sửa đề, 2 nghiệm phân biệt nhé 

Để pt có 2 nghiệm pb thì \(\Delta>0\)

\(\Delta=16-4\left(-m^2-5\right)=16+4m^2+20=4m^2+36>0\forall m\)

Theo Vi et : \(\hept{\begin{cases}x_1+x_2=-\frac{b}{a}=-4\\x_1x_2=\frac{c}{a}=-m^2-5\end{cases}}\)

mà \(\left(x_1+x_2\right)^2=16\Rightarrow x_1^2+x_2^2=16-2\left(-m^2-5\right)=2m^2+26\)

bình phương 2 hệ thức có dạng \(\left(x_1-x_2\right)^2=16\Rightarrow x_1^2+x_2^2-2x_1x_2=16\)

\(\Leftrightarrow2m^2+26-2\left(-m^2-5\right)=16\)

\(\Leftrightarrow4m^2+36=16\Leftrightarrow4m^2=-20\Leftrightarrow m^2=-5\)vô lí 

\(x-2\sqrt{x-2}-5=0\)

\(\Leftrightarrow x-2-2\sqrt{x-2}+1-4=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2-4=0\)

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

\(\Leftrightarrow\left(\sqrt{x-2}-3\right)\left(\sqrt{x-2}+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-2}-3=0\\\sqrt{x-2}+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=11\\x\in\theta\end{cases}}}\)

thank ~

Đặt \(x_1=x;x_2=y\)

\(\frac{x}{y^2}+\frac{y}{x^2}=m-1\Leftrightarrow\frac{x^3+y^3}{x^2y^2}=m-1\)

\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)=\left(m-1\right)\left(xy\right)^2\)

thay vi et vô thooi :)) 

Ta có: \(\sqrt{x^2-2x}=\sqrt{x^2-2x}.1\le\frac{x^2-2x+1}{2}\)

\(\Rightarrow\frac{x-1}{\sqrt{x^2-2x}}\ge\frac{2\left(x-1\right)}{\left(x-1\right)^2}=\frac{2}{x-1}\)

\(A=x+\frac{x-1}{\sqrt{x^2-2x}}\ge x-1+\frac{2}{x-1}+1\ge2\sqrt{\left(x-1\right).\frac{2}{x-1}}+1=1+2\sqrt{2}\)

Dấu \(=\)xảy ra khi \(\hept{\begin{cases}x^2-2x=1\\x-1=\frac{2}{x-1}\\x>2\end{cases}}\Leftrightarrow x=1+\sqrt{2}\).

\(\left(\sqrt{x}+\sqrt{y}\right)^2-\sqrt{3}\left(\sqrt{x}+\sqrt{y}\right)-\sqrt{xy}\)

\(=x+2\sqrt{xy}+y-\sqrt{3x}-\sqrt{3y}-\sqrt{xy}\)

\(=\left(\sqrt{x}-\sqrt{y}\right)^2-\sqrt{3x}-\sqrt{3y}+3\sqrt{xy}\)

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

\(=\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{3x}-1\right)\left(\sqrt{3y}-1\right)-1\)\(\ge-1\forall x,y\)

Dấu "=" xảy ra <=> x=y=1/3

\(\sqrt{xy}+\sqrt{x}+\sqrt{y}\ge3\)

ÁP DỤNG BĐT COSI
\(\sqrt{xy}+\sqrt{x}+\sqrt{y}\le\frac{x+y}{2}+\frac{x+1}{2}+\frac{y+1}{2}=x+y+1\ge3=>x+y\ge2\)

\(P\ge\frac{\left(x+y\right)^2}{x+y}=2\left(cosi\right)\) vậy min P=2 <=> x=y=1

                      Bài làm :

Ta có :

\(\left(\sqrt{x}+1\right)\left(\sqrt{y}+1\right)\ge4\)

\(\Leftrightarrow\sqrt{xy}+\sqrt{y}+\sqrt{x}+1\ge4\)

\(\Leftrightarrow\sqrt{xy}+\sqrt{x}+\sqrt{y}\ge3\)

Áp dụng BĐT cosi cho các số không âm ; ta được :

\(3\le\sqrt{xy}+\sqrt{x}+\sqrt{y}\le\frac{x+y}{2}+\frac{x+1}{2}+\frac{y+1}{2}=x+y+1\)

\(\Rightarrow x+y\ge2\)

Ta có :

\(P=\frac{x^2}{y}+\frac{y^2}{x}\ge\frac{\left(x+y\right)^2}{x+y}=x+y\)

\(\Rightarrow P\ge2\)

Dấu "=" xảy ra khi x=y=1

Vậy MinP = 2 <=> x=y=1

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

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

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

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

- \(x+y=1\Rightarrow0\le x,y\le1\).

\(P=\frac{1+\sqrt{x}}{2+\sqrt{y}}+\frac{2+\sqrt{x}}{1+\sqrt{y}}\ge\frac{1}{2+\sqrt{y}}+\frac{2}{1+\sqrt{y}}\ge\frac{1}{2+1}+\frac{2}{1+1}=\frac{4}{3}\)

Dấu \(=\)xảy ra tại \(x=0,y=1\).

\(P=\frac{1+\sqrt{x}}{2+\sqrt{y}}+\frac{2+\sqrt{x}}{1+\sqrt{y}}\le\frac{1+\sqrt{x}}{2}+\frac{2+\sqrt{x}}{1}\le\frac{1+1}{2}+\frac{2+1}{1}=4\)

Dấu \(=\)xảy ra tại \(x=1,y=0\).

trình bày đầy đủ :

Ta có BĐT sau: \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)( x,y >0 )

CM: \(\Leftrightarrow\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\)

Áp dụng bđt cô si cho 2 số dương x,y ta có:

\(x+y\ge2\sqrt{xy}\)

\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\)

\(\Rightarrow\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\)( đúng )

Áp dụng bđt trên ta có: 

\(P=\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\ge\frac{4}{2\sqrt{2}}=\sqrt{2}\)

Dấu "=" xảy ra <=> \(a=b=\sqrt{2}\)

Vậy MIN P= \(\sqrt{2}\)\(a=b=\sqrt{2}\)

\(bđtcosi\)

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\ge\frac{4}{2\sqrt{2}}=\sqrt{2}\)

Dấu = xảy ra <=> a=b=\(\sqrt{2}\)

Min P=\(\sqrt{2}\)<=>a=b=\(\sqrt{2}\)