\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)ĐK x>=0 x khác -1

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

b/ x =\(\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{3+2\sqrt{3}+1}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)

\(\Rightarrow\sqrt{x}=\frac{\sqrt{3}+1}{2}\)

Em thay vào tính nhé!

c) với x>1

A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}.\sqrt{x}=\frac{x+\sqrt{x}}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\)

Áp dụng bất đẳng thức Cosi 

A\(\ge2\sqrt{2}+3\)

Xét dấu bằng xảy ra ....

dấu bằng xảy ra khi nào v ạ ??

A)

Đặt \(\sqrt{1+2x}=a; \sqrt{1-2x}=b\) (\(a,b>0\) )

\(\Rightarrow \left\{\begin{matrix} a^2+b^2=2\\ a^2-b^2=4x=\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} 2a^2=2+\sqrt{3}\rightarrow 4a^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\\ 2b^2=2-\sqrt{3}\rightarrow 4b^2=4-2\sqrt{3}=(\sqrt{3}-1)^2\end{matrix}\right.\)

\(\Rightarrow a=\frac{\sqrt{3}+1}{2}; b=\frac{\sqrt{3}-1}{2}\)

\(\Rightarrow ab=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}=\frac{1}{2}; a-b=1\)

Có:

\(A=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+ab^2}{(1+a)(1-b)}\)

\(=\frac{2-ab(a-b)}{1+(a-b)-ab}=\frac{2-\frac{1}{2}.1}{1+1-\frac{1}{2}}=1\)

B)

\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)

\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\)

\(\rightarrow 4(x^2-1)=\frac{a}{b}+\frac{b}{a}-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)

\(\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\) do $a>b$

T có: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)

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

Ta có \(ab+bc+ca=2abc\)

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)

Đặt \(\left\{{}\begin{matrix}x=\dfrac{1}{a}\\y=\dfrac{1}{b}\\z=\dfrac{1}{c}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y+z=2\\P=\dfrac{x^3}{\left(2-x\right)^2}+\dfrac{y^3}{\left(2-y\right)^3}+\dfrac{z^3}{\left(2-z\right)^2}\end{matrix}\right.\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\dfrac{x^3}{\left(2-x\right)^2}+\dfrac{2-x}{8}+\dfrac{2-x}{8}\ge3\sqrt[3]{\dfrac{x^3}{64}}=\dfrac{3x}{4}\)

Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y^3}{\left(2-y\right)^2}+\dfrac{2-y}{8}+\dfrac{2-y}{8}\ge\dfrac{3y}{4}\\\dfrac{z^3}{\left(2-z\right)^2}+\dfrac{2-z}{8}+\dfrac{2-z}{8}\ge\dfrac{3z}{8}\end{matrix}\right.\)

\(\Rightarrow P+\dfrac{12-2\left(x+y+z\right)}{8}\ge\dfrac{3}{4}\left(x+y+z\right)\)

\(\Rightarrow P\ge\dfrac{1}{2}\)

Dấu " = " xảy ra khi \(x=y=z=\dfrac{2}{3}\)

Ta có : \(ab+bc+ca=2abc\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+y+z=2\\P=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^3}+\frac{z^3}{\left(2-z\right)^2}\end{cases}}\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\)

Tương tự ta có :

\(\hept{\begin{cases}\frac{y^3}{\left(2-y\right)^2}+\frac{2-y}{8}+\frac{2-y}{8}\ge\frac{3y}{4}\\\frac{z^3}{\left(2-z\right)^2}+\frac{2-z}{8}+\frac{2-z}{8}\ge\frac{3z}{8}\end{cases}}\)

\(\Rightarrow P+\frac{12-2\left(x+y+z\right)}{8}\ge\frac{3}{4}\left(x+y+z\right)\)

\(\Rightarrow P\ge\frac{1}{12}\)

Dấu " = " xảy ra khi \(x=y=z=\frac{2}{3}\)

Ta có : \(ab+bc+ca=2abc\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+y+z=2\\P=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^3}+\frac{z^3}{\left(2-z^2\right)}\end{cases}}\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\)

Tương tự ta có : \(\hept{\begin{cases}\frac{y^3}{\left(2-y\right)^2}+\frac{2-y}{8}+\frac{2-y}{8}\ge\frac{3y}{4}\\\frac{z^3}{\left(2-z\right)^2}+\frac{2-z}{8}+\frac{2-z}{8}\ge\frac{3z}{8}\end{cases}}\)

\(\Rightarrow P+\frac{12-2\left(x+y+z\right)}{8}\ge\frac{3}{4}\left(x+y+z\right)\)

\(\Rightarrow P\ge\frac{1}{2}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{2}{3}\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\dfrac{a^3}{\left(1-a\right)^2}+\dfrac{1-a}{8}+\dfrac{1-a}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\)

Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{\left(1-b\right)^2}+\dfrac{1-b}{8}+\dfrac{1-b}{8}\ge\dfrac{3b}{4}\\\dfrac{c^3}{\left(1-c\right)^2}+\dfrac{1-c}{8}+\dfrac{1-c}{8}\ge\dfrac{3c}{4}\end{matrix}\right.\)

\(\Rightarrow P+\dfrac{6-2\left(a+b+c\right)}{8}\ge\dfrac{3}{4}\left(a+b+c\right)\)

\(\Rightarrow P\ge\dfrac{1}{4}\)

Vậy \(P_{min}=\dfrac{1}{4}\)

Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)

đó đâu phải BĐT cauchy-Schwarz đâu bạn ơi

Ta có : \(ab+bc+ca=2abc\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+y+z=2\\P=\frac{x^3}{\left(2-x\right)^2}\end{cases}+\frac{y^3}{\left(2-y\right)^3}+\frac{z^3}{\left(2-z\right)^2}}\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\)

Tương tự ta có : \(\hept{\begin{cases}\frac{y^3}{\left(2-y\right)^2}+\frac{2-y}{8}+\frac{2-y}{8}\ge\frac{3y}{4}\\\frac{z^3}{\left(2-z\right)^2}+\frac{2-z}{8}+\frac{2-z}{8}\ge\frac{3z}{8}\end{cases}}\)

\(\Rightarrow P+\frac{12-2\left(x+y+z\right)}{8}\ge\frac{3}{4}\left(x+y+z\right)\)

\(\Rightarrow P\ge\frac{1}{2}\)

Dấu " = " xảy ra khi \(x=y=z=\frac{2}{3}\)