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23 tháng 7 2023

\(a,\dfrac{8y}{3x^2}.\dfrac{9x^2}{4y^2}=\dfrac{72x^2y}{12x^2y^2}=\dfrac{6}{y}\\b,\dfrac{3x+x^2}{x^2+x+1}.\dfrac{3x^3-3}{x+3}=\dfrac{x\left(x+3\right)3\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x+3\right)}=3x\left(x-1\right)=3x^2-3x \)

\(c,\dfrac{2x^2+4}{x-3}.\dfrac{3x+1}{x-1}.\dfrac{6-2x}{x^2+2}=\dfrac{2\left(x^2+2\right)\left(3x+1\right)2\left(3-x\right)}{\left(x-3\right)\left(x-1\right)\left(x^2+2\right)}=\dfrac{-4\left(3x+1\right)}{x-1}=\dfrac{-12x-4}{x-1}\)

\(d,\dfrac{2x^2}{3y^3}:\left(-\dfrac{4x^3}{21y^2}\right)=\dfrac{-2x^2.21y^2}{3y^3.4x^3}=\dfrac{-42x^2y^2}{12x^3y^3}=\dfrac{-7}{2xy}\)

\(e,\dfrac{2x+10}{x^3-64}:\dfrac{\left(x+5\right)^2}{2x-8}=\dfrac{2\left(x+5\right)}{\left(x-4\right)\left(x^2+4x+16\right)}.\dfrac{2\left(x-4\right)}{\left(x+5\right)^2}=\dfrac{4}{\left(x+5\right)\left(x^2+4x+16\right)}=\dfrac{4}{x^3+9x^2+16x+80}\)

\(f,\dfrac{1}{x+y}\left(\dfrac{x+y}{xy}-x-y\right)-\dfrac{1}{x^2}:\dfrac{y}{x}=\dfrac{1}{x+y}\left(\dfrac{\left(x+y\right)\left(1-xy\right)}{xy}\right)-\dfrac{x}{x^2y}=\dfrac{1-xy}{xy}-\dfrac{x}{x^2y}=\dfrac{-x^2y}{x^2y}=-1\)

23 tháng 7 2023

\(a,\dfrac{x^2-9}{x-2}:\dfrac{x-3}{x}\\ =\dfrac{\left(x-3\right)\left(x+3\right)}{x-2}\times\dfrac{x}{x-3}\\ =\dfrac{x\left(x+3\right)}{\left(x-2\right)}\)

\(b,\dfrac{x}{z^2}.\dfrac{xz}{y^3}:\dfrac{x^3}{yz}\\ =\dfrac{x}{z^2}.\dfrac{xz}{y^3}.\dfrac{yz}{x^3}=\dfrac{x^2yz^2}{z^2y^3x^3}=\dfrac{1}{xy^2}\)

\(c,\dfrac{2}{x}-\dfrac{2}{x}:\dfrac{1}{x}+\dfrac{4}{x}.\dfrac{x^2}{2}\\ =\dfrac{2}{x}-\dfrac{2}{x}\times\dfrac{x}{1}+\dfrac{4x^2}{2x}\\ =\dfrac{2}{x}-\dfrac{2}{1}+2x\\ =\dfrac{2-2x+2x^2}{x}\)

23 tháng 7 2023

a) \(\dfrac{x^2-9}{x-2}:\dfrac{x-3}{x}\)

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

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

b) \(\dfrac{x}{z^2}\cdot\dfrac{xz}{y^3}:\dfrac{x^3}{yz}\)

\(=\dfrac{x}{z^2}\cdot\dfrac{xz}{y^3}\cdot\dfrac{yz}{x^3}\)

\(=\dfrac{1}{xy^2}\)

c) \(\dfrac{2}{x}-\dfrac{2}{x}:\dfrac{1}{x}+\dfrac{4}{x}\cdot\dfrac{x^2}{2}\)

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

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

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

\(=\dfrac{6}{x}-2\)

23 tháng 7 2023

\(a,=\dfrac{4y.5x^3}{3x^2.2y^3}=\dfrac{20x^3y}{6x^2y^3}=\dfrac{10x}{3y^2}\\ b,=\dfrac{\left(x-1\right)^2.x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2.x.\left(x+1\right)}{\left(x-1\right)^2.\left(x+1\right)}=x\)

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

23 tháng 7 2023

\(a,=\dfrac{2\left(2x^2+1\right).\left(3x+2\right).2\left(2-x\right)}{\left(x-2\right)\left(x-4\right)\left(2x^2+1\right)}=\dfrac{-4.\left(3x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{-4\left(3x+2\right)}{x-4}\\ b,=\dfrac{\left(x+3\right).\left(x+2\right)}{x.\left(x+3\right)^2}\times\dfrac{x\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+3\right)\left(x+2\right)x\left(x+3\right)}{x.\left(x+3\right)^2.\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\)

22 tháng 7 2023

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

\(b,\dfrac{1}{x+5}-\dfrac{1}{x-5}+\dfrac{2x}{x^2-25}\\ =\dfrac{1}{x+5}-\dfrac{1}{x-5}+\dfrac{2x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{x-5-x-5+2x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2x-10}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2}{x+5}\)

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

18 tháng 2 2022

\(\left(\dfrac{x}{x+1}+\dfrac{x-1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\) \(\left(đk:x\ne0;-1\right)\)

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

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

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

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

b: \(=\dfrac{x\left(x-y\right)+y\left(x+y\right)}{\left(x+y\right)\left(x-y\right)}\)

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

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

d: \(=\dfrac{\left(x+2\right)\cdot y-x\left(y-2\right)}{xy\left(x+y\right)}\)

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

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

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

g: \(=\dfrac{-2x+x+3-x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2x+6}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2}{x+3}\)

11 tháng 11 2017

Nguyễn Ngọc Thanh Trúc đề là gì

11 tháng 11 2017

thực hiện phép tính

`a, a/(a-3) - 3/(a+3) = (a(a+3) - 3(a-3))/(a^2-9)`

`= (a^2+9)/(a^2-9)`

`b, 1/(2x) + 2/x^2 = x/(2x^2) + 4/(2x^2) = (x+4)/(2x^2)`

`c, 4/(x^2-1) - 2/(x^2+x) = (4x)/(x(x-1)(x+1)) - (2(x-1))/(x(x+1)(x-1))`

`= (2x+2)/(x(x-1)(x+1)`

`= 2/(x(x-1))`

23 tháng 7 2023

a) \(\dfrac{3a^2}{10b^3}\cdot\dfrac{15b}{9a^4}\)

\(=\dfrac{3a^2\cdot15b}{10b^3\cdot9a^4}\)

\(=\dfrac{1\cdot3}{2\cdot b^2\cdot3\cdot a^2}=\dfrac{3}{6a^2b^2}\)

b) \(\dfrac{x-3}{x^2}\cdot\dfrac{4x}{x^2-9}\)

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

\(=\dfrac{\left(x-3\right)\cdot4x}{x^2\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{4}{x\left(x+3\right)}\)

c) \(\dfrac{a^2-6x+9}{a^2+3a}\cdot\dfrac{2a+6}{a-3}\)

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

\(=\dfrac{\left(a-3\right)^2\cdot2\cdot\left(a+3\right)}{a\left(a+3\right)\left(a-3\right)}\)

\(=\dfrac{2\left(a-3\right)}{a}\)

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

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

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

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

=))) để r xem