\(\left(m+1\right)^2\ge4m\)

\(\Leftrightarrow m^2+2m+1\ge4m\)

\(\Leftrightarrow m^2-2m+1\ge0\Leftrightarrow\left(m-1\right)^2\ge0\)

\(m^2+n^2+2\ge2\left(m+n\right)\)

\(\Leftrightarrow m^2-2m+1+n^2-2n+1\ge0\)

\(\Leftrightarrow\left(m-1\right)^2+\left(n-1\right)^2\ge0\)

làm câu đầu trước nha :

<=> m²+2m+1>=4m

<=>m²-2m+1>=0

<=>(m-1)²>=0 ( điều phải chứng minh

Đặt A=√(3+√(3+.....>>>A^2=3+√3+√3....=3+A

>>>A^2-A-3=0>>>A=(1+√13)/2

>>>M=3-A/6-A=(7-√13)/18

Áp dụng BĐT AM-GM ta có:

\(xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel:

\(S=\frac{1}{x^2+y^2}+\frac{3}{4xy}=\frac{1}{x^2+y^2}+\frac{2}{4xy}+\frac{1}{4xy}\)

\(=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{4xy}\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\frac{1}{4xy}\)

\(\ge\frac{\left(1+1\right)^2}{\left(x+y\right)^2}+\frac{1}{4\cdot\frac{1}{4}}=4+1=5\)

Xảy ra khi \(x=y=\frac{1}{2}\)

Vì 1-x-2x^2>=0>>>2x^2-x-1<=0>>>-1<=x<=1/2

F(x)=1/2(x+2√(1-2x)(x+1)<=1/2(x+1-2x+x+1)(BĐT Cô-si)

<=1/2.2=1.

Dấu= xảy ra khi 1-2x=x+1 khi x=0(TM)

 Học tốt!

ĐKXĐ : \(x\le2\)

\(A=x-\sqrt{2-x}\)

\(\Rightarrow2-A=2-x+\sqrt{2-x}\)

\(\Leftrightarrow2-A+\frac{1}{4}=2-x+\sqrt{2-x}+\frac{1}{4}\)

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

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

Vì \(\sqrt{2-x}+\frac{1}{2}\ge\frac{1}{2}\Rightarrow\left(\sqrt{2-x}+\frac{1}{2}\right)^2\ge\frac{1}{4}\)

\(\Leftrightarrow\frac{9}{4}-A\ge\frac{1}{4}\Leftrightarrow A\le\frac{9}{4}-\frac{1}{4}=2\) có max là 2

Dấu "=" xảy ra \(\sqrt{2-x}=0\Rightarrow x=2\)

Vậy \(A_{max}=2\) tại \(x=2\)

Điều kiện : \(x\ge0;x\ne4;x\ne9\)

\(A=\left(\frac{1}{1+\sqrt{x}}\right):\left[\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right]\)

\(A=\frac{1}{1+\sqrt{x}}:\left[\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)

\(A=\frac{1}{1+\sqrt{x}}:\left[\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)

\(A=\frac{1}{1+\sqrt{x}}:\left[\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)

\(A=\frac{1}{1+\sqrt{x}}:\left[\frac{x-9-\left(x-4\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)

\(A=\frac{1}{1+\sqrt{x}}:\left[\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)

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

A=(x​+x​+y​y−xy​​):(xy​+yx​+xy​−xy​−xy​x+y​)

=\frac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\frac{x\left(\sqrt{xy}-x\right)\sqrt{xy}+y\left(\sqrt{xy}+y\right)\sqrt{xy}-\left(x+y\right)\left(\sqrt{xy}+y\right)\left(\sqrt{xy}-x\right)}{\sqrt{xy}\left(\sqrt{xy}+y\right)\left(\sqrt{xy}-x\right)}=x​+y​x+xy​+y−xy​​:xy​(xy​+y)(xy​−x)x(xy​−x)xy​+y(xy​+y)xy​−(x+y)(xy​+y)(xy​−x)​

=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\frac{x^2y-x^2\sqrt{xy}+xy^2+y^2\sqrt{xy}-y^2\sqrt{xy}+x^2\sqrt{xy}}{xy^2-x^2y}=x​+y​x+y​:xy2−x2yx2y−x2xy​+xy2+y2xy​−y2xy​+x2xy​​

=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{xy^2-x^2y}{xy^2+x^2y}=x​+y​x+y​.xy2+x2yxy2−x2y​

=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{xy\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)}{xy\left(x+y\right)}=x​+y​x+y​.xy(x+y)xy(y​−x​)(x​+y​)​

=\sqrt{y}-\sqrt{x}=y​−x​