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 ~ 

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

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

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

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

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

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

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

a: \(B=\dfrac{\sqrt{x}}{x+\sqrt{x}}:\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)

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

\(=\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b: B=2/7

=>\(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{2}{7}\)

=>\(2\left(x+\sqrt{x}+1\right)=7\sqrt{x}\)

=>\(2x+2\sqrt{x}-7\sqrt{x}+2=0\)

=>\(2x-5\sqrt{x}+2=0\)

=>\(\left(2\sqrt{x}-1\right)\cdot\left(\sqrt{x}-2\right)=0\)

=>\(\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)

\(A=\sqrt{\left(x+\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{7}}{2}\right)^2}+\sqrt{\left(\dfrac{1}{2}-x\right)^2+\left(\dfrac{\sqrt{7}}{2}\right)^2}\)

\(A\ge\sqrt{\left(x+\dfrac{1}{2}+\dfrac{1}{2}-x\right)^2+\left(\sqrt{7}\right)^2}=2\sqrt{2}\)

\(A_{min}=2\sqrt{2}\) khi \(x+\dfrac{1}{2}=\dfrac{1}{2}-x\Leftrightarrow x=0\)

Bạn cũng có thể bình phương A lên