\(x^2+2\left(x+1\right)^2+3\left(x-2\right)^2+4\left(x+3\right)^2\)

\(=x^2+2\left(x^2+2x+1\right)+3\left(x^2-4x+4\right)+4\left(x^2+6x+9\right)\)

\(=x^2+2x^2+4x+2+3x^2-12x+12+4x^2+24x+36\)

\(=10x^2+16x+50\)

\(x^2+2\left(x+1\right)^2+3\left(x+2\right)^2+4\left(x+3\right)^2\)

\(=x^2+2\left(x^2+2x+1\right)+3\left(x^2+4x+4\right)+4\left(x^2+6x+9\right)\)

\(=x^2+2x^2+4x+2+3x^2+12x+12+4x^2+24x+36\)

\(=10x^2+40x+50\)

10x²+40x+50

\(x^2+2\left(x+1\right)^2+3\left(x+2\right)^2+4\left(x+3\right)^2\)

\(=x^2+2\left(x^2+2x+1\right)+3\left(x^2+4x+4\right)+4\left(x^2+6x+9\right)\)

\(=10x^2+40x+50\)

\(=\left(x^2+10x+25\right)+\left(9x^2+30x+25\right)\)

\(=\left(x+5\right)^2+\left(3x+5\right)^2\)

\(x^2+2\left(x+1\right)^2+3\left(x+2\right)^2+4\left(x+3\right)^2\)

\(=x^2+2\left(x^2+2x+1\right)+3\left(x^2+4x+4\right)+4\left(x^2+6x+9\right)\)

\(=x^2+2x^2+4x+2+3x^2+12x+12+4x^2+24x+36\)

\(=10x^2+40x+50\)

\(=\left(9x^2+30x+25\right)+\left(x^2+10x+25\right)\)

\(=\left(3x+2\right)^2+\left(x+5^2\right)\)

Bài 2 :

a ) \(A=\left(a+b+c\right)^2+a^2+b^2+c^2\)

\(A=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)

\(A=\left(a^2+2ab+b^2\right)+\left(a^2+2ac+c^2\right)+\left(b^2+2bc+c^2\right)\)

\(A=\left(a+b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2\)

đề dài v~

1.

a) \(f\left(x\right)=5x^2-2x+1\)

\(5f\left(x\right)=25x^2-10x+5\)

\(5f\left(x\right)=\left(25x^2-10x+1\right)+4\)

\(5f\left(x\right)=\left(5x-1\right)^2+4\)

Mà  \(\left(5x-1\right)^2\ge0\)

\(\Rightarrow5f\left(x\right)\ge4\)

\(\Leftrightarrow f\left(x\right)\ge\frac{4}{5}\)

Dấu " = " xảy ra khi :

\(5x-1=0\Leftrightarrow x=\frac{1}{5}\)

Vậy ....

b)  \(P\left(x\right)=3x^2+x+7\)

\(3P\left(x\right)=9x^2+3x+21\)

\(3P\left(x\right)=\left(9x^2+3x+\frac{1}{4}\right)+\frac{83}{4}\)

\(3P\left(x\right)=\left(3x+\frac{1}{2}\right)^2+\frac{83}{4}\)

Mà  \(\left(3x+\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow3P\left(x\right)\ge\frac{83}{4}\)

\(\Leftrightarrow P\left(x\right)\ge\frac{83}{12}\)

Dấu "=" xảy ra khi :

\(3x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{6}\)

Vậy ...

c)  \(Q\left(x\right)=5x^2-3x-3\)

\(5Q\left(x\right)=25x^2-15x-15\)

\(\Leftrightarrow5Q\left(x\right)=\left(25x^2-15x+\frac{9}{4}\right)-\frac{69}{4}\)

\(\Leftrightarrow5Q\left(x\right)=\left(5x-\frac{3}{2}\right)^2-\frac{69}{4}\)

Mà  \(\left(5x-\frac{3}{2}\right)^2\ge0\)

\(\Rightarrow5Q\left(x\right)\ge\frac{-69}{4}\)

\(\Leftrightarrow Q\left(x\right)\ge-\frac{69}{20}\)

Dấu "=" xảy ra khi :

\(5x-\frac{3}{2}=0\Leftrightarrow x=0,3\)

Vậy ...

2.

a)  \(f\left(x\right)=-3x^2+x-2\)

\(-3f\left(x\right)=9x^2-3x+6\)

\(-3f\left(x\right)=\left(9x^2-3x+\frac{1}{4}\right)+\frac{23}{4}\)

\(-3f\left(x\right)=\left(3x-\frac{1}{2}\right)^2+\frac{23}{4}\)

Mà  \(\left(3x-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow-3f\left(x\right)\ge\frac{23}{4}\)

\(\Leftrightarrow f\left(x\right)\le\frac{23}{12}\)

Dấu "=" xảy ra khi :

\(3x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{6}\)

Vậy ...

b)  \(P\left(x\right)=-x^2-7x+1\)

\(-P\left(x\right)=x^2+7x-1\)

\(-P\left(x\right)=\left(x^2+7x+\frac{49}{4}\right)-\frac{53}{4}\)

\(-P\left(x\right)=\left(x+\frac{7}{2}\right)^2-\frac{53}{4}\)

Mà  \(\left(x+\frac{7}{2}\right)^2\ge0\)

\(\Rightarrow-P\left(x\right)\ge-\frac{53}{4}\)

\(\Leftrightarrow P\left(x\right)\le\frac{53}{4}\)

Dấu "=" xảy ra khi :

\(x+\frac{7}{2}=0\Leftrightarrow x=-\frac{7}{2}\)

Vậy ...

c)  \(Q\left(x\right)=-2x^2+x-8\)

\(-2Q\left(x\right)=4x^2-2x+16\)

\(-2Q\left(x\right)=\left(4x^2-2x+\frac{1}{4}\right)+\frac{63}{4}\)

\(-2Q\left(x\right)=\left(2x-\frac{1}{2}\right)^2+\frac{63}{4}\)

Mà :  \(\left(2x-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow-2Q\left(x\right)\ge\frac{63}{4}\)

\(\Leftrightarrow Q\left(x\right)\le-\frac{63}{8}\)

Dấu "=" xảy ra khi :

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

Vậy ...

3/

Ta có x + y + z = 0

<=> (x + y + z)² = 0

<=> x² + y² + z² + 2xy + 2xz + 2yz = 0

<=> x² + y² + z² + 2(xy + xz + yz) = 0

Thay xy + yz + xz = 0 vào BT trên

=> x² + y² + z² = 0

Mà \(x^2\ge0;y^2\ge0;z^2\ge0\Rightarrow x^2+y^2+z^2\ge0\)

=> x = y = z (đpcm)