đấy là tích không phải tổng nhé

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

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

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

\(\Leftrightarrow\left(x^2-1\right)A=3\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

\(\Leftrightarrow\left(x^2-1\right)A=3\left(x^{16}-1\right)\left(x^{16}+1\right)\)

\(\Leftrightarrow\left(x^2-1\right)A=3\left(x^{32}-1\right)\)

\(\Leftrightarrow A=\frac{3\left(x^{32}-1\right)}{x^2-1}\)

\(a,-x^2-x+2\)

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

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

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

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

\(b,x^3-x^2-4x^2-8x-4\)

\(=\left(x^3-x^2\right)-\left(4x^2+8x+4\right)\)

\(=\left(x^3-x^2\right)-\left[\left(2x\right)^2+2.2x.2+2^2\right]\)

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

\(=-x^2\left(x+1\right)-\left[2\left(x+1\right)\right]^2\)

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

\(=\left(x+1\right)\left[-x^2-4\left(x+1\right)\right]\)

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

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

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

Q=x² - 2xy + y² - 12x + 12y + 36 + 5y² + 10y + 5 + 1976

Q=(x - y)² - 2.(x - y).6 + 6² + 5(y² + 2y + 1) + 1976

Q=(x - y - 6)² +5.(y + 1)² + 1976 (≥ 1976 > 0 ∀ x,y ∈ R)

Vậy biểu thức Q luôn nhận giá trị dương với mọi số thực x,y

12 + 26 = 38 

Cái kia chịu nhak 

Hk tốt 

>..<

Trả lời :

Anh có Củi

Em có Lửa

Hai chúng Mình cùng Đốt

12 + 26 = 38

\(\left(5x-4\right)\left(2x+3\right)=10x^2+15x-8x-12=10x^2+7x-12\)

\(b,\frac{x-4}{x-2}+\frac{5x-8}{x-2}=\frac{x-4+5x-8}{x-2}=\frac{6\left(x-2\right)}{x-2}=6\)

\(c,\frac{x-9}{x^2-9}-\frac{3}{x^2+3x}=\frac{x-9}{\left(x+3\right)\left(x-3\right)}-\frac{3}{x\left(x+3\right)}\)

\(=\frac{x^2-9x}{x\left(x+3\right)\left(x-3\right)}-\frac{3x-9}{x\left(x+3\right)\left(x-3\right)}=\frac{x^2-9x-3x+9}{x\left(x+3\right)\left(x-3\right)}=\frac{x^2-6x+9}{x\left(x+3\right)\left(x-3\right)}\)

\(=\frac{x-3}{x\left(x+3\right)}\)

    CÂU 1 :

 a, ( 5x-4 ) ( 2x + 3 )

=  10x + 15x -8x -12

= 17x - 12 

 b, \(\frac{x-4}{x-2}\)+ \(\frac{5x-8}{x-2}\)

= \(\frac{x-4+5x-8}{x-2}\)

= \(\frac{6x-12}{x-2}\)

= \(\frac{6\left(x-2\right)}{x-2}\)

= 6

 c, \(\frac{x-9}{x^2-9}\)- \(\frac{3}{x^2+3x}\)

= \(\frac{x-9}{\left(x-3\right)\left(x+3\right)}\)- \(\frac{3}{x\left(x+3\right)}\)

= \(\frac{\left(x-9\right).x}{x\left(x-3\right).\left(x+3\right)}\)- \(\frac{3.\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\)

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

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

= \(\frac{x^2-12x+9}{x\left(x-3\right)\left(x+3\right)}\)