Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(\dfrac{\left(x-1\right)^2}{x^2+x+1}+\dfrac{2x^2-4x-1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right)\cdot\dfrac{x^2+1}{x+1}\)
\(=\dfrac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{x+1}\)
\(=\dfrac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{x+1}=\dfrac{x^2+1}{x+1}\)
\(A=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2017}-\dfrac{1}{x+2018}\)
\(A=\dfrac{1}{x}-\dfrac{1}{x+2018}=\dfrac{2018}{x\left(x+2018\right)}\)
\(B=\dfrac{1}{4}\left(\dfrac{1}{x\left(x+2\right)}-\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}-\dfrac{1}{\left(x+4\right)\left(x+6\right)}+...+\dfrac{1}{\left(x+96\right)\left(x+98\right)}-\dfrac{1}{\left(x+98\right)\left(x+100\right)}\right)\)
\(B=\dfrac{1}{4}\left(\dfrac{1}{x\left(x+2\right)}-\dfrac{1}{\left(x+98\right)\left(x+100\right)}\right)=\dfrac{1}{4}\left(\dfrac{x^2+198x+9800-x^2-2x}{x\left(x+2\right)\left(x+98\right)\left(x+100\right)}\right)\)
\(B=\dfrac{196x+9800}{4x\left(x+2\right)\left(x+98\right)\left(x+100\right)}\)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+....+\dfrac{1}{\left(x+2017\right)\left(x+2018\right)}\\ =\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2017}-\dfrac{1}{x+2018}\\ =\dfrac{1}{x}-\dfrac{1}{x+2018}\\ =\dfrac{2018}{x\left(x+2018\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2016}-\dfrac{1}{x+2017}+\dfrac{1}{x+2017}-\dfrac{1}{x+2018}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+2018}\)
\(=\dfrac{2018}{x\left(x+2018\right)}\)
\(E=\left[\left(\dfrac{3}{x+1}-\dfrac{x}{x^2+2x+1}\right):\dfrac{2x^2+3x}{x^2+7x}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)
\(=\left[\left(\dfrac{3}{x+1}-\dfrac{x}{\left(x+1\right)^2}\right):\dfrac{2x^2+3x}{x^2+7x}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)
\(=\left[\left(\dfrac{2x+3}{\left(x+1\right)^2}\right):\dfrac{2x^2+3x}{x^2+7x}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)
\(=\left[\dfrac{x^2+7x}{x\left(x+1\right)^2}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)
\(=\dfrac{2x\left(2x+5\right)}{x\left(x+1\right)^2}.\dfrac{x^2+x}{3x+1}\)
\(=\dfrac{2x\left(2x+5\right)}{x\left(x+1\right)^2}.\dfrac{x^2+x}{3x+1}=\dfrac{2x\left(2x+5\right)}{\left(x+1\right)\left(3x+1\right)}\)
a)
\(\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{x+1-x}{x\left(x+1\right)}=\dfrac{1}{x\left(x+1\right)}\left(đpcm\right)\)
b)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{x+5}\\ =\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}\\ =\dfrac{1}{x}\)
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
a: \(\Leftrightarrow x^3-3x^2+3x-1-x^3+2x^2-x=5x\left(2-x\right)-11\left(x+2\right)\)
=>-x^2+2x-1=10x-5x^2-11x-22
=>-x^2+2x-1=-5x^2-x-22
=>4x^2+3x+21=0
=>PTVN
b: \(\Leftrightarrow\left(x+10\right)\left(x+4\right)+3\left(x+4\right)\left(x-2\right)=4\left(x+10\right)\left(x-2\right)\)
=>x^2+14x+40+3(x^2+2x-8)=4(x^2+8x-20)
=>x^2+14x+40+3x^2+6x-24=4x^2+32x-80
=>20x+16=32x-80
=>-12x=-96
=>x=8
c: \(\Leftrightarrow6\left(x-3\right)+7\left(x-5\right)=13x+4\)
=>6x-18+7x-35=13x+4
=>-53=4(loại)
d: =>3(2x-1)-5(x-2)=3(x+7)
=>6x-3-5x+10=3x+21
=>3x+21=x+7
=>x=-7
e: =>x^3-6x^2+12x-8-x^3-3x^2-3x-1=-9x^2+1
=>-9x^2+9x-9=-9x^2+1
=>9x=10
=>x=10/9
\(\left[\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}.\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right]:\dfrac{x-y}{x}\)
= \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\cdot\dfrac{x^3-y^3}{xy}\right)\cdot\dfrac{x}{x-y}\)
= \(\dfrac{\left(x^2-y^2\right)\left(x+y\right)-x^3+y^3}{xy\left(x+y\right)}\cdot\dfrac{x}{x-y}\)
= \(\dfrac{xy\left(x-y\right)}{y\left(x+y\right).\left(x-y\right)}\)
= \(\dfrac{x}{x+y}\)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+2017\right)\left(x+2018\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{\left(x+1\right)}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+...+\dfrac{1}{x+2017}-\dfrac{1}{x+2018}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+2018}\)
\(=\dfrac{2018}{x\left(x+2018\right)}\)
Dạng này mình làm suốt rồi, bạn cứ yên tâm.
Cảm ơn nhìu!!!^ ^