Áp dụng BĐT Bunhia- cốp -xki ta có

\(M=\left(\sqrt{a}+\sqrt{b}\right)^2\le\left(1^2+1^2\right)\left(a+b\right)\le2\)

Vậy maxM =2 \(\Leftrightarrow a=b=\frac{1}{2}\)

a) ĐKXĐ: \(x>0\)

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

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

\(A=x-\sqrt{x}=2\)

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

\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)(do \(\sqrt{x}+1\ge1>0\))

b) \(A=x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)>0\)(do \(x>1\))

\(\Leftrightarrow A=x-\sqrt{x}=\left|A\right|\)

c) \(A=x-\sqrt{x}=\left(x-\sqrt{x}+\dfrac{1}{4}\right)-\dfrac{1}{4}\)

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

\(minA=-\dfrac{1}{4}\Leftrightarrow\sqrt[]{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)

\(a,A=\dfrac{x\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1\left(x>0\right)\\ A=\dfrac{x\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-2\sqrt{x}-1+1\\ A=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\\ A=2\Leftrightarrow x-\sqrt{x}-2=0\\ \Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}=2\left(\sqrt{x}>0\right)\\ \Leftrightarrow x=4\left(tm\right)\)

\(b,x>1\Leftrightarrow\sqrt{x}-1>0\\ \Leftrightarrow\left|A\right|=\left|x-\sqrt{x}\right|=\left|\sqrt{x}\left(\sqrt{x}-1\right)\right|=\sqrt{x}\left(\sqrt{x}-1\right)=A\left(\sqrt{x}>0\right)\)

\(c,A=x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\\ A_{min}=-\dfrac{1}{4}\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)

\(\sqrt{2a^2+ab+2b^2}=\sqrt{\frac{5}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}\ge\frac{5}{4}\left(a+b\right)\)

Tương tự cộng vế theo vế thì 

\(M\ge\frac{5}{4}\left(2a+2b+2c\right)=\frac{5}{2}\left(a+b+c\right)=\frac{5}{2}\cdot2019\)

Dấu "=" xảy ra tại \(a=b=c=\frac{2019}{3}\)

bài 4 có trên mạng nha chị.tí e làm cách khác

bài 5 chị tham khảo bđt min cop ski r dùng svác là ra ạ.giờ e coi đá bóng,coi xong nghĩ tiếp ạ.

e nhầm đoạn này r

\(\sqrt{2a^2+ab+2b^2}\ge\frac{\sqrt{5}}{2}\left(a+b\right)\) rồi cộng lại thì 

\(M\ge\frac{\sqrt{5}}{2}\left(2a+2b+2c\right)=\sqrt{5}\cdot2019\) ạ

Chắc lần này sẽ không nhầm nhưng hướng là thế ạ.

Bài 5 cần gì dùng Mincopxki chi cho mệt nhỉ?

\(\left(x^2+\frac{1}{x^2}\right)\left[2^2+\left(\frac{1}{2}\right)^2\right]\ge\left(2x+\frac{1}{2x}\right)^2\)

Do đó: \(\sqrt{x^2+\frac{1}{x^2}}\ge\frac{2x+\frac{1}{2x}}{\sqrt{2^2+\frac{1}{2^2}}}=\frac{4x+\frac{1}{x}}{\sqrt{17}}\)

Tương tự rồi cộng lại rồi dùng Cauchy-Schwarz

Lời giải:
TXĐ: $[-1;1]$

$y'=\frac{1}{2\sqrt{x+1}}-\frac{1}{2\sqrt{1-x}}+\frac{x}{2}$

$y'=0\Leftrightarrow x=0$

$f(0)=2$;

$f(1)=f(-1)=\sqrt{2}+\frac{1}{4}$
Vậy $f_{\min}=2; f_{\max}=\frac{1}{4}+\sqrt{2}$

a . ta có : \(1\le1+\sqrt{2-x}\Rightarrow GTNN=1\)

\(-2\le\sqrt{x-3}-2\Rightarrow GTNN=-2\)

b. \(0\le\sqrt{4-x^2}\le2\)

\(\sqrt{2x^2-x+3}=\sqrt{2\left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\frac{23}{8}}=\sqrt{2\left(x-\frac{1}{4}\right)^2+\frac{23}{8}}\ge\frac{\sqrt{46}}{4}\)

vậy \(GTNN=\frac{\sqrt{46}}{4}\)

ta có : \(0\le-x^2+2x+5=-\left(x-1\right)^2+6\le6\)

\(\Rightarrow1-\sqrt{6}\le1-\sqrt{-x^2+2x+5}\le1\)Vậy \(\hept{\begin{cases}GTNN=1-\sqrt{6}\\GTLN=1\end{cases}}\)

\(A^2=\left(\sqrt{1-x}+\sqrt{1+x}\right)^2\le\left(1^2+1^2\right)\left(1-x+1+x\right)=4\\ \Leftrightarrow A\le2\\ A_{max}=2\Leftrightarrow1-x=1+x\Leftrightarrow x=0\\ A^2=2+2\sqrt{1-x^2}\ge2\\ \Leftrightarrow A\ge\sqrt{2}\\ A_{min}=\sqrt{2}\Leftrightarrow1-x^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy \(\sqrt{2}\le A\le2\)

Theo đề bài, ta có:

\(x^3+y^3=x^2-xy+y^2\)

hay \(\left(x^2-xy+y^2\right)\left(x+y-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2-xy+y^2=0\\x+y=1\end{cases}}\)

+ Với \(x^2-xy+y^2=0\Rightarrow x=y=0\Rightarrow P=\frac{5}{2}\)

+ với \(x+y=1\Rightarrow0\le x,y\le1\Rightarrow P\le\frac{1+\sqrt{1}}{2+\sqrt{0}}+\frac{2+\sqrt{1}}{1+\sqrt{0}}=4\)

Dấu đẳng thức xảy ra <=> x=1;y=0 và \(P\ge\frac{1+\sqrt{0}}{2+\sqrt{1}}+\frac{2+\sqrt{0}}{1+\sqrt{1}}=\frac{4}{3}\)

Dấu đẳng thức xảy ra <=> x=0;y=1

Vậy max P=4 và min P =4/3