\(\sqrt{x^2-2x+2}+\sqrt{x^2-4x+8}\)

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

\(\ge\sqrt{\left(x-1+2-x\right)^2+\left(1+2\right)^2}=\sqrt{10}\)

Dấu "=" xảy ra <=> \(\dfrac{x-1}{1}=\dfrac{2-x}{2}\Leftrightarrow x=\dfrac{4}{3}\)

\(M=A+B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}+3}=\dfrac{\sqrt{x}+2\sqrt{x}}{\sqrt{x}+3}=\dfrac{3\sqrt{x}}{\sqrt{x}+3}\left(x\ge0\right)\)

`M=A+B`

`=sqrtx/(sqrtx+3)+(2sqrtx)/(sqrtx+3)`

`=(sqrtx+2sqrtx)/(sqrtx+3)`

`=(3sqrtx)/(sqrtx+3)`

Với \(x\ge0\), ta có:

\(M=A+B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}+3}\) \(=\dfrac{3\sqrt{x}}{\sqrt{x}+3}\) \(=\dfrac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) \(=\dfrac{3x-9\sqrt{x}}{x-9}\)

#Cho mình sửa lại chút nhé! Nãy lag tí :)))

Đề thiếu rồi em, biết ... nó phải bằng cái gì đó chứ?

Lời giải:

$A=\frac{10\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+4)}-\frac{(2\sqrt{x}-3)(\sqrt{x}-1)}{(\sqrt{x}+4)(\sqrt{x}-1)}-\frac{(\sqrt{x}+1)(\sqrt{x}+4)}{(\sqrt{x}-1)(\sqrt{x}+4)}$

$=\frac{10\sqrt{x}-(2\sqrt{x}-3)(\sqrt{x}-1)-(\sqrt{x}+1)(\sqrt{x}+4)}{(\sqrt{x}+4)(\sqrt{x}-1)}$

$=\frac{-3x+10\sqrt{x}-7}{(\sqrt{x}+4)(\sqrt{x}-1)}$

$=\frac{-(\sqrt{x}-1)(3\sqrt{x}-7)}{(\sqrt{x}+4)(\sqrt{x}-1)}=\frac{7-3\sqrt{x}}{\sqrt{x}+4}$

\(\sqrt{\left(120-11\right)^2}+\sqrt{\left(10-\sqrt{120}\right)^2}\)

\(=120-11+10+\sqrt{120}\)

\(=\sqrt{120}\left(\sqrt{120}+1\right)-1\)

\(a,=\left(120-11\right)+\left|10-\sqrt{120}\right|=109+\sqrt{120}-10=99+2\sqrt{30}\\ b,=\sqrt{\left(\sqrt{x+1}+1\right)^2-\left(\sqrt{x+1}+1\right)^2}=\sqrt{0}=0\)

ta có: \(\sqrt{x}+2\sqrt{y}=10=>\left(\sqrt{x}+2\sqrt{y}\right)^2=100\)

áp dụng BDT Bunhia 

\(\sqrt{x}+2\sqrt{y}\le\sqrt{\left(1+2^2\right)\left(x+y\right)}\)

\(=>100\le5\left(x+y\right)=>x+y\ge\dfrac{100}{5}=20\)

\(a.A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\left(x\ge0;x\ne1\right)\)

Để : \(A=\dfrac{2}{7}\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{7}\)

\(\Leftrightarrow x+\sqrt{x}-6=0\)

\(\Leftrightarrow x-2\sqrt{x}+3\sqrt{x}-6=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=0\)

\(\Leftrightarrow x=4\left(TM\right)\)

\(b.A^2=\left(\dfrac{2}{x+\sqrt{x}+1}\right)^2=\dfrac{4}{\left(x+\sqrt{x}+1\right)^2}\left(1\right)\)

\(2A=2.\dfrac{2}{x+\sqrt{x}+1}=\dfrac{4}{x+\sqrt{x}+1}\left(2\right)\)

Mà : \(x+\sqrt{x}+1\le\left(x+\sqrt{x}+1\right)^2\left(3\right)\)

Từ \(\left(1;2;3\right)\Rightarrow2A\ge A^2\)