Ta có: \(a^2+b^2+c^2+d^2\ge a\left(b+c+d\right)\)

<=> \(4a^2+4b^2+4c^2+4d^2\ge4ab+4ac+4ad\)

<=> \(\left(a^2-4ab+4b^2\right)+\left(a^2-4ac+4c^2\right)+\left(a^2-4ad+4d^2\right)+a^2\ge0\)

<=> \(\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+a^2\ge0\)luôn đúng 

Vậy \(a^2+b^2+c^2+d^2\ge a\left(b+c+d\right)\) đúng 

Dấu "=" xảy ra <=> a = 0; a - 2b = 0; a - 2c = 0; a - 2d = 0 <=> a = b = c = d = 0 

Bài 2:

Ta có: M = a²+ab+b² -3a-3b-3a-3b +2001

=> 2M = ( a² + 2ab + b²) -4.(a+b) +4 + (a² -2a+1)+(b² -2b+1) + 3996

2M= ( a+b-2)² + (a-1)² +(b-1)² + 3996

=> Min_M = 1998 tại a=b=1

Câu 3: 

Ta có: P= x² +xy+y² -3.(x+y) + 3

=> 2P = ( x² + 2xy +y²) -4.(x+y) + 4 + (x² -2x+1) +(y² -2y+1)

2P = ( x+y-2)² +(x-1)²+(y-1)²

=> Min_P = 0 tại x=y=1

ĐK: \(x\ge\frac{1}{2}\)

\(\hept{\begin{cases}x\left(2x-2y-1\right)=3\left(y+2\right)\left(1\right)\\3y+6\sqrt{2x-1}=y^2-x+23\left(2\right)\end{cases}}\)

pt (1) <=> \(2x^2-2xy-x-3y-6=0\)

<=> \(2x^2-x\left(2y+1\right)-\left(3y+6\right)=0\)

có \(\Delta=\left(2y+1\right)^2+4\left(3y+6\right)=4y^2+28y+49=\left(2y+7\right)^2\)

=> (1) có hai nghiệm: \(\orbr{\begin{cases}x_1=\frac{\left(2y+1\right)-\left(2y+7\right)}{4}=-\frac{3}{2}\left(loai\right)\\x_2=\frac{\left(2y+1\right)+\left(2y+7\right)}{4}=y+2\end{cases}}\)

+) Với \(x=y+2\) thế vào (2) ta có: 

\(3y+6\sqrt{2\left(y+2\right)-1}=y^2-\left(y+2\right)+23\)

<=> \(6\sqrt{2y+3}=y^2-4y+21\)

ĐK: \(y\ge-\frac{3}{2}\)

\(6\sqrt{2y+3}=y^2-4y+21\)

<=> \(6\sqrt{2y+3}-2y-12=y^2-6y+9\)

<=> \(\frac{2\left(9\left(2y+3\right)-\left(y+6\right)^2\right)}{3\sqrt{2y+3}+y+6}-\left(y-3\right)^2=0\)

<=> \(\frac{-2\left(y-3\right)^2}{3\sqrt{2y+3}+y+6}-\left(y-3\right)^2=0\)

<=> \(\left(y-3\right)^2\left(\frac{-2}{3\sqrt{2y+3}+y+6}-1\right)=0\)

<=> y - 3 = 0 

<=> y = 3 thỏa mãn 

khi đó x = y + 2 = 3 + 2 = 5 thỏa mãn

Kết luận:...

ĐK: \(2x+3\ge0\Leftrightarrow x\ge-\frac{3}{2}\)

pt <=> \(x^2-2x+1=4\left(2x+3\right)+4\sqrt{2x+3}+1\)

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

TH1: x - 1 = \(2\sqrt{2x+3}+1\)

<=> \(2\sqrt{2x+3}=x-2\)

<=> \(\hept{\begin{cases}x\ge2\\4\left(2x+3\right)=x^2-4x+4\end{cases}}\Leftrightarrow x=6+2\sqrt{11}\)

TH2: 1-x = \(2\sqrt{2x+3}+1\)

<=> \(-x=2\sqrt{2x+3}\)

<=> \(\hept{\begin{cases}x\le0\\x^2=4x+12\end{cases}}\Leftrightarrow x=-2\)không thỏa mãn ĐK

Kết luận:...

ĐK: \(\hept{\begin{cases}x^2-1\ge0\\x^4-x^2+1\ge0\end{cases}}\)(@@)

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

<=> \(3\sqrt{x^2-1}+x^2-\sqrt{x^4-x^2+1}=0\)

<=> \(3\sqrt{x^2-1}+\frac{x^4-x^4+x^2-1}{x^2+\sqrt{x^4-x^2+1}}=0\)

<=> \(3\sqrt{x^2-1}+\frac{x^2-1}{x^2+\sqrt{x^4-x^2+1}}=0\)

<=> \(\sqrt{x^2-1}\left(3+\frac{\sqrt{x^2-1}}{x^2+\sqrt{x^4-x^2+1}}\right)=0\)

<=> \(\sqrt{x^2-1}=0\)

<=> x = 1 hoặc x = -1 thỏa mãn (@@) 

Kết luận:...