\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)

Đặt \(x^2+5x+4=t\)

\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)

\(=t.\left(t+2\right)-15\)

\(=t^2+2t+1-16\)

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

\(=\left(t-3\right)\left(t+5\right)\)

\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)

Ta có :

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-15\)

\(=\left[x\left(x+4\right)+1\left(x+4\right)\right]\left[x\left(x+3\right)+2\left(x+3\right)\right]-15\)

\(=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)-15\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)

\(=\left(x^2+5x+4\right)\left[\left(x^2+5x+4\right)+2\right]-15\)(1)

Đặt \(x^2+5x+4=y\)thì (1) trở thành :

\(y\left(y+2\right)-15\)

\(=y^2+2y-15\)

\(=y^2+5y-3y-15\)

\(=\left(y^2+5y\right)-\left(3y+15\right)\)

\(=y\left(y+5\right)-3\left(y+5\right)\)

\(=\left(y-3\right)\left(y+5\right)\)(2)

Thay \(y=x^2+5x+4\)thì (2) trở thành:

\(\left(x^2+5x+4-3\right)\left(x^2+5x+4+5\right)\)

\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)

\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)

\(=\frac{x^2\left(x+1\right)-4\left(x+1\right)}{x^3+x^2+7x^2+7x+10x+10}\)

\(=\frac{\left(x^2-4\right)\left(x+1\right)}{\left(x+1\right)\left(x^2+7x+10\right)}\)

\(=\frac{x^2-4}{x^2+7x+10}\)

\(=\frac{x^2-4}{x^2+5x+2x+10}\)

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

\(=\frac{x-2}{x+5}\)

tui 2k5 nè , kb ko ?

ok bn

nhưng lần sau ko đăng linh tinh nx nha

#G2k6#

ko đnagư câu hỏi linh tinh

x²-y²=1998

<=> (y+x)(x-y)=1998 bạn phân tích số đó ra thừa số nguyên tố sao

cho y+x>=x-y nha 

Áp dụng tính chất tổng hiệu:

a-b=n;a+b=h(a>b)

=> a=(h+n):2;b=h-a

để ccm nha

xem lại đề nha, k thể ab>ac đc 

Nó tối giản mà bạn.