a: Thay \(x=3+2\sqrt{2}\) vào A, ta được:

\(A=\dfrac{3+2\sqrt{2}-\sqrt{2}-1+2}{\sqrt{2}+1+3}=\dfrac{4+\sqrt{2}}{4+\sqrt{2}}=1\)

\(b,B=\dfrac{x-4+2\sqrt{x}+6-3\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ B=\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\\ c,M=B:A=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}+3}{x-\sqrt{x}+2}=\dfrac{\sqrt{x}+1}{x-\sqrt{x}+2}\\ M=\dfrac{x-\sqrt{x}+2-x+2\sqrt{x}-1}{x-\sqrt{x}+2}\\ M=1-\dfrac{x-2\sqrt{x}+1}{x-\sqrt{x}+2}=1-\dfrac{\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+2}\)

Ta có \(\left(\sqrt{x}-1\right)^2\ge0;x-\sqrt{x}+2=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\)

Do đó \(\dfrac{\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+2}\ge0\)

\(\Leftrightarrow M=1-\dfrac{\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+2}\le1-0=1\)

Vậy \(M_{max}=1\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(tm\right)\)

a: Xét ΔABE và ΔACD có

AB=AC

\(\widehat{BAE}\) chung

AE=AD

Do đó: ΔABE=ΔACD

b: Ta có: ΔABE=ΔACD

nên BE=CD

c: Xét ΔDBC và ΔECB có 

DB=EC

DC=EB

BC chung

Do đó: ΔDBC=ΔECB

Suy ra: \(\widehat{KCB}=\widehat{KBC}\)

hay ΔKBC cân tại K

d: Xét ΔABK và ΔACK có 

AB=AC

BK=CK

AK chung

Do đó: ΔABK=ΔACK

Suy ra: \(\widehat{BAK}=\widehat{CAK}\)

hay AK là tia phân giác của góc BAC

ĐKXĐ: \(x\ge1\)

Đặt \(\left\{{}\begin{matrix}\sqrt[]{x-1}=a\ge0\\\sqrt[3]{2-x}=b\end{matrix}\right.\) \(\Rightarrow a^2+b^3=1\)

Ta được hệ: 

\(\left\{{}\begin{matrix}a+b=1\\a^2+b^3=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=1-a\\a^2+b^3=1\end{matrix}\right.\)

\(\Rightarrow a^2+\left(1-a\right)^3=1\)

\(\Leftrightarrow a^3-4a^2+3a=0\)

\(\Leftrightarrow a\left(a-1\right)\left(a-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt[]{x-1}=0\\\sqrt[]{x-1}=1\\\sqrt[]{x-1}=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=10\end{matrix}\right.\)

Để \(B\subset A\) \(\Leftrightarrow m-2\le-1\Leftrightarrow m\le1\)

Để \(B\subset A\) \(\Leftrightarrow4\le m+1\) \(\Leftrightarrow m\ge3\)