(x - y - z)^2 

= (x - y - z)(x - y - z)

= x^2 - xy - xz - xy + y^2 + zy - zx + zy + z^2

= xY2 + y^2 + z^2  - 2xy - 2xz + 2zy (đpcm)

(x-y-z)(x-y-z) nhân phân phối vô 

\(VT=x\sqrt{y}+\frac{1}{2}y\sqrt{4\left(2x+2y\right)}\le\frac{x\left(y+1\right)}{2}+\frac{1}{2}y\left(\frac{4+2x+2y}{2}\right)\)

\(=\frac{2xy+2x}{4}+\frac{4y+2xy+2y^2}{4}=\frac{2\left(x+2y\right)+4xy+2y^2}{4}\)

\(=\frac{2\left(x+2y\right)+\frac{2}{3}.3y\left(2x+y\right)}{4}\le\frac{2\left(x+2y\right)+\frac{2}{3}\left(\frac{2\left(x+2y\right)}{2}\right)^2}{4}\le3\) (*)

Đẳng thức xảy ra khi x= y = 1.

Is that true? Bài  này khó nhằn đấy, Đối với mình việc nhìn ra chỗ  (*) ko dễ chút nào, chả biết có tính sai gì ko nữa..

a, x^2 + 2x - 8

= x^2 -2x + 4x - 8

= x(x - 2) + 4(x - 2)

= (x + 4)(x - 2)

b, x^2 + 5x + 6

= x^2 + 2x + 3x + 6

= x(x + 2) + 3(x + 2)

= (x + 3)(x + 2)

a/ \(x^2+2x-8\)

\(=x^2+4x-2x-8\)

\(=\left(x^2+4x\right)-\left(2x+8\right)\)

\(=x\left(x+4\right)-2\left(x+4\right)\)

\(=\left(x-2\right)\left(x+4\right)\)

b/ \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

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

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

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

\(\left(x-2\right)\left(x-4\right)\left(x-6\right)\left(x-8\right)+16\)

\(=\left[\left(x-2\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x-6\right)\right]+16\)

\(=\left(x^2-10x+16\right)\left(x^2-10x+24\right)+16\)(1) 

Đặt \(x^2-10x+20=t\)thay vào (1) ta được : 

\(\left(t-4\right)\left(t+4\right)+16\)

\(=t^2-16+16\)

\(=t^2\)Thay \(t=x^2-10x+20\)ta được :

\(\left(x^2-10x+20\right)^2\)

\(=\left(x^2-2.5.x+25-25+20\right)^2\)

\(=\left[\left(x-5\right)^2-5\right]^2\)

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

Mk ko chép lại đề nha !

\(4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(2x+255=0\)

\(2x=-255\)

\(x=\frac{-225}{2}\)

(2x - 1)² + (x + 3)² - 5(x + 7)(x - 7)      = 0

4x² - 4x + 1 + x² + 6x + 9 - 5x² + 245 = 0

                                             2x + 255 = 0

                                             2x           = -255

                                               x           = -127,5.

\(2x^2+3x-5\)

\(=2x^2+5x-2x-5\)

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

\(=x\left(2x+5\right)-\left(2x+5\right)\)

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

\(\Rightarrow2x^2+3x-5\)\(=\left(x-1\right)\left(2x+5\right)\)