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22 tháng 11 2017

 = (x+1).(x+3)-(1-x).(x-3)+2x.(1-x)/(x-3).(x+3)

 = x^2+4x+3+x^2-4x+3+2x-2x^2/(x+3).(x-3)

 = 2x+6/(x+3).(x-3) = 2.(x+3)/(x+3).(x-3) = 2/x-3

k mk nha

22 tháng 11 2017

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

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

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

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

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

24 tháng 11 2019

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

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

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

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

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

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

\(=\frac{2}{1+2x}\)

25 tháng 4 2020

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

b) \(\frac{x}{x-3}-\frac{5x^2+27}{x^2-9}+\frac{x-9}{x+3}\)

\(=\frac{x}{x-3}-\frac{5x^2+27}{\left(x-3\right)\left(x+3\right)}+\frac{x-9}{x+3}\)

\(=\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{5x^2+27}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-9\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}-\frac{5x^2+27}{\left(x-3\right)\left(x+3\right)}+\frac{x^2-12x+27}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x^2+3x-\left(5x^2+27\right)+x^3-12x+27}{\left(x-3\right)\left(x+3\right)}\)

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

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

\(=\frac{-3x}{x-3}\)

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 ??

28 tháng 3 2022

`Answer:`

\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{2x}{9-x}\right):\left(\frac{\sqrt{x}-1}{x-3\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\left(ĐK:x>0;x\ne9;x\ne25\right)\)

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

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

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

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

\(=-\sqrt{x}.\frac{\sqrt{x}}{5-\sqrt{x}}\)

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

12 tháng 8 2020

a, \(3x\left(x^3-2x\right)=3x^4-6x^2\)

b, \(\frac{4y^3}{7x^2}.\frac{14x^3}{y}=\frac{56x^3y^3}{7x^2y}\)tự chia nhé

c, \(\frac{x^2-9}{2x+6}:\frac{3-x}{2}=\frac{x^2-9}{2x+6}.\frac{2}{3-x}=\frac{2x^2-18}{6x-2x^2+18-6x}=\frac{2x^2-18}{-2x^2+18}=1\)

Làm đc 2 bài đầu chưa, t làm câu cuối cho, hai câu đầu dễ í mà

2 tháng 8 2023

a) \(\dfrac{3}{4xy}+\dfrac{5x}{2x^2z}+\dfrac{7}{6yz^2}\) (MSC: \(12x^2yz^2\))

\(=\dfrac{3\cdot3xz^2}{4xy\cdot3xz^2}+\dfrac{5x\cdot6yz}{2x^2z\cdot6yz}+\dfrac{7\cdot2x^2}{6yz^2\cdot2x^2}\)

\(=\dfrac{9xz^2}{12x^2yz^2}+\dfrac{30xyz}{12x^2yz^2}+\dfrac{14x^2}{12x^2yz^2}\)

\(=\dfrac{9xz^2+30xyz+14x^2}{12x^2yz^2}\)

\(=\dfrac{x\left(9z^2+30yz+14x\right)}{12x^2yz^2}\)

\(=\dfrac{9z^2+30yz+14x}{12x^2yz^2}\)

b) \(\dfrac{x^2}{x^2+3x}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

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

\(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x+3}{x+3}+\dfrac{3}{x}\)

\(=1+\dfrac{3}{x}\)

\(=\dfrac{x}{x}+\dfrac{3}{x}\)

\(=\dfrac{x+3}{x}\)

a: \(=\dfrac{3\cdot3\cdot xz^2+5x\cdot6\cdot y+7\cdot x^2\cdot2}{12x^2yz^2}=\dfrac{9xz^2+30xy+14x^2}{12x^2yz^2}\)

\(=\dfrac{9z^2+30y+14x}{12xyz^2}\)

b: \(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}=1+\dfrac{3}{x}=\dfrac{x+3}{x}\)