Học tại trường Chưa có thông tin
Đến từ Chưa có thông tin , Chưa có thông tin
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Số lượng câu trả lời 15
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Điểm SP 25

Người theo dõi (5)

anaxi
dũng đào
Huyền Trang
Xích U Lan

Đang theo dõi (0)


Câu trả lời:

a, \(\frac{5x+14}{3x\left(x-5\right)}-\frac{x-14}{3x\left(5-x\right)}\)

= \(\frac{5x+14}{3x\left(x-5\right)}+\frac{x-14}{3x\left(x-5\right)}\)

= \(\frac{5x+14+x-14}{3x\left(x-5\right)}\)

= \(\frac{6x}{3x\left(x-5\right)}\)

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

--------------------------------------------------

b, \(\frac{1}{x+3}-\frac{1}{x-4}+\frac{1}{x^2-x-12}\)

= \(\frac{1}{x+3}-\frac{1}{x-4}+\frac{1}{x^2-4x+3x-12}\)

= \(\frac{1}{x+3}-\frac{1}{x-4}+\frac{1}{x\left(x-4\right)+3\left(x-4\right)}\)

= \(\frac{1}{x+3}-\frac{1}{x-4}+\frac{1}{\left(x+3\right)\left(x-4\right)}\)

= \(\frac{x-4-x-3+1}{\left(x+3\right)\left(x-4\right)}\)

= \(\frac{-6}{\left(x+3\right)\left(x-4\right)}\)

--------------------------------------------------

c, \(x^3+1+\frac{x^4-3x^2+2}{x^2-1}\)

= \(x^3+1+\frac{x^4-x^2-2x^2+2}{x^2-1}\)

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

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

= x3 +1 +x2 - 2

= x3 + x2 - 1

----------------------------------------------------

d, \(\frac{1}{x^2+3x+2}-\frac{2x}{x^3+4x^2+4x}+\frac{1}{x^2+5x+6}\)

= \(\frac{1}{x^2+x+2x+2}-\frac{2x}{x\left(x^2+4x+4\right)}+\frac{1}{x^2+2x+3x+6}\)

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

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

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

= \(\frac{x^2+3x+2x+6-x^2-3x-x-6+x^2+2x+x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

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

---------------------------------------------------

e, \(\frac{1}{x^2+x-2}+\frac{1}{x^2-x-2}+\frac{1+x}{\left(x+1\right)^2-\left(x+3\right)}\)

= \(\frac{1}{x^2-x+2x-2}+\frac{1}{x^2+x-2x-2}+\frac{1+x}{x^2+2x+1-x-3}\)

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

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

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

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

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

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

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

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

= \(\frac{x^3+2x^2-3x-6}{\left(x^2-1\right)\left(x^2-4\right)}\)

Chúc bạn học tốt!