\(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)2xyz\)

\(=\left[xy\left(x+y\right)+xyz\right]+\left[yz\left(y+z\right)+xyz\right]+xz\left(x+z\right)\)

\(=xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+z\right)\)

\(=\left(xy+yz\right)\left(x+y+z\right)+xz\left(x+z\right)\)

\(=y\left(x+z\right)\left(x+y+z\right)+xz\left(x+z\right)\)

\(=\left(x+z\right)\left[y\left(x+y+z\right)+xz\right]=\left(x+z\right)\left(xy+y^2+yz+xz\right)\)

\(=\left(x+z\right)\left[y\left(x+y\right)+z\left(x+y\right)\right]\)

\(=\left(x+z\right)\left(z+y\right)\left(y+x\right)\)

\(=\left(x+y\right)\left(y+z\right)\left(z+x\right).\)

Phức tạp. Cs cách nào nhanh kkk?

\(G=\frac{x^2-1}{x^2+1}=\frac{x^2+1-2}{x^2+1}\)

\(=1-\frac{2}{x^2+1}\)

Ta có: \(x^2\ge0\)

\(\Rightarrow x^2+1\ge1\)

\(\Rightarrow\frac{2}{x^2+1}\le2\)

\(\Rightarrow-\frac{2}{x^2+1}\ge-2\)

\(\Rightarrow1-\frac{2}{x^2+1}\ge-1\)

Vậy \(G_{min}=-1\Leftrightarrow x^2=0\Leftrightarrow x=0\)

a) y² - x² - 6x - 9 = y² - (x² + 6x + 9) = y² - (x + 3)² = (y - x - 3)(x + y + 3)

b) x² - y² - 2y - 1 = x² - (y² + 2y + 1) = x² - (y + 1)² = (x - y - 1)(x + y + 1)

c) 3x²(xy - 2y) - 15(xy - 2y) = 3y(x - 2)(x² - 5)

6) c) x³ - x² + x = 1

<=> x³ - x² + x - 1 = 0

<=> (x³ - x²) + (x - 1) = 0

<=> x² (x - 1) + (x - 1) = 0

<=> (x - 1) (x² + 1) = 0

=> x - 1 = 0 hoặc x² + 1 = 0

* x - 1 = 0 => x = 1

* x² + 1 = 0 => x² = -1 => x = -1

Vậy x = 1 hoặc x = -1

Bài 5: 

a) Đặt   \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(\Rightarrow8A=3^{32}-1\)

\(\Rightarrow A=\frac{3^{32}-1}{8}\)

b) (7x+6)² + (5-6x)² - (10-12x)(7x+6)

=(7x+6)² + (5-6x)² - 2(5-6x)(7x+6)

\(=\left(7x+6-5+6x\right)^2\)

\(=\left(13x+1\right)^2\)

\(A=12\left(x-1\right)^2+\frac{8x}{y}=12y^2+\frac{8x}{y}=12y^2+\frac{8\left(1-y\right)}{y}\) (chú ý cái giả thiết =>x = 1-y)

\(=12y^2+\frac{8}{y}-8=12y^2+\frac{4}{y}+\frac{4}{y}-8\ge3\sqrt[3]{12y^2.\frac{4}{y}.\frac{4}{y}}-8\)

\(=3\sqrt[3]{192}-8=12\sqrt[3]{3}-8\)

Không chắc lắm.

minimize 3(2x-2)^2 +8x/y ,x&gt;0,y&gt;0,x+y=1  Theo Wolfram Alpha thì đáp số (giá trị Min) của t là đúng nha Khua Kít:)

1) 

=3(x-y)+(x-y)(x+y)

=(x-y)(3x+3y)

2)

=x^2+2x+x+2

=x(x+2)+x+2

=(x+1)(x+2).

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

\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)

\(=\left(x-y\right)\left(3+x+y\right)\)

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

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

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

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