K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

16 tháng 9 2017

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

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

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

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

\(=\dfrac{-\left(x^3-1\right)}{x\left(x^3-1\right)}=\dfrac{-1}{x}\)

1: Sửa đề: 2/x+2

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

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

=>4x-3=-3x-6

=>7x=-3

=>x=-3/7(nhận)

2: \(\Leftrightarrow\dfrac{\left(3x+1\right)\left(3-x\right)+\left(3+x\right)\left(1-3x\right)}{\left(1-3x\right)\left(3-x\right)}=2\)

=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)

=>-6x^2+6=2(3x^2-10x+3)

=>-6x^2+6=6x^2-20x+6

=>-12x^2+20x=0

=>-4x(3x-5)=0

=>x=5/3(nhận) hoặc x=0(nhận)

3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)

=>x*19/6=35/12

=>x=35/38

23 tháng 9 2023

\(a,\dfrac{x^2+2}{x^3+1}-\dfrac{1}{x+1}\left(ĐKXĐ:x\ne-1\right)\\ =\dfrac{x^2+2-\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\\ c,\dfrac{1}{2-2x}-\dfrac{3}{2+2x}+\dfrac{2x}{x^2-1}\\ =\dfrac{-1}{2\left(x-1\right)}-\dfrac{3}{2\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\left(ĐKXĐ:x\ne\pm1\right)\\ =\dfrac{-1\left(x+1\right)-3\left(x-1\right)+2x.2}{2\left(x+1\right)\left(x-1\right)}\\ =\dfrac{-x-1-3x+3+4x}{2\left(x+1\right)\left(x-1\right)}=\dfrac{2}{2\left(x+1\right)\left(x-1\right)}=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)

23 tháng 9 2023

sáng sớm mà chăm thế anh 8h còn đi học mà :> 

26 tháng 4 2021

đk: \(_{x+1\ne0\Leftrightarrow x\ne-1}\)\(\dfrac{1-x}{x+1}+3=\dfrac{2x-3}{x+1}\Leftrightarrow\dfrac{1-x}{x+1}+\dfrac{3\left(x+1\right)}{x+1}=\dfrac{2x+3}{x-1}\Leftrightarrow1-x+3x+3-2x-3=0\Leftrightarrow-2x+1=0\Leftrightarrow-2x=-1\Leftrightarrow x=0,5\)

6 tháng 12 2021

\(a,=\dfrac{x^2-2+2-x}{x\left(x-1\right)^2}=\dfrac{x\left(x-1\right)}{x\left(x-1\right)^2}=\dfrac{1}{x-1}\\ b,=\dfrac{6x-3+6x^2-6x+2x^2+1}{2x\left(2x-1\right)}=\dfrac{8x^2-2}{2x\left(2x-1\right)}\\ =\dfrac{2\left(2x-1\right)\left(2x+1\right)}{2x\left(2x-1\right)}=\dfrac{2x+1}{x}\\ c,=\dfrac{x^3+x^2+x+2x-2+4x^2-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^3+5x^2+3x-3}{x^3-1}\)

m: \(=\left(\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{x-1}{2\left(x+1\right)}\right)\cdot\dfrac{2x}{x+1}-\dfrac{3}{x-1}\)

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

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

p: \(=\left(\dfrac{-\left(x+2\right)}{x-2}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{-x^2\left(x-2\right)}{x\left(x-3\right)}\)

\(=\dfrac{-x^2-4x-4+4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)

\(=\dfrac{4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{x-3}=\dfrac{-4x^2\left(x-2\right)}{\left(x+2\right)\left(x-3\right)}\)

Giải các phương trình có chứa ẩn ở mẫu sau: a, \(\dfrac{x-3}{x-2}+\dfrac{x+2}{x}=2\) b, \(\left(x-2\right)\left(\dfrac{2}{3}x-6\right)=0\) d, \(\dfrac{x}{x+1}-\dfrac{2x-3}{x-1}=\dfrac{2x+3}{x^2-1}\) f, \(\dfrac{x-1}{x}+\dfrac{x-2}{x+1}=2\) g, \(\dfrac{x}{x-1}+\dfrac{x-1}{x}=2\) h, \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\) i, \(\dfrac{2}{x+1}-\dfrac{3}{x-1}=5\) j, \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\) k, \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x-3}=1\) l,...
Đọc tiếp

Giải các phương trình có chứa ẩn ở mẫu sau:

a, \(\dfrac{x-3}{x-2}+\dfrac{x+2}{x}=2\)

b, \(\left(x-2\right)\left(\dfrac{2}{3}x-6\right)=0\)

d, \(\dfrac{x}{x+1}-\dfrac{2x-3}{x-1}=\dfrac{2x+3}{x^2-1}\)

f, \(\dfrac{x-1}{x}+\dfrac{x-2}{x+1}=2\)

g, \(\dfrac{x}{x-1}+\dfrac{x-1}{x}=2\)

h, \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)

i, \(\dfrac{2}{x+1}-\dfrac{3}{x-1}=5\)

j, \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)

k, \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x-3}=1\)

l, \(\dfrac{2}{x+1}-\dfrac{1}{xx-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)

m, \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)

n, \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)

o, \(\dfrac{x-2}{x+2}+\dfrac{3}{x-2}=\dfrac{x^2-11}{x^2-4}\)

p, \(\dfrac{x+4}{x+1}+\dfrac{x}{x-1}=\dfrac{2x^2}{x^2-1}\)

z, \(\dfrac{2x}{x-1}+\dfrac{4}{x^2+2x-3}=\dfrac{2x-5}{x+3}\)

q, \(\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{7x^2-3x}{9-x^2}\)

r, \(\dfrac{1}{x-3}+2=\dfrac{5}{x-1}+x\)

s, \(\dfrac{2}{x^2+4x-21}=\dfrac{3}{x-3}\)

3
24 tháng 6 2017

Phân thức đại số

Phân thức đại số

1 tháng 6 2018

rảnh vãi

b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)

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

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

\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)

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

c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)

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

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

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

a) Ta có: \(\left(x+1\right)\left(2x-3\right)-3\left(x-2\right)=2\left(x-1\right)^2\)

\(\Leftrightarrow2x^2-3x+2x-3-3x+6=2\left(x^2-2x+1\right)\)

\(\Leftrightarrow2x^2-4x+3-2x^2+4x-2=0\)

\(\Leftrightarrow1=0\)(vô lý)

Vậy: \(S=\varnothing\)

21 tháng 2 2021

Ai giúp vs