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NV
7 tháng 8 2021

ĐKXĐ: \(\left\{{}\begin{matrix}2x+5\ne0\\x-2\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{5}{2}\\x\ne2\end{matrix}\right.\)

D

24 tháng 4 2022

để pt được xác định thì :

\(x-2\ne0;x^2-1\ne0\)

=>\(\left\{{}\begin{matrix}x\ne2\\x\ne-1\\x\ne1\end{matrix}\right.\)

Vậy chọn B

16 tháng 5 2023

Điều kiện xác định là `{(x-3 ne 0),(x(x-3) ne 0):}`

                 `<=>{(x ne 3),(x ne 0):}`

      `=>bb A`

16 tháng 5 2023

ĐCXĐ: \(\left\{{}\begin{matrix}x\ne0\\x-3\ne0\end{matrix}\right.\)\(\left\{{}\begin{matrix}x\ne0\\x\ne3\end{matrix}\right.\)

24 tháng 5 2022

Với `x \ne -5,x \ne -1` có:

`A=[x+2]/[x+5]+[-5x-1]/[x^2+6x+5]-1/[1+x]`

`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+5)(x+1)]`

`A=[x^2+x+2x+2-5x-1-x-5]/[(x+5)(x+1)]`

`A=[x^2-3x-4]/[(x+5)(x+1)]`

`A=[(x-4)(x+1)]/[(x+5)(x+1)]`

`A=[x-4]/[x+5]`

24 tháng 5 2022

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

AH
Akai Haruma
Giáo viên
24 tháng 11 2018

a)

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

b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)

c)

\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)

d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)

AH
Akai Haruma
Giáo viên
24 tháng 11 2018

e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)

f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)

g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)

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

16 tháng 3 2023

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

\(=\dfrac{-3x+6-2x-4+4x}{x^2-4}\)

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

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

\(=-\dfrac{1}{x+2}\left(x\ne2;x\ne-2\right)\)

16 tháng 3 2023

\(\dfrac{-3}{x+2}-\dfrac{2}{x-2}+\dfrac{4x}{x^2-4}\left(x\ne\pm2\right)\)

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

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

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

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

20 tháng 12 2022

Phân thức `(7x^2 - x)/(x^2 - 9)` được xác đinh khi

`x^2-9≠0`

`(x-3).(x+3)≠0`

`=> x≠3; x ≠ -3`

`=> C`

a: Khi x=1 thì\(P=\dfrac{1-2}{1+2}=\dfrac{-1}{2}\)

b: \(=\dfrac{3x+6+5x-6+2x^2-4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2+4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x}{x-2}\)

c: \(P=A\cdot B=\dfrac{2x}{x-2}\cdot\dfrac{x-2}{x+1}=\dfrac{2x}{x+1}\)

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

P<=2

=>x+1>0

=>x>-1