1\right)\left(3x+1\right)}+|3x+1|=1\)

Dùng bảng xét dấu mà xét nốt bạn nhé