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

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

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

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

\(A=\sqrt{x-1-2\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}\)

\(A=\sqrt{x-2-2\sqrt{x-2}+1}+\sqrt{x-2-6\sqrt{x-2}+9}\)

\(A=\sqrt{\left(\sqrt{x-2}-1\right)^2}+\sqrt{\left(\sqrt{x-2}-3\right)^2}\)

\(A=\left|\sqrt{x-2}-1\right|+\left|\sqrt{x-2}-3\right|\)

\(A=\left|\sqrt{x-2}-1\right|+\left|3-\sqrt{x-2}\right|\)

\(A\ge\left|\sqrt{x-2}-1+3-\sqrt{x-2}\right|=\left|2\right|=2\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(\sqrt{x-2}-1\right)\left(3-\sqrt{x-2}\right)\ge0\)

TH1 : \(\hept{\begin{cases}\sqrt{x-2}-1\ge0\\3-\sqrt{x-2}\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge3\\x\le11\end{cases}\Leftrightarrow}3\le x\le11}\)

TH2 : \(\hept{\begin{cases}\sqrt{x-2}-1\le0\\3-\sqrt{x-2}\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le3\\x\ge11\end{cases}}}\) ( loại ) 

Vậy GTNN của \(A\) là \(2\) khi \(3\le x\le11\)

Chúc bạn học tốt ~ 

1:

a: =x^2-7x+49/4-5/4

=(x-7/2)^2-5/4>=-5/4

Dấu = xảy ra khi x=7/2

b: =x^2+x+1/4-13/4

=(x+1/2)^2-13/4>=-13/4

Dấu = xảy ra khi x=-1/2

e: =x^2-x+1/4+3/4=(x-1/2)^2+3/4>=3/4

Dấu = xảy ra khi x=1/2

f: x^2-4x+7

=x^2-4x+4+3

=(x-2)^2+3>=3

Dấu = xảy ra khi x=2

2:

a: A=2x^2+4x+9

=2x^2+4x+2+7

=2(x^2+2x+1)+7

=2(x+1)^2+7>=7

Dấu = xảy ra khi x=-1

b: x^2+2x+4

=x^2+2x+1+3

=(x+1)^2+3>=3

Dấu = xảy ra khi x=-1

\(x=\dfrac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)\)

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

\(y=\sqrt{\left(\sqrt{6}-1\right)^2}=\sqrt{6}-1\)

\(\Rightarrow x-y=1\Rightarrow P=1\)

\(B=x-2020-\sqrt{x-2020}+\dfrac{1}{4}+\dfrac{8079}{4}\)

\(B=\left(\sqrt{x-2020}-\dfrac{1}{2}\right)^2+\dfrac{8079}{4}\ge\dfrac{8079}{4}\)

\(B_{min}=\dfrac{8079}{4}\) khi \(x=\dfrac{8081}{4}\)

Lời giải:

$2T=2x-2\sqrt{x-1}-6\sqrt{x+7}+56$

$=[(x-1)-2\sqrt{x-1}+1]+[(x+7)-6\sqrt{x+7}+9]+40$

$=(\sqrt{x-1}-1)^2+(\sqrt{x+7}-3)^2+40\geq 40$

$\Rightarrow T\geq 20$

Vậy $T_{\min}=20$. Giá trị này đạt tại \(\left\{\begin{matrix} \sqrt{x-1}-1=0\\ \sqrt{x+7}-3=0\end{matrix}\right.\Leftrightarrow x=2\)

a: \(M-\dfrac{3}{2}=\dfrac{x+7}{\sqrt{x}+3}-\dfrac{3}{2}\)

\(=\dfrac{2x+14-3\sqrt{x}-9}{2\left(\sqrt{x}+3\right)}\)

\(=\dfrac{2x-3\sqrt{x}+5}{2\left(\sqrt{x}+3\right)}>0\)

=>M>3/2

b: \(M=\dfrac{x-9+16}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}\)

\(=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6>=2\cdot\sqrt{\dfrac{16}{\sqrt{x}+3}\cdot\left(\sqrt{x}+3\right)}-6=2\cdot4-6=2\)

Dấu = xảy ra khi (căn x+3)^2=16

=>căn x+3=4

=>x=1