\(\hept{\begin{cases}\left(x+1\right)+\sqrt{x}+\sqrt{y+1}=2\\\left(y+1\right)+\sqrt{y}+\sqrt{x+1}=2\end{cases}}\)         ĐK:  \(\hept{\begin{cases}x\ge0\\y\ge0\end{cases}}\)

Lấy pt (1) - (2) Ta được

\(\left(x+1\right)-\left(y+1\right)+\sqrt{x}-\sqrt{y}+\left(\sqrt{y+1}-\sqrt{x+1}\right)=0\)

\(\Leftrightarrow\left(x-y\right)+\left(\sqrt{x}-\sqrt{y}\right)+\frac{\left(y+1\right)-\left(x+1\right)}{\sqrt{y+1}+\sqrt{x+1}}=0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\left(\sqrt{x}-\sqrt{y}\right)-\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y+1}+\sqrt{x+1}}=0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+1-\frac{\sqrt{x}+\sqrt{y}}{\sqrt{y+1}+\sqrt{x+1}}\right)=0\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}=\sqrt{a}\cdot\sqrt{3a+b}+\sqrt{b}\cdot\sqrt{3b+a}\)

\(\le\sqrt{\left(a+b\right)\left(3a+b+3b+a\right)}=2\left(a+b\right)\)

\(\Rightarrow\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\frac{a+b}{2\left(a+b\right)}=\frac{1}{2}\)

Xảy ra khi \(a=b\)

Thắng Nguyễn ơi, bài này dùng cô si được ko bạn

b)\(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)

\(pt\Leftrightarrow\frac{4}{x}+\sqrt{x-\frac{1}{x}}-\sqrt{\frac{3}{2}}=x+\sqrt{2x-\frac{5}{x}}-\sqrt{\frac{3}{2}}\)

\(\Leftrightarrow\left(\frac{4}{x}-x\right)+\frac{x-\frac{1}{x}-\frac{3}{2}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}=\frac{2x-\frac{5}{x}-\frac{3}{2}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}\)

\(\Leftrightarrow\frac{-\left(x-2\right)\left(x+2\right)}{x}+\frac{\frac{\left(x-2\right)\left(2x+1\right)}{2x}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}-\frac{\frac{\left(x-2\right)\left(4x+5\right)}{2x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{-\left(x+2\right)}{x}+\frac{\frac{\left(2x+1\right)}{2x}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}-\frac{\frac{\left(4x+5\right)}{2x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}\right)=0\)

Pt trong ngoặc VN suy ra x=2

a)\(x^2+3\sqrt{x^2-1}=\sqrt{x^4-x^2+1}\)

\(\Leftrightarrow x^2+3\sqrt{x^2-1}-1=\sqrt{x^4-x^2+1}-1\)

\(\Leftrightarrow\frac{x^2\left(3\sqrt{x^2-1}+1\right)}{3\sqrt{x^2-1}+1}+\frac{9\left(x^2-1\right)-1}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2+1-1}{\sqrt{x^4-x^2+1}+1}\)

\(\Leftrightarrow\frac{9x^2-10+3x^2\sqrt{x^2-1}+x^2}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2}{\sqrt{x^4-x^2+1}+1}\)

\(\Leftrightarrow\frac{\sqrt{x^2-1}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}=\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}\)

\(\Leftrightarrow\frac{\sqrt{\left(x-1\right)\left(x+1\right)}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(\frac{\frac{1}{\sqrt{x^2-1}}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2}{\sqrt{x^4-x^2+1}+1}\right)=0\)

pt trong căn vô nghiệm

suy ra x=1; x=-1

x+y = x +1

y = 1

3y = 3 

x - 3 = 2

x = 5

\(\Rightarrow\hept{\begin{cases}x=5\\y=1\end{cases}}\)

X = 5 , Y = 1 

ĐK: m#0

pt có 2 nghiệm phân biệt

<=> \(\Delta'>0\)

<=> 2m^2 - 2m +4 > 0 (luôn đúng)

Ta có

(x₁-x₂)²=9

<=> (x₁+x₂)² - 4x₁x₂ = 9

Đến đây áp dụng hệ thức Viet là ra

\(m=-4+4\sqrt{2}\) hoặc   \(m=-4-4\sqrt{2}\)

\(a,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)

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

\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)

\(=\sqrt{xy}\)

\(b,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)

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

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

Chúc bạn học giỏi 

Kết bạn với mình nha 

\(5=\sqrt{25}\)

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

\(5>\sqrt{3+1}\)

\(5=\sqrt{25}\)

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

Vậy \(5>\sqrt{3+1}\)

Nếu đúng thì bấm đúng cho mình nhé