thực hiện phép tính :
a) \(\frac{1}{x-1}+\frac{1}{x+1}+\frac{2}{x^2-1}\)
b)\(\frac{x+1}{2x-2}+\frac{x+2}{3x-3}\)
c)\(\frac{x}{x+1}-\frac{1}{x^2+1}\)
giải hô e vs ạ <3 e cảm ơn !!!
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
11 tháng 12 2019
\(a)=\frac{-2\left(x+3\right)}{x\left(1-3x\right)}.\frac{1-3x}{x\left(x+3\right)}\)
\(=\frac{-2}{x^2}\)
\(b)=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}\)
\(=x\left(x-3\right)\)
\(c)=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)
\(=\frac{\left(x+3\right).x}{x\left(x-1\right)\left(x+1\right)}-\frac{1.\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)-\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+3}{x+1}\)
# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha
S
14 tháng 12 2018
\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x^2+x}\)
b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep
c, tt
d, cx r
14 tháng 12 2018
a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)
\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)
\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)
\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)
PA
1 tháng 12 2016
\(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right)\times\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\)
\(=\left[\frac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right]\times\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x^2-x+1\right)-3+3\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\times\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2-x+1-3+3x+3}{x+1}\times\frac{3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{x+1}\times\frac{3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{3\left(x+1\right)^2}{\left(x+1\right)\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{3x}{x\left(x+2\right)}-\frac{2x-2}{x\left(x+2\right)}\)
\(=\frac{3x-2x+2}{x\left(x+2\right)}\)
\(=\frac{x+2}{x\left(x+2\right)}\)
\(=\frac{1}{x}\)
BN
3 tháng 12 2019
a) \(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}=\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x.x}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}=\frac{0}{x\left(x-3\right)}=0\)
b) \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)
\(=\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10+8}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\frac{1\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}-\frac{4\left(3x-2\right)}{\left(3x+2\right)\left(3x-2\right)}-\frac{-10x+8}{\left(3x-2\right)\left(3x+2\right)}\)
\(\frac{3x+2-12x+2+10x-8}{\left(3x-2\right)\left(3x+2\right)}=\frac{x-4}{\left(3x-2\right)\left(3+2\right)}\)
BN
3 tháng 12 2019
c) \(\frac{4a^2-3a+5}{a^3-1}-\frac{1-2a}{a^2+a+1}-\frac{6}{a-1}\)
\(=\frac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{2a-1}{a^2+a+1}-\frac{6}{a-1}\)
\(=\frac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\frac{6\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{4a^2-3a+5+2a^2-2a-a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{-12}{\left(a-1\right)\left(a^2+a+1\right)}\)
d) \(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}=\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}=\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{x^2-6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{\left(x-3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{x-3y}{x\left(x+3y\right)}\)
e) \(\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)
\(=\frac{3x-2}{\left(x-1\right)^2}-\frac{6}{\left(x-1\right)\left(x+1\right)}-\frac{3x-2}{\left(x+1\right)^2}\)
\(=\frac{\left(3x+2\right)\left(x+1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}-\frac{6\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x-1\right)\left(x+1\right)}-\frac{\left(3x-2\right)\left(x-1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}\)
\(=\frac{3x^3+6x^2+3x+2x^2+4x+2-6x^2+6-3x^3+6x^2-3x+2x^2-4x+2}{\left(x-1\right)^2\left(x+1\right)^2}\)
\(=\frac{8x^2+10}{\left(x-1\right)^2\left(x+1\right)^2}\)
f) \(\frac{5}{a+1}-\frac{10}{a-\left(a^2+1\right)}-\frac{15}{a^3+1}=\frac{5a^2}{a^3+1}+\frac{10}{a^3+1}-\frac{15}{a^3+1}\)
\(=\frac{5a^2+10-15}{a^3+1}=\frac{5a^2-5}{a^3+1}\)
SN
6 tháng 11 2016
mk ko biết làm
xin lỗi bn nhae
xin lỗi vì đã ko giúp được bn
chcus bn học gioi!
nhae@@@
AT
16 tháng 3 2020
a, \(\frac{4x+1}{2}-\frac{3x+2}{3}=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{12x+3-6x-4}{6}=\frac{6x-1}{6}\)
b, \(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}=\frac{x+3}{\left(x-1\right)\left(x+2\right)}-\frac{1}{x\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)}{x\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1}{x\left(x-1\right)\left(x+1\right)}=\frac{\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+1}{x\left(x-1\right)}\)
16 tháng 3 2020
\(\frac{4x+1}{2}-\frac{3x+2}{3}\)
\(=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{6x-1}{6}\)
tương tự đến hết nha a hay cj gì đps !
22 tháng 4 2020
\(a,\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
\(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{4x}{2\left(x+1\right)\left(x-3\right)}\)
\(x\left(x+1\right)+x\left(x-3\right)=4x\)
\(x^2+x+x^2-3x=4x\)
\(2x^2-2x=4x\)
\(2x^2-2x-4x=0\)
\(2x\left(x-3\right)=0\)
\(2x=0\Leftrightarrow x=0\)
hoặc
\(x-3=0\Leftrightarrow x=3\)
NN
22 tháng 4 2020
b) \(ĐKXĐ:x\ne\pm4\)
\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
\(\Leftrightarrow5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)
\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)
\(\Rightarrow5\left(x^2-16\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)
\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)
\(\Leftrightarrow5x^2-2x^2-3x^2+9x-11x=4-4+80-96\)
\(\Leftrightarrow-2x=-16\)\(\Leftrightarrow x=8\)( thoả mãn ĐKXĐ )
Vậy tập nghiệm của phương trình là: \(S=\left\{8\right\}\)
a) \(\frac{1}{x-1}+\frac{1}{x+1}+\frac{2}{x^2-1}=\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+1+x-1+2}{\left(x-1\right)\left(x+1\right)}=\frac{2x+2}{\left(x-1\right)\left(x+1\right)}=\frac{2}{x-1}\)
b) \(\frac{x+1}{2x-2}+\frac{x+2}{3x-3}=\frac{x+1}{2\left(x-1\right)}+\frac{x+2}{3\left(x-1\right)}=\frac{3\left(x+1\right)}{6\left(x-1\right)}+\frac{2\left(x+2\right)}{6\left(x-1\right)}\)
\(=\frac{3x+3+2x+4}{6x-6}=\frac{5x+7}{6x-6}\)
c) \(\frac{x}{x+1}-\frac{1}{x^2+1}=\frac{x\left(x^2+1\right)}{\left(x+1\right)\left(x^2+1\right)}-\frac{x+1}{\left(x+1\right)\left(x^2+1\right)}=\frac{x^3+x-x-1}{\left(x+1\right)\left(x^2+1\right)}\)
\(=\frac{x^3-1}{x^3+x^2+x+1}\)