Xét \(2A=2\sqrt{x-2}+4\sqrt{x+1}+4038-2x\)     (Đk:\(x\ge2\))

     \(2A=-\left[\left(x-2\right)-2\sqrt{x-2}+1\right]-\left[\left(x+1\right)-4\sqrt{x+1}+2\right]+4042\)

   \(2A=-\left(\sqrt{x-2}-1\right)^2-\left(\sqrt{x+1}-2\right)^2+4042\le4042\)

\(\Leftrightarrow A\le2021\)

\(\Rightarrow Amax=2021\) khi x=3   (tm)Tự đăng câu hỏi xong tự trả lời (T-T)         

Đặt A=\(\dfrac{\sqrt{x}-1}{x+\sqrt{x}+2}\)\(\Rightarrow Ax+A\sqrt{x}+2A-\sqrt{x}+1=0\)

\(\Leftrightarrow Ax+\sqrt{x}\left(A-1\right)+2A+1=0\)

\(\Delta=\left(A-1\right)^2-4A\left(2A+1\right)=A^2-2A+1-8A^2-4A\)\(=-7A^2-6A+1\ge0\)

\(\Rightarrow-1\le A\le\dfrac{1}{7}\)

Vậy Max A là \(\dfrac{1}{7}\)

Dâu"=" xảy ra \(\Leftrightarrow A=\dfrac{1}{7}\)

\(\Leftrightarrow7\sqrt{x}-7=x+\sqrt{x}+2\)

\(\Leftrightarrow x-6\sqrt{x}+9=0\)

\(\Leftrightarrow\left(\sqrt{x}-3\right)^2=0\Leftrightarrow x=9\)

1.

\(2P=2\sqrt{x-2}+4\sqrt{x+1}-2x+4016\)

\(=-\left(x-2-2\sqrt{x-2}+1\right)-\left(x+1-4\sqrt{x+1}+4\right)+4020\)

\(=-\left(\sqrt{x-2}-1\right)^2-\left(\sqrt{x+1}-2\right)^2+4020\)

2.

\(\sqrt{u}+\sqrt{v}=7\Rightarrow u+v+2\sqrt{uv}=49\)

\(\Rightarrow u+v+2\sqrt{6}=49\Rightarrow u+v=49-2\sqrt{6}\)

\(\Rightarrow\left|u-v\right|=\sqrt{\left(u-v\right)^2}=\sqrt{\left(u+v\right)^2-4uv}=\sqrt{\left(49-2\sqrt{6}\right)^2-4.6}=...\)

3.

\(\left(a-2\right)^2+\left(b-1\right)^2=545\)

\(P=23\left(a-2\right)+4\left(b-1\right)+2063\)

\(\Rightarrow\left(P-2063\right)^2=\left[23\left(a-2\right)+4\left(b-1\right)\right]^2\le\left(23^2+4^2\right)\left[\left(a-2\right)^2+\left(b-1\right)^2\right]\)

lm tiếp hộ e câu 3 với

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b: Ta có: \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

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

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

c: Ta có: \(x+\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ

\(\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}>0\forall x\)

Lời giải:

ĐK: $x\geq 0; x\neq 1$

\(A=\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}.\frac{(\sqrt{x}-1)^2}{2}-\frac{\sqrt{x}+2}{(\sqrt{x}-1)^2}.\frac{(\sqrt{x}-1)^2}{2}\)

\(=\frac{(\sqrt{x}-2)(\sqrt{x}-1)}{2(\sqrt{x}+1)}-\frac{\sqrt{x}+2}{2}=\frac{(\sqrt{x}-2)(\sqrt{x}-1)-(\sqrt{x}+2)(\sqrt{x}+1)}{2(\sqrt{x}+1)}=\frac{-6\sqrt{x}}{2(\sqrt{x}+2)}=\frac{-3\sqrt{x}}{\sqrt{x}+2}\)

Vì $x\geq 0$ nên $3\sqrt{x}\geq 0; \sqrt{x}+2>0$

$\Rightarrow \frac{3\sqrt{x}}{\sqrt{x}+2}\geq 0$

$\Rightarrow A\leq 0$ hay $A_{\max}=0$ khi $x=0$

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

Do \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\2\sqrt{x}+1>0\end{matrix}\right.\) \(\Rightarrow\dfrac{\sqrt{x}}{2\sqrt{x}+1}\ge0\)

\(\Rightarrow A\le1\)

\(A_{max}=1\) khi \(x=0\)