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4 tháng 1 2021

a, \(\dfrac{x^2+y^2}{4\left(x+y\right)}+\dfrac{2xy}{4\left(x+y\right)}\)=\(\dfrac{x^2+2xy+y^2}{4\left(x+y\right)}\)   = \(\dfrac{\left(x+y\right)^2}{4\left(x+y\right)}\)  =\(\dfrac{x+y}{4}\) 

a. \(\dfrac{x^2+y^2}{4\left(x+y\right)}+\dfrac{2xy}{4\left(x+y\right)}\)

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

\(=\dfrac{\left(x+y\right)^2}{4\left(x+y\right)}\)

\(=\dfrac{x+y}{4}\)

b. \(\dfrac{x+5}{2x-2}-\dfrac{4}{x^2-1}:\dfrac{2}{x+1}\)

\(=\dfrac{x+5}{2\left(x-1\right)}-\dfrac{4}{\left(x+1\right)\left(x-1\right)}:\dfrac{2}{x+1}\)

\(=\dfrac{x+5}{2\left(x-1\right)}-\dfrac{2}{x-1}\)

\(=\dfrac{x+5}{2\left(x-1\right)}-\dfrac{4}{2\left(x-1\right)}\)

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

29 tháng 6 2017

Biến đổi các biểu thức hữu tỉ. Giá trị của phân thức

Biến đổi các biểu thức hữu tỉ. Giá trị của phân thức

24 tháng 6 2017

Phân thức đại số

Phân thức đại số

6 tháng 7 2018

ĐKXĐ: \(x\ne\pm y\)

\(A=\dfrac{x^2}{\left(x-y\right)^2.\left(x+y\right)}-\dfrac{2xy^2}{x^4-2x^2y^2+y^4}+\dfrac{7^2}{\left(x^2-y^2\right)\left(x+y\right)}\)

\(A=\dfrac{x^2}{\left(x-y\right)^2.\left(x+y\right)}-\dfrac{2xy^2}{\left(\left(x+y\right).\left(x-y\right)\right)^2}+\dfrac{49}{\left(x+y\right)^2.\left(x-y\right)}\)

\(A=\dfrac{x^2}{\left(x-y\right)^2.\left(x+y\right)^{ }}-\dfrac{2xy^2}{\left(x-y\right)^2.\left(x+y\right)^2}+\dfrac{49}{\left(x+y\right)^2.\left(x-y\right)}\)

\(A=\dfrac{x^2.\left(x+y\right)-2xy^2+49.\left(x-y\right)}{\left(x-y\right)^2.\left(x+y\right)^2}\)

\(A=\dfrac{x^3+x^2y-2xy^2+49x-49y}{\left(x-y\right)^2.\left(x+y\right)^2}\)

6 tháng 7 2018

ĐKXĐ: \(x\ne\pm1\)

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

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

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

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

13 tháng 12 2021

\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)

21 tháng 4 2017

Giải bài 23 trang 46 Toán 8 Tập 1 | Giải bài tập Toán 8

21 tháng 4 2017

Giải bài 23 trang 46 Toán 8 Tập 1 | Giải bài tập Toán 8

27 tháng 11 2017

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

26 tháng 11 2023

\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\left(dkxd:x\ne0,x\ne5\right)\\ =\dfrac{3x-x-1}{x\left(x-5\right)}=\dfrac{2x-1}{x^2-5x}\)

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\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\left(dkxd:x\ne0,y\ne-2\right)\\ =\dfrac{8}{4}.\dfrac{15x^2.x^3}{3x^2}=10x^3\)

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\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\left(dkxd:x\ne1,x\ne-1\right)\\ =\dfrac{8\left(y-1\right)}{3\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)^2}{4\left(y-1\right)^3}\\ =\dfrac{2\left(x-1\right)}{3\left(x+1\right)\left(y-1\right)^2}\)