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30 tháng 11 2016

\(MTC:\left(x-3\right)^2\left(x^2+3x+9\right)\)

\(\frac{x}{x^3-27}=\frac{x}{\left(x-3\right)\left(x^2+3x+9\right)}=\frac{x\left(x-3\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)

\(\frac{2x}{x^2-6x+9}=\frac{2x}{\left(x-3\right)^2}=\frac{2x\left(x^2+3x+9\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)

\(\frac{1}{x^2+3x+9}=\frac{\left(x-3\right)^2}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)

\(MTC:2\left(x-1\right)\left(x+1\right)\)

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

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

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

\(MTC:2\left(x+1\right)\left(x^2-x+1\right)\)

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

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

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

29 tháng 11 2019

Làm ngắn gọn thôi nhé :v

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

\(A=\frac{x^5-3x^4-3x^3+11x^2-6x}{x^5-8x^2+22x^2-24x+9}\)

\(A=\frac{x^4-3x^3-3x^2+11x-6}{x^4-8x^3+22x^2-24x+9}\)

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

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

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

\(B=\frac{-x^4-4x^3+16x+16}{-x^4+8x^2-16}\)

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

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

\(C=\frac{1+x}{3-x}-\frac{1-2x}{3+x}-\frac{x\left(1-x\right)}{9-x^2}\)

\(C=\frac{1+x}{3-x}-\left(\frac{1-2x}{3+x}\right)-\frac{x\left(1-x\right)}{9-x^2}\)

\(C=\frac{10x}{-x^2+9}\)

\(D=\frac{5}{2x^2+6x}-\frac{4-3x^2}{x^2-9}-3\)

\(D=\frac{5}{2x^2+6x}-\left(\frac{4-3x^2}{x^2-9}\right)-3\)

\(D=\frac{51x^2+138x-45}{2x^4+6x^2-18x^2-54x}\)

\(D=\frac{3\left(17x-5\right)\left(x+3\right)}{2x\left(x+3\right)\left(x+3\right)\left(x-2\right)}\)

\(D=\frac{51x-15}{2x^3-18x}\)

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

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

\(E=\frac{10x^4-10}{x^6-3x^4+3x^2-1}\)

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

\(E=\frac{10x^2+10}{x^4-2x+1}\)

a)có khả năng sai đề bài

b)Liệu có sai đề bài không

c)\(=\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)(phân số cuối có âm vì (1-x)=-(x-1)

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

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

d)\(=\frac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{4xy}{\left(x-2y\right)\left(x+2y\right)}\)

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

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

22 tháng 5 2021

\(\frac{x-3}{5}-\frac{2x-1}{10}=\frac{x+1}{2}+\frac{1}{4}\)

\(< =>\frac{\left(x-3\right).4}{20}-\frac{\left(2x-1\right).2}{20}=\frac{\left(x+1\right).10}{20}+\frac{5}{20}\)

\(< =>4x-12-4x+2=10x+10+5\)

\(< =>10x=-10-10-5=-25\)

\(< =>x=-\frac{25}{10}=-\frac{5}{2}\)

22 tháng 5 2021

\(\frac{x+3}{2}-\frac{2x-1}{3}-1=\frac{x+5}{5}\)

\(< =>\frac{\left(x+3\right).15}{30}-\frac{\left(2x-1\right).10}{30}-\frac{30}{30}=\frac{\left(x+5\right).5}{30}\)\(< =>15x+45-20x+10-30=5x+25\)

\(< =>-5x+25=5x+25< =>10x=0< =>x=0\)

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

13 tháng 9 2016

Trong khó thế...

13 tháng 9 2016

ừa hi

14 tháng 2 2020

Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\) 

 \(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)

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

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

\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\) 

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

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

\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)

5 tháng 5 2018
Sai đề câu a phải không
NV
16 tháng 11 2019

Đề bài yêu cầu gì bạn?