\(x=2018-2\sqrt{2018}+1=\left(\sqrt{2018}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{2018}-1\)

\(\Rightarrow P=\frac{\sqrt{2018}-1}{\sqrt{2018}-1+1}=\frac{\sqrt{2018}-1}{\sqrt{2018}}=\frac{2018-\sqrt{2018}}{2018}\)

Câu 1:

Áp dụng BĐT Cô-si:

\(A=\sqrt{\left(2-x\right)\left(2+x\right)}\le\frac{2-x+2+x}{2}=2\)

Dấu "=" xảy ra \(\Leftrightarrow2-x=2+x\Leftrightarrow x=0\)

Câu 2:

\(B=\sqrt{-x^2+x+\frac{1}{4}}\)

\(B=\sqrt{-\left(x^2-x-\frac{1}{4}\right)}\)

\(B=\sqrt{-\left(x^2-x+\frac{1}{4}-\frac{1}{2}\right)}\)

\(B=\sqrt{-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\right]}\)

\(B=\sqrt{\frac{1}{2}-\left(x-\frac{1}{2}\right)^2}\le\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2}\)

\(a\text{) }x^2+\sqrt{x+2019}=2019\left(x\ge-2019\right)\\ \Leftrightarrow x^2+x+\frac{1}{4}=\left(x+2019\right)-\sqrt{x+2019}+\frac{1}{4}\\ \Leftrightarrow\left(x+\frac{1}{2}\right)^2=\left(\sqrt{x+2019}-\frac{1}{2}\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\sqrt{x+2019}-\frac{1}{2}\left(1\right)\\x+\frac{1}{2}=\frac{1}{2}-\sqrt{x+2019}\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x+\frac{1}{4}=\sqrt{x+2019}\\ ĐK:x\ge-\frac{1}{4}\\ \Leftrightarrow\left(x+\frac{1}{4}\right)^2=x+2019\\ \Leftrightarrow\left(x+\frac{1}{4}\right)^2=x+2019\\ \Leftrightarrow x^2+\frac{1}{2}x+\frac{1}{16}-x-2019=0\\ \Leftrightarrow x^2-\frac{1}{2}x+\frac{1}{16}-2019=0\\ \Leftrightarrow\left(x-\frac{1}{4}\right)^2-2019=0\\ \Leftrightarrow\left(x-\frac{1}{4}-\sqrt{2019}\right)\left(x-\frac{1}{4}+\sqrt{2019}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{4}-\sqrt{2019}=0\\x-\frac{1}{4}+\sqrt{2019}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4\sqrt{2019}+1}{4}\left(T/m\right)\\x=\frac{-\sqrt{2019}+1}{4}\left(K^o\text{ }T/m\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow x=-\sqrt{x+2019}\\ ĐK:-2019\le x\le0\\ \Leftrightarrow x^2=x+2019\\ \Leftrightarrow x^2-x-2019=0\\ \Leftrightarrow x^2-x-2019=0\\ \Leftrightarrow.....\)

\(b\text{) }x+\sqrt{2-x^2}+x\sqrt{2-x^2}=3\)

\(Đặt\text{ }\sqrt{2-x^2}=y\)

\(\Rightarrow\left\{{}\begin{matrix}x+y+xy=3\\x^2+y^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+2y+2xy=6\\2x^2+2y^2=4\end{matrix}\right.\\\Leftrightarrow2x^2+2y^2-\left(2x+2y+2xy\right)=-2\\ \Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-2y+1\right)+\left(x^2-2xy+y^2\right)=0\\ \Leftrightarrow\left(x-1\right)^2+\left(y-1\right)^2+\left(x-y\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-1=0\\x-y=0\end{matrix}\right.\Leftrightarrow x=1\)

Vậy....

a) Đặt $\sqrt{x+1}=a; \sqrt{9-x}=b$ thì bài toán trở thành:

Tìm max, min của $f(a,b)=a+b$ với $a,b\geq 0$ và $a^2+b^2=10$Ta có:

$f^2(a,b)=(a+b)^2=a^2+b^2+2ab=10+2ab\geq 10$ do $ab\geq 0$

$\Rightarrow f(a,b)\geq \sqrt{10}$ hay $f_{\min}=\sqrt{10}$

Mặt khác: $f^2(a,b)=(a+b)^2\leq 2(a^2+b^2)=20$ (theo BĐT AM-GM)

$\Rightarrow f(a,b)\leq \sqrt{20}=2\sqrt{5}$ hay $f_{\max}=2\sqrt{5}$

b) 

Đặt $\sqrt{x}=a; \sqrt{2-x}=b$ thì bài toán trở thành:

Tìm max, min của $f(a,b)=a+b+ab$ với $a,b\geq 0$ và $a^2+b^2=2$. Ta có:

$f(a,b)=\sqrt{(a+b)^2}+ab=\sqrt{a^2+b^2+2ab}+ab=\sqrt{2+2ab}+ab\geq \sqrt{2}$ do $ab\geq 0$

Vậy $f_{\min}=\sqrt{2}$

Lại có, theo BĐT AM-GM:

$f(a,b)=\sqrt{2+2ab}+ab\leq \sqrt{2+a^2+b^2}+\frac{a^2+b^2}{2}=\sqrt{2+2}+\frac{2}{2}=3$

Vậy $f_{\max}=3$

c) Đặt $\sqrt{8-x^2}=a$ thì bài toán trở thành tìm max, min của:

$f(x,a)=x+a+ax$ với $x,a\geq 0$ và $x^2+a^2=8$. Bài này chuyển về y hệt  như phần b. 

$f_{\min}=2\sqrt{2}$

$f_{\max}=8$

d) Tương tự:

$f_{\min}=2$ khi $x=\pm 2$

$f_{\max}=2+2\sqrt{2}$ khi $x=0$

ta có: \(2P=2x^2-2x\sqrt{y}+2x+2y-2\sqrt{y}+2\)

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

\(2P=\left(x-\sqrt{y}\right)^2+\left(x+1\right)^2+\left(\sqrt{y}-1\right)^2\ge0\forall x,y\)

\(\Rightarrow P\ge0\forall x,y\)

dấu = xảy ra khi \(\left\{\begin{matrix}x=\sqrt{y}\\x=-1\\\sqrt{y}=1\end{matrix}\right.\)(có gì đó sai sai)

chờ Em hai mươi năm :v

Lời giải:

Xét số hạng tổng quát:

\(\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n+1})}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}\)

\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Do đó:

\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2019}}-\frac{1}{\sqrt{2020}}\)

\(=1-\frac{1}{\sqrt{2020}}\)