Ta có : \(x^2-2\left(m-1\right)x+m^2+m+1=0\left(a=1;b=-2m+2;c=m^2+m+1\right)\)

\(\Delta=\left(-2m+2\right)^2-4\left(m^2+m+1\right)=4m^2+4-4m^2-4m-4=-4m< 0\)

Nếu \(-4m< 0\Leftrightarrow m>0\) chắc ĐK là vậy.

Theo hệ thức Vi et ta có : \(x_1+x_2=2m+2;x_1x_2=m^2+m+1\)

Theo bài ra ta có : \(x_1^2+x_2^2=4x_1x_2-2\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=4x_1x_2-2\) Thay vao ta có pt mới : 

\(\Leftrightarrow\left(2m+2\right)^2-4\left(m^2+m+1\right)=4\left(m^2+m+1\right)-2\)

\(\Leftrightarrow4m+4-4m^2-m-1=4m^2+4m+4-2\)

\(\Leftrightarrow3m+3-4m^2=4m^2+4m+2\)

\(\Leftrightarrow-m+1-8m^2=0\) Ta có : \(\left(-1\right)^2-4\left(-8\right)=1+32=33>0\)

\(x_1=\frac{1-\sqrt{33}}{-16};x_2=\frac{1+\sqrt{33}}{-16}\)

Tớ ngu ! tớ nhận. 

Sửa từ dòng 4 trở lên.

\(\Leftrightarrow4m^2+4-4m^2-m-1=4m^2+4m+4-2\)

\(\Leftrightarrow3-m=4m^2+4m+2\)

\(\Leftrightarrow3-m-4m^2-4m-2=0\)

\(\Leftrightarrow1-5m-4m^2=0\)Ta có : \(\left(-5\right)^2-4\left(-4\right)=25+16=41>0\)

\(x_1=\frac{5-\sqrt{41}}{-4};x_2=\frac{5+\sqrt{41}}{-4}\)

Trả lời 

\(\sqrt{x^2+2x+1}+\sqrt{x^2+4x+4}=3\)

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

\(\Leftrightarrow\left|x+1\right|+\left|x+2\right|=3\)

\(\Leftrightarrow x+1+x+2=3\)

\(\Leftrightarrow2x+3=3\)

\(\Leftrightarrow2x=0\)

\(\Leftrightarrow x=0\)

Vậy \(x=0\)

\(\sqrt{x^2+2x+1}+\sqrt{x^2+4x+4}=3\)

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

\(\Leftrightarrow x+1+x+2=3\Leftrightarrow2x+3=3\)

\(\Leftrightarrow2x=0\Leftrightarrow x=0\)

Với \(a>0,b>0,a\ne b\)

\(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)

\(=\)\(\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{\sqrt{a}-\sqrt{b}}{a-b}+\frac{1}{a-b}\)

\(=\frac{1}{\sqrt{a}+\sqrt{b}}-\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{a-b}=\frac{1}{a-b}\)

Gọi \(T=\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\)

Áp dụng Bất đẳng thức Cauchy Schwarz dạng engel ta có :

\(T=\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=\frac{1^2}{1+x}+\frac{1^2}{1+y}+\frac{1^2}{1+z}\ge\frac{\left(1+1+1\right)^2}{3+x+y+z}\)

\(< =>T=\frac{9}{3+7}=\frac{9}{10}\)

Đẳng thức xảy ra khi và chỉ khi \(x=y=z=\frac{7}{3}\)

Vậy \(Min_T=\frac{9}{10}\)khi \(x=y=z=\frac{7}{3}\)

hóng cách khác :))

Mình làm như thế này nè:

Áp dụng BĐT AM - GM ta dễ có:

\(\frac{1}{x+1}+\frac{9\left(x+1\right)}{100}\ge2\sqrt{\frac{1}{x+1}\cdot\frac{9\left(x+1\right)}{100}}=\frac{3}{5}\)

Tương tự:\(\frac{1}{y+1}+\frac{9\left(y+1\right)}{100}\ge\frac{3}{5};\frac{1}{z+1}+\frac{9\left(z+1\right)}{100}\ge\frac{3}{5}\)

Cộng lại:

\(T+\frac{9\left(x+y+z\right)+27}{100}\ge\frac{9}{5}\Leftrightarrow T\ge\frac{9}{10}\)

Đẳng thức xảy ra tại \(x=y=z=\frac{7}{3}\)