\(Q=\sqrt{x+3}+\sqrt{10-x}\)

\(\Leftrightarrow Q^2=\left(\sqrt{x+3}+\sqrt{10-x}\right)^2\le\left(1^2+1^2\right)\left[\left(\sqrt{x+3}\right)^2+\left(\sqrt{10-x}\right)^2\right]\)

\(\Leftrightarrow Q^2\le2\left(x+3+10-x\right)=2.13=26\)

\(\Leftrightarrow Q\le\sqrt{26}\)

\(maxQ=\sqrt{26}\Leftrightarrow x+3=10-x\Leftrightarrow x=\dfrac{7}{2}\)

Áp dụng BĐT Bunhiacopski:

\(Q=\sqrt{x+3}+\sqrt{10-x}\\ \Leftrightarrow Q^2=\left(\sqrt{x+3}+\sqrt{10-x}\right)^2\le\left(1^2+1^2\right)\left(x+3+10-x\right)=2\cdot13=26\\ \Leftrightarrow Q\le\sqrt{26}\\ Q_{max}=\sqrt{26}\Leftrightarrow x+3=10-x\Leftrightarrow x=\dfrac{7}{2}\)

\(\left\{{}\begin{matrix}x\ge3\\E=\sqrt{x-3}+x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\E=x-3+3+\sqrt{x-3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\E=\left(\sqrt{x-3}+\dfrac{1}{4}\right)^2+\dfrac{11}{4}\end{matrix}\right.\)

\(\sqrt{x-3}\ge0\Rightarrow\sqrt{x-3}+\dfrac{1}{4}\ge\dfrac{1}{4}\Rightarrow\left(\sqrt{x-3}+\dfrac{1}{4}\right)^2\ge\dfrac{1}{4}\Rightarrow\left(\sqrt{x-3}+\dfrac{1}{4}\right)^2+\dfrac{11}{4}\ge\dfrac{1}{4}+\dfrac{11}{4}=3\)Đẳng thức khi x =3

GTNNE =3 khi x =3

\(x\ge3\)

\(E=x-3+\sqrt{x-3}+\dfrac{1}{4}+\dfrac{11}{4}\)

\(E=\left(\sqrt{x-3}+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)

\(\sqrt{x-3}\ge0\Rightarrow E\ge\left(\dfrac{1}{2}\right)^2+\dfrac{11}{4}=\dfrac{12}{4}=3\)

GTNN E =3 khi x= 3

Chắc dùng Mincowski

`A=(1/(x-sqrtx)+1/(sqrtx-1)):(sqrtx+1)/(sqrtx-1)^2`

`=((sqrtx+1)/(x-sqrtx)).(sqrtx-1)^2/(sqrtx+1)`

`=(sqrtx-1)^2/(x-sqrtx)`

`=(sqrtx-1)/sqrtx`

Đặt \(A=\dfrac{2}{\sqrt{x}+3}\)

Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\Rightarrow\dfrac{2}{\sqrt{x}+3}\le\dfrac{2}{3}\)

\(\Rightarrow A_{max}=\dfrac{2}{3}\) khi \(x=0\) 

GTLN ak. bạn có nhầm đề k vậy, bạn xem lại đề đi.

mình k ak

bạn giúp mình phân tích cái kia ra là đc

k tìm dc GTLN nhà bạn, bạn xem lại đề đi