\(A=\sqrt{\left(x-2\right)\left(x-1\right)x\left(x+1\right)+5}\)

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

Đặt \(t=x^2-x\) ta đc:

\(A=\sqrt{\left(t-2\right)t+5}=\sqrt{t^2-2t+5}\)

\(=\sqrt{\left(t-1\right)^2+4}\ge\sqrt{4}=2\)

Dấu = khi \(t=1\Leftrightarrow x^2-x=1\Leftrightarrow x=\pm\frac{1}{2}+\frac{\sqrt{5}}{2}\)

Vậy....

b)\(B=\sqrt{x^2-4x+4}+\sqrt{x^2+6x+9}\)

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

\(=\left|x-2\right|+\left|x+3\right|\)

Áp dụng Bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(\left|x-2\right|+\left|x+3\right|=\left|x-2\right|+\left|-x-3\right|\ge\left|x-2+\left(-x\right)-3\right|=5\)

Dấu = khi \(\left(x-2\right)\left(x+3\right)\ge0\)\(\Rightarrow-3\le x\le2\)

\(\Rightarrow\hept{\begin{cases}-3\le x\le2\\\left(x+3\right)\left(x-2\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\x=2\end{cases}}\)

Vậy....

a)\(A=\sqrt{25}-\sqrt{x^2-4x+4}\)

\(=5-\sqrt{\left(x-2\right)^2}\)

Thấy: \(\sqrt{\left(x-2\right)^2}\ge0\)\(\Rightarrow-\sqrt{\left(x-2\right)^2}\le0\)

\(\Rightarrow A=5-\sqrt{\left(x-2\right)^2}\le5\)

Khi \(x=2\)

b)Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\):

\(B=\sqrt{\left(x-5\right)^2}+\sqrt{\left(x-6\right)^2}\)

\(=\left|x-5\right|+\left|x-6\right|\)\(=\left|x-5\right|+\left|6-x\right|\)

\(\ge\left|x-5+6-x\right|=1\)

Khi \(5\le x\le6\)

a) \(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)

\(P=\left[\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(P=\left[\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{-4x-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{-\left(\sqrt{x}-3\right)}\)

\(P=\dfrac{-4\sqrt{x}\cdot\sqrt{x}}{-\left(\sqrt{x}-3\right)}\)

\(P=\dfrac{4x}{\sqrt{x}-3}\)

b) \(P=\dfrac{4x}{\sqrt{x}-3}\)

\(P=4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\)

Theo BĐT côsi ta có:

\(P\ge\sqrt{\dfrac{4\left(\sqrt{x}-3\right)\cdot36}{\sqrt{x}-3}}+24=36\)

Vậy: \(P_{min}=36\Leftrightarrow x=36\) 

3 câu này bạn áp dụng cái này nhé.

`a^2 >=0 forall a`.

`|a| >=0 forall a`.

`1/a` xác định `<=> a ne 0`.

a: P=(x+30)^2+(y-4)^2+1975>=1975 với mọi x,y

Dấu = xảy ra khi x=-30 và y=4

b: Q=(3x+1)^2+|2y-1/3|+căn 5>=căn 5 với mọi x,y

Dấu = xảy ra khi x=-1/3 và y=1/6

c: -x^2-x+1=-(x^2+x-1)

=-(x^2+x+1/4-5/4)

=-(x+1/2)^2+5/4<=5/4

=>R>=3:5/4=12/5

Dấu = xảy ra khi x=-1/2

Ta có : \(\left(x-1\right)^2\ge0\)

            \(\left|y-5\right|\ge0\)

            \(\sqrt{z-4}\ge0\)

Để có được \(Min_A\Leftrightarrow\hept{\begin{cases}x-1=0\\y-5=0\\z-4=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=5\\z=4\end{cases}}\)

\(\Leftrightarrow A=1^2+0+0+0+2020=2021\)

Vậy \(Min_A=2021\Leftrightarrow\left(x;y;z\right)=\left(1;5;4\right)\)

1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4 
--> Pmin=4 khi x=4

2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1

=> M=2x²-8x+\(\sqrt{x^2-4x+5}\)+6=2(t²-5)+t+6

<=> M=2t²+t-4\(\ge\)2.1²+1-4=-1

M_min=-1 khi t=1 hay x=2