Áp dụng bđt Cô-si: 

\(2.1.\sqrt{1-x}+x\le2.\dfrac{1+1-x}{2}+x=2\)

Dấu "=" xảy ra khi và chỉ khi \(\sqrt{1-x}=1\) <=> x = 0

\(2.1.\sqrt{1-x}+x\le1+1-x+x=2\)

Dấu "=" xảy ra khi \(1=1-x\Rightarrow x=0\)

\(x+\dfrac{16}{x-1}\\ =x-1+\dfrac{16}{x-1}+1\)

Áp dụng BĐT Cô-si ta có:
\(x-1+\dfrac{16}{x-1}+1\\ \ge2\sqrt{\left(x-1\right).\dfrac{16}{x-1}}+1\\ =2\sqrt{16}+1\\ =9\)

Dấu "=" xảy ra

 \(\Leftrightarrow x-1=\dfrac{16}{x-1}\\ \Leftrightarrow\left(x-1\right)^2=16\\ \Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

xin lỗi bn mik mới học lớp 6 thôi

Bài 1:

\(P=x\sqrt{3-x^2}=\sqrt{x^2}\cdot\sqrt{3-x^2}\)

\(=\sqrt{x^2\left(3-x^2\right)}\)\(\le\frac{x^2+3-x^2}{2}=\frac{3}{2}\)

Dấu = khi \(x=\sqrt{\frac{3}{2}}\)

Vậy Max_P=\(\frac{3}{2}\Leftrightarrow x=\sqrt{\frac{3}{2}}\)

\(y=\frac{5x}{x^2+4}\le\frac{5x}{2\sqrt{x^2.4}}=\frac{5}{4}\)

Dấu "=" xảy ra khi \(x=2\)

\(y=\frac{x^2}{\left(x^2+\frac{3}{2}+\frac{3}{2}\right)^3}\le\frac{x^2}{\left(3\sqrt[3]{x^2.\frac{3}{2}.\frac{3}{2}}\right)^3}=\frac{4x^2}{243x^2}=\frac{4}{243}\)

Dấu "=" xảy ra khi \(x=\frac{\sqrt{6}}{2}\)

Đặt \(\sqrt{1+a^2}+\sqrt{1-a^2}=x\Rightarrow\sqrt{2}\le x\le2\)

\(x^2=2+2\sqrt{1-a^4}\Rightarrow\sqrt{1-a^4}=\dfrac{x^2-2}{2}\)

\(\Rightarrow\dfrac{x^2-2}{2}+\left(b+1\right)x+b-4\le0\)

\(\Rightarrow x^2+2\left(b+1\right)x+2b-10\le0\)

\(\Rightarrow x^2+2x-10\le-2b\left(x+1\right)\)

\(\Rightarrow-2b\ge\dfrac{x^2+2x-10}{x+1}\)

\(\Rightarrow-2b\ge\max\limits_{\left[\sqrt{2};2\right]}f\left(x\right)\) với \(f\left(x\right)=\dfrac{x^2+2x-10}{x+1}\)

Xét trên \(\left[\sqrt{2};2\right]\) ta có:

\(f\left(x\right)=\dfrac{3x^2+6x-30}{3\left(x+1\right)}=\dfrac{3x^2+8x-28-2\left(x+1\right)}{3\left(x+1\right)}=\dfrac{\left(3x+14\right)\left(x-2\right)}{3\left(x+1\right)}-\dfrac{2}{3}\le-\dfrac{2}{3}\)

\(\Rightarrow-2b\ge-\dfrac{2}{3}\Rightarrow b\le\dfrac{1}{3}\)

Vậy \(b_{max}=\dfrac{1}{3}\)