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\(\dfrac{1}{x}-\dfrac{1}{x+1}\)
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Những câu hỏi liên quan
VD
16 tháng 6 2023
= (x+1/2)(x-1/2 x + 1/4)
=(x+1/2)(1/2x + 1/4)
=x(1/2x + 1/4) + 1/2(1/2x+1/4)
=1/2 x^2 + 1/4 x + 1/4 x + 1/8
= x^2/2 + 1/2 x + 1/8
H9
HT.Phong (9A5)
CTVHS
16 tháng 6 2023
\(\left(x-\dfrac{1}{2}\right)\left(x-\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
\(=\left(x+\dfrac{1}{2}\right)\left(\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
\(=\dfrac{1}{2}x^2+\dfrac{1}{4}x+\dfrac{1}{4}x+\dfrac{1}{8}\)
\(=\dfrac{1}{2}x^2+\dfrac{1}{2}x+\dfrac{1}{8}\)
25 tháng 12 2021
Nhận thấy \(x^3-x=x\left(x^2-1\right)=x\left(x-1\right)\left(x+1\right)\)
\(\dfrac{3}{x}-\dfrac{x}{x-1}-\dfrac{x^2}{x+1}-\dfrac{x^2-3}{x^3-x}\\ =\dfrac{3x^2-3-x^3-x^2-x^4+x^3-x^2+3}{x\left(x-1\right)\left(x+1\right)}\\ =\dfrac{-x^4+x^2}{x\left(x-1\right)\left(x+1\right)}=\dfrac{-x^2\left(x-1\right)\left(x+1\right)}{x\left(x-1\right)\left(x+1\right)}=-x\)
K
1
8 tháng 12 2021
\(=\dfrac{3x^2+5x+14+x^2-1-4x^2+4x-4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{9x+9}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{9}{x^2-x+1}\)
DT
4
NH
6 tháng 1 2023
\(\dfrac{x-1}{x-y}+\dfrac{1-y}{x-y}\\ =\dfrac{x-1+1-y}{x-y}\\ =\dfrac{x-y}{x-y}\\ =1\)
6 tháng 1 2023
\(\dfrac{x-1}{x-y}+\dfrac{1-y}{x-y}=\dfrac{\left(x-1\right)+\left(1-y\right)}{x-y}=\dfrac{x-1+1-y}{x-y}=\dfrac{x-y-1+1}{x-y}=\dfrac{x-y}{x-y}=1\)
HH
1
28 tháng 12 2021
\(\dfrac{x+1}{x^2-4}:\dfrac{x+1}{x+2}=\dfrac{x+1}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+1}{x+2}=\dfrac{x+1}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{x+1}=x-2\)
HM
1
3 tháng 1 2021
Đặt \(x^2+1=a\)
Ta có: \(\dfrac{1}{x^2-x+1}-\dfrac{x^2+2}{x^2+1}+1\)
\(=\dfrac{1}{a-x}+\dfrac{a+1}{a}+1\)
\(=\dfrac{a}{a\left(a-x\right)}+\dfrac{\left(a+1\right)\left(a-x\right)}{a\left(a-x\right)}+\dfrac{a\left(a-x\right)}{a\left(a-x\right)}\)
\(=\dfrac{a+a^2-ax+a-x+a^2-ax}{a\left(a-x\right)}\)
\(=\dfrac{2a^2+2a-2ax-x}{a\left(a-x\right)}\)
\(=\dfrac{2\left(x^2+1\right)^2+2\left(x^2+1\right)-2x\left(x^2+1\right)-x}{\left(x^2+1\right)\left(x^2+1-x\right)}\)
\(=\dfrac{2\left(x^4+2x^2+1\right)+2x^2+2-2x^3-2x-x}{\left(x^2+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{2x^4+4x^2+2+2x^2+2-2x^3-3x}{\left(x^2+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{2x^4-2x^3+6x^2-3x+4}{\left(x^2+1\right)\left(x^2-x+1\right)}\)
26 tháng 2 2021
Ta có: \(\left(\dfrac{x+2}{3x}+\dfrac{2}{x+1}-3\right):\dfrac{2-4x}{x+1}\cdot\dfrac{3x}{x^2-3x-1}\)
\(=\left(\dfrac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\dfrac{6x}{3x\left(x+1\right)}-\dfrac{9x\left(x+1\right)}{3x\left(x+1\right)}\right):\dfrac{2-4x}{x+1}\cdot\dfrac{3x}{x^2-3x-1}\)
\(=\dfrac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}\cdot\dfrac{x+1}{2-4x}\cdot\dfrac{3x}{x^2-3x-1}\)
\(=\dfrac{-8x^2+2}{3x\left(x+1\right)}\cdot\dfrac{x+1}{2-4x}\cdot\dfrac{3x}{x^2-3x-1}\)
\(=\dfrac{-2\left(4x^2-1\right)}{3x\cdot2\cdot\left(1-2x\right)}\cdot\dfrac{3x}{x^2-3x-1}\)
\(=\dfrac{2\left(1-2x\right)\left(2x+3\right)}{6x\left(1-2x\right)}\cdot\dfrac{3x}{x^2-3x-1}\)
\(=\dfrac{2x+3}{x^2-3x-1}\)
18 tháng 11 2021
\(=\left(x-1\right)\left(x+1\right)\cdot\dfrac{x+1-x+1-x^2+1}{\left(x-1\right)\left(x+1\right)}\left(x\ne\pm1\right)\\ =3-x^2\)
MC: x(x + 1)
1/x - 1/(x + 1)
= (x + 1)/[x(x + 1)] - x/[x(x + 1)]
= (x + 1 - x)/[x(x + 1)]
= 1/[x(x + 1)]