DQ
3
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
18 tháng 12 2016
a,\(A=\frac{6x+12}{\left(x+2\right)\left(2x-6\right)}=\frac{6\left(x+2\right)}{2\left(x+2\right)\left(x-3\right)}=\frac{3}{x-3}\)
b, Giá trị của x để phân thức có giá trị bằng (-2) :
\(\frac{3}{x-3}=-2\Rightarrow x=1,5\)
N
11 tháng 12 2018
a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)
CH
Cô Hoàng Huyền
Admin
VIP
27 tháng 9 2017
a) \(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6}{1-x}\)
\(=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(x-1\right)\left(2x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{4x^2-3x+17+2x^2-x-2x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{-12x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=-\frac{12}{x^2+x+1}\)
b) \(\frac{1}{x^2-x+1}-\frac{x^2+2}{x^3+1}+1=\frac{x+1-x^2-2+x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x-x^2+x^3}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x}{x+1}\)
c) \(N=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{a}{a\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{1+b}{b+1+bc}+\frac{abc^2}{ac+abc^2+abc}\)
\(N=\frac{1+b}{b+1+bc}+\frac{abc^2}{ac\left(1+bc+b\right)}\)
\(N=\frac{1+b}{b+1+bc}+\frac{bc}{1+bc+b}\)
\(N=\frac{1+b+bc}{b+1+bc}\)
\(N=1.\)
TT
1 tháng 3 2020
a) Ta thấy x=-2 thỏa mãn ĐKXĐ của B.
Thay x=-2 và B ta có :
\(B=\frac{2\cdot\left(-2\right)+1}{\left(-2\right)^2-1}=\frac{-3}{3}=-1\)
b) Rút gọn :
\(A=\frac{3x+1}{x^2-1}-\frac{x}{x-1}\)
\(=\frac{3x+1-x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{-x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)
Xấu nhỉ ??
Sửa đề : \(B=\frac{x+3}{x+1}-\frac{2x-1}{1-x}+\frac{x+7}{x^2-1}\)
\(=\frac{x+3}{x+1}+\frac{2x-1}{x-1}+\frac{x+7}{x^2-1}\)
\(=\frac{\left(x+3\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{\left(2x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{x+7}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^2-x+3x-3+2x^2+2x-x-1+x+7}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^2+4x+3}{\left(x+1\right)\left(x-1\right)}=\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x-1\right)}=\frac{x+3}{x-1}\)
\(B=\frac{x+3}{x+1}-\frac{2x-1}{1-x}+\frac{x+7}{x^2-1}\)
\(=\frac{x+3}{x+1}+\frac{2x-1}{x-1}+\frac{x+7}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{\left(x+3\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{\left(2x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{x+7}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^2+2x-3}{\left(x+1\right)\left(x-1\right)}+\frac{2x^2+x-1}{\left(x+1\right)\left(x-1\right)}+\frac{x+7}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^2+2x-3+2x^2+x-1+x+7}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{3x^2+4x+3}{\left(x+1\right)\left(x-1\right)}\)