1) Áp dụng bđt Cauchy cho 3 số dương ta có

 \(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)

\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)

\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)

Cộng (1);(2);(3) theo vế ta được

\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)

\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)

\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)

2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)

Dấu"=" khi a = 4b

nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)

Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)

Đặt \(\sqrt{a+b}=t>0\) ta được

\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)

\(=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\)

Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)

nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)

khi đó a + b = 1

mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)

\(M=A\cdot B=\dfrac{x}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)

=>\(M=\dfrac{x}{\sqrt{x}+2}\)

=>\(M=\dfrac{x-4+4}{\sqrt{x}+2}=\sqrt{x}-2+\dfrac{4}{\sqrt{x}+2}\)

=>\(M=\sqrt{x}+2+\dfrac{4}{\sqrt{x}+2}-4\)

=>\(M>=2\cdot\sqrt{\left(\sqrt{x}+2\right)\cdot\dfrac{4}{\sqrt{x}+2}}-4=0\)

Dấu '=' xảy ra khi \(\sqrt{x}+2=\sqrt{4}=2\)

=>\(\sqrt{x}=0\)

=>x=0(nhận)

a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}-2-\sqrt{x}+3}\)

\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)

\(=\dfrac{-3}{\sqrt{x}+3}\)

b: Thay \(x=7-2\sqrt{6}\) vào A, ta được:

\(A=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-7+2\sqrt{6}-5\left(\sqrt{6}+1\right)-1}\)

\(=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-8+2\sqrt{6}-5\sqrt{6}-5}\)

\(=\dfrac{-3\sqrt{6}+3}{13+3\sqrt{6}}=\dfrac{93-48\sqrt{6}}{115}\)

a) ĐKXĐ: \(x>0\)

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

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

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

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

\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)(do \(\sqrt{x}+1\ge1>0\))

b) \(A=x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)>0\)(do \(x>1\))

\(\Leftrightarrow A=x-\sqrt{x}=\left|A\right|\)

c) \(A=x-\sqrt{x}=\left(x-\sqrt{x}+\dfrac{1}{4}\right)-\dfrac{1}{4}\)

\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(minA=-\dfrac{1}{4}\Leftrightarrow\sqrt[]{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)

\(a,A=\dfrac{x\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1\left(x>0\right)\\ A=\dfrac{x\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-2\sqrt{x}-1+1\\ A=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\\ A=2\Leftrightarrow x-\sqrt{x}-2=0\\ \Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}=2\left(\sqrt{x}>0\right)\\ \Leftrightarrow x=4\left(tm\right)\)

\(b,x>1\Leftrightarrow\sqrt{x}-1>0\\ \Leftrightarrow\left|A\right|=\left|x-\sqrt{x}\right|=\left|\sqrt{x}\left(\sqrt{x}-1\right)\right|=\sqrt{x}\left(\sqrt{x}-1\right)=A\left(\sqrt{x}>0\right)\)

\(c,A=x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\\ A_{min}=-\dfrac{1}{4}\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)

ĐKXĐ: \(x\ge0;x\ne1\)

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

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

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

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

b.

\(M=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\ge1-\dfrac{2}{0+1}=-1\)

\(M_{min}=-1\) khi \(x=0\)

\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)

A = \(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)  ; Đk \(x\) \(\ge\) 0

A = 2  - \(\dfrac{3}{\sqrt{x}+1}\) 

\(\sqrt{x}\) \(\ge\) 0 => \(\sqrt{x}\) + 1 \(\ge\) 1

\(\dfrac{3}{\sqrt{x}+1}\) \(\le\) \(\dfrac{3}{1}\)

\(\dfrac{3}{\sqrt{x}+1}\) \(\le\) 3

- \(\dfrac{3}{\sqrt{x}+1}\) \(\ge\) -3

2 - \(\dfrac{3}{\sqrt{x}+1}\) \(\ge\) 2 - 3

2 - \(\dfrac{3}{\sqrt{x}+1}\) \(\ge\) -1

A(min ) = -1  dấu bằng xảy ra khi  \(\sqrt{x}\) = 0 hay \(x\) = 0

a: ĐKXĐ: x>0; x<>1

\(Q=\dfrac{x+\sqrt{x}+\sqrt{x}}{x-1}:\dfrac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\)

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

\(=\dfrac{x}{\sqrt{x}-1}\)

b: Q>2

=>Q-2>0

=>\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)

=>căn x-1>0

=>x>1

a) ĐK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(Q=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\)

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

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

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

\(=\dfrac{x}{\sqrt{x}-1}\)

b) Q>2 <=> \(\dfrac{x}{\sqrt{x}-1}>2\Leftrightarrow x>2\sqrt{x}-2\)

\(\Leftrightarrow x-2\sqrt{x}+2>0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+1>0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2\ge0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1\le0\\\sqrt{x}-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le1\end{matrix}\right.\)

KL:.....