Có:x+y =1 => (x+y)² = 1 => x² + y² = 1-2xy

 \(\frac{x}{y+1}+\frac{y}{x+1}=\frac{x\left(x+1\right)+y\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}=\frac{x^2+x+y^2+y}{yx+y+x+1}=\frac{1-2xy+1}{yx+2}\)\(=\frac{2-2xy}{2+yx}\)

Vì x,y không âm

=> \(-xy\le xy\)

=> \(-2xy\le xy\)

=>\(2-2xy\le2+xy\)

=> \(\frac{2-2xy}{2+xy}\le1\)                 

=> đpcm

Ta có: \(\frac{4}{\sqrt{5}-\sqrt{3}}-\sqrt{12}\)

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

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

\(=2\sqrt{5}+2\sqrt{3}-2\sqrt{3}\)

\(=2\sqrt{5}\)

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

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

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

\(=2\sqrt{5}+2\sqrt{3}-2\sqrt{3}\) 

\(=2\sqrt{5}\)           

Đặt \(D=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(\Leftrightarrow D^2=8+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)

\(\Leftrightarrow D^2=8+2\sqrt{16-10-2\sqrt{5}}\)

\(\Leftrightarrow D^2=8+2\sqrt{6-2\sqrt{5}}\)

\(\Leftrightarrow D^2=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(\Leftrightarrow D^2=8+2\left(\sqrt{5}-1\right)\)

\(\Leftrightarrow D^2=6+2\sqrt{5}\)

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

\(\Rightarrow D=\sqrt{5}+1\)

Thay vào ta tính được: \(A=\sqrt{5}+1-\sqrt{5}=1\)

Vậy A = 1

       Bài làm :

Ta có :

\(\sqrt{\left(a-b\right)^2}=\left|a-b\right|\)

\(\sqrt{\left(a-b\right)^2}\) =  a  - b

cái này địa lý lớp 5 hả

câu này lớp 9 mà

Điều kiện: x>0, x khác 1

Ta có: \(M=\frac{x\sqrt{x-1}}{x-\sqrt{x}}-\frac{x\sqrt{x+1}}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)

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

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

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

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

Cách 1:

Với mọi x, ta có:

\(x^2+3x+3=\left(x+\frac{3}{2}\right)^2+\frac{3}{4}>0;2x^2+3x+2=2\left(x+\frac{3}{4}\right)^2+\frac{7}{8}>0\)

Do đó: \(\sqrt[3]{x^2+3x+3}>0;\sqrt[3]{2x^2+3x+2}>0\)

Áp dụng bất đẳng thức Co-si cho 3 số:

\(\sqrt[3]{x^2+3x+3}=\sqrt[3]{\left(x^2+3x+3\right).1.1}\le\frac{x^2+3x+3+1+1}{3}=\frac{x^2+3x+5}{3}\)

\(\sqrt[3]{2x^2+3x+2}=\sqrt[3]{\left(2x^2+3x+2\right).1.1}\le\frac{2x^2+3x+4}{3}\)

\(\Rightarrow6x^2+12x+8\le\frac{x^2+3x+5}{3}+\frac{2x^2+3x+4}{3}=x^2+2x+3\)

\(\Rightarrow5x^2+10x+5\le0\Rightarrow5\left(x+1\right)^2\le0\Rightarrow x=-1\)

Vậy nghiệm của phương trình là x=-1

Cách 2:

Đặt \(a=\sqrt[3]{x^2+3x+3}>0;b=\sqrt[3]{2x^2+3x+2}>0\)

Phương trình trở thành: \(a+b=2a^3+2b^3-2\)

Lại có: \(\left(a+b\right)\left(a-b\right)^2\ge0,\forall a>0,b>0\Rightarrow2a^3+2b^3\ge\frac{1}{2}\left(a+b\right)^3\)

\(\Rightarrow a+b\ge\frac{1}{2}\left(a+b\right)^3-2\Leftrightarrow\left(a+b-2\right)\left[\left(a+b\right)^2+2\left(a+b\right)+2\right]\le0\)

\(\Leftrightarrow a+b\le2\)

Từ phương trình ban đầu ta còn có: \(a+b=6\left(x+1\right)^2+2\ge2\Rightarrow a+b=2\Rightarrow x=-1\)