để hàm số đồng biến thì \(k^2-2k-3>0\Leftrightarrow\left(k+1\right)\left(k-3\right)>0\Leftrightarrow\orbr{\begin{cases}k>3\\k< -1\end{cases}}\)

để hàm số nghịch biến thì \(k^2-2k-3< 0\Leftrightarrow\left(k+1\right)\left(k-3\right)< 0\Leftrightarrow-1< k< 3\)

Bạn vào link tham khảo :

https://hoidap247.com/cau-hoi/1226651

# Hok tốt !

\(x+y=1\Rightarrow\hept{\begin{cases}1-x=y\\1-y=x\end{cases}}\)

thay vào A ta được : \(A=\frac{1-y}{\sqrt{y}}+\frac{1-x}{\sqrt{x}}\)

\(\Rightarrow A=\frac{1}{\sqrt{y}}-\sqrt{y}+\frac{1}{\sqrt{x}}-\sqrt{x}\)

\(\Rightarrow A=\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)-\left(\sqrt{x}+\sqrt{y}\right)\)

áp dụng \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ta có : \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\ge\frac{4}{\sqrt{x}+\sqrt{y}}\)

áp dụng \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\) ta có : \(\left(\sqrt{x}+\sqrt{y}\right)^2\le2\left(\sqrt{x}^2+\sqrt{y}^2\right)=2\)

\(\Rightarrow\sqrt{x}+\sqrt{y}\le\sqrt{2}\)

\(\Rightarrow A\ge\sqrt{8}-\sqrt{2}=\sqrt{2}\)

dấu = xảy ra khi a=y=1/2

\(\sqrt{\frac{2ab}{a^2+b^2}}=\frac{\sqrt{2ab\left(a^2+b^2\right)}}{a^2+b^2}\)  mà \(a^2+b^2\ge2ab\) \(\Rightarrow\sqrt{\frac{2ab}{a^2+b^2}}\ge\frac{2ab}{a^2+b^2}\)

tương tự ta có : \(\sqrt{\frac{2bc}{c^2+b^2}}\ge\frac{2bc}{c^2+b^2}\) và \(\sqrt{\frac{2ac}{a^2+c^2}}\ge\frac{2ac}{c^2+a^2}\)

\(\Rightarrow VT\ge\frac{2ab}{a^2+b^2}+\frac{2bc}{b^2+c^2}+\frac{2ac}{a^2+c^2}\)

\(\Rightarrow VT+3\ge\frac{2ab}{a^2+b^2}+1+\frac{2bc}{b^2+c^2}+1+\frac{2ca}{c^2+a^2}+1\)

\(\Rightarrow VT+3\ge\frac{\left(a+b\right)^2}{a^2+b^2}+\frac{\left(b+c\right)^2}{b^2+c^2}+\frac{\left(c+a\right)^2}{c^2+a^2}\)

áp dụng \(\frac{x^2}{y}+\frac{m^2}{n}+\frac{p^2}{q}\ge\frac{\left(x+m+p\right)^2}{y+n+q}\) ta đc :

\(VT+3\ge\frac{\left(a+b+b+c+c+a\right)^2}{a^2+b^2+c^2+a^2+b^2+c^2}\)

\(\Rightarrow VT+3\ge\frac{4\left(a+b+c\right)^2}{2\left(a^2+b^2+c^2\right)}\)

\(\Rightarrow VT+3\ge\frac{2\left[a^2+b^2+c^2+2\left(ab+bc+ca\right)\right]}{a^2+b^2+c^2}\)

\(\Rightarrow VT+3\ge\frac{2\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}+\frac{4\left(ab+bc+ca\right)}{a^2+b^2+c^2}\)

\(\Rightarrow VT+3\ge\frac{4\left(ab+bc+ca\right)}{a^2+b^2+c^2}+2\)

\(\Rightarrow VT\ge\frac{4\left(ab+bc+ca\right)}{a^2+b^2+c^2}-1\left(đpcm\right)\)

dấu = xảy ra khi a=b=c > 0

vì x²+y²+z²=1 mà x²+y²+z²>=xy+yz+xz suy ra 1>= xy+yz+xz

x²+y²+z²=1 suy ra (x-y)²=1-2xy-z² ,(y-z)²=1-2yz-x²,(x-z)²=(x-z)²=1-2xz-y²

\(\sqrt{3}+\frac{1}{2\sqrt{3}}[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2]=\)

\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)(do (x-y)²=1-2xy-z²(y-z)²=1-2yz-x²,(x-z)²=(x-z)²=1-2xz-y²)

theo bdt cosi ta có:

\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)

\(\le\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2z\sqrt{2xy}+2y\sqrt{2xz}+2x\sqrt{2yz}\right)]\)

\(\le\sqrt{3}+\frac{1}{2\sqrt{3}}[3-3\sqrt[3]{\left(2z\sqrt{2xy}.2y\sqrt{2xz}.2x\sqrt{2yz}\right)}\)

\(=\sqrt{3}+\frac{\sqrt{3}}{2}[1-2\sqrt{2}.\sqrt[3]{xyz^2}]\)\(=\sqrt{3}\left(1+\frac{1}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)=\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\)

suy ra 

\(\frac{x+y+z}{xy+yz+xz}\ge3.\sqrt[3]{xyz}\left(doxy+yz+xz\le1\right)\)

ta giả sử:

\(3\sqrt[3]{xyz}\ge\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\Leftrightarrow\sqrt{3}\ge\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\) mà \(\sqrt{3}>\frac{3}{2}\)

suy ra \(\frac{3}{2}\ge\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\)(luôn đúng) suy ra điều giả sử trên là đúng

hay \(3\sqrt[3]{xyz}\ge\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\)

mà \(\frac{x+y+z}{xy+yz+xz}\ge3.\sqrt[3]{xyz}\),\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)\(\le\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\)

suy ra \(\frac{x+y+z}{xy+yz+xz}\ge\)\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)

suy ra \(\frac{x+y+z}{xy+yz+xz}\ge\)\(\sqrt{3}+\frac{1}{2\sqrt{3}}[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2]\)(đpcm)

em mới có lớp 8, nếu em làm sai cho em xin lỗi nha anh

bạn ơi đk: 1 trong 3 số x,y,z là >=0 còn lại là >0 thì nó vẫn ra điều trên

dòng suy ra cuối cùng mình ghi lộn, phải là -8+2021 = 2013 mới đúng :v

\(a^2+\frac{8}{a^2}+\frac{b^2}{8}=8\)

\(\Leftrightarrow8a^4+64+a^2b^2=64a^2\)

\(\Leftrightarrow a^2b^2=64a^2-8a^4-64\)

\(\Leftrightarrow a^2b^2=-8\left(a^4-8a^2+8\right)\)

\(\Leftrightarrow a^2b^2=-8\left[\left(a^2-4\right)^2-8\right]\)

\(\Leftrightarrow a^2b^2=-8\left(a^2-4\right)^2+64\le64\)

\(\Leftrightarrow-8\le ab\le8\)

\(\Rightarrow A\ge-8\sqrt{6}+2021\)

dấu = xảy ra \(\Leftrightarrow\hept{\begin{cases}ab=-8\\a^2-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}b=\pm4\\a=\pm2\end{cases}}\)

a, ĐK : x >= 0 

 \(x+3\sqrt{x}=\sqrt{x}\left(\sqrt{x}+3\right)\)

b, ĐK : x >= 0 

 \(x-9\sqrt{x}=\sqrt{x}\left(\sqrt{x}-9\right)\)

c,ĐK : x >= 0 

 \(-2x+10\sqrt{x}=-2\sqrt{x}\left(\sqrt{x}-5\right)\)

d, ĐK : x >= 0 

\(-5x+15\sqrt{x}=-5\sqrt{x}\left(\sqrt{x}-3\right)\)