A =\(\frac{1}{\left(3x\right)}+\frac{1}{6x}+\frac{1}{12x}+\frac{1}{24x}+\frac{1}{48x}+\frac{1}{96x}\)
Hãy tính kết quả của A
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3 tháng 2 2022
a: \(A=\left(\dfrac{2\left(2x+1\right)}{2\left(2x+4\right)}-\dfrac{x}{3x-6}-\dfrac{2x^3}{3x^3-12x}\right):\dfrac{6x+13x^2}{24x-12x^2}\)
\(=\left(\dfrac{2x+1}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^3}{3x\left(x^2-4\right)}\right):\dfrac{x\left(13x+6\right)}{x\left(24-12x\right)}\)
\(=\left(\dfrac{2x+1}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^2}{3\left(x-2\right)\left(x+2\right)}\right):\dfrac{13x+6}{-12\left(x-2\right)}\)
\(=\dfrac{3\left(2x+1\right)\left(x-2\right)-2x\left(x+2\right)-4x^2}{6\left(x+2\right)\left(x-2\right)}\cdot\dfrac{-12\left(x-2\right)}{13x+6}\)
\(=\dfrac{3\left(2x^2-3x-2\right)-2x^2-4x-4x^2}{x-2}\cdot\dfrac{-2}{13x+6}\)
\(=\dfrac{6x^2-9x-6-6x^2-4x}{x-2}\cdot\dfrac{-2}{13x+6}\)
\(=\dfrac{-\left(13x+6\right)\cdot\left(-2\right)}{\left(13x+6\right)\left(x-2\right)}=\dfrac{2}{x-2}\)
b: Để A>0 thì x-2>0
hay x>2
Để A>-1 thì A+1>0
\(\Leftrightarrow\dfrac{2+x-2}{x-2}>0\)
=>x/x-2>0
=>x>2 hoặc x<0
DJ
1
30 tháng 3 2020
\(\frac{12x+1}{6x-2}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(9x^2-1\right)}\)
đkxđ \(x\ne\pm\frac{1}{3}\)
\(\Leftrightarrow\frac{12x+1}{2\left(3x-1\right)}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(3x-1\right)\left(3x+1\right)}\)
\(\Leftrightarrow\frac{\left(24x+2\right)\left(3x+1\right)}{4\left(3x-1\right)\left(3x+1\right)}-\frac{\left(36x-20\right)\left(3x-1\right)\left(3x+1\right)}{4\left(3x-1\right)\left(3x+1\right)}=\frac{-36x^2+10x-9}{4\left(3x-1\right)\left(3x+1\right)}\)
\(\Leftrightarrow72x^2+6x+24x+2-108x^2+60x+36x-20-108x+36x^2+9=0\)
\(\Leftrightarrow18x-9=0\)
\(\Leftrightarrow18x=9\)
\(\Leftrightarrow x=\frac{1}{2}\left(tmđk\right)\)
QA
1
28 tháng 4 2020
a) ĐKXĐ: \(x\notin\pm\frac{1}{3}\)
Ta có: \(\frac{12x+1}{6x-2}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(9x^2-1\right)}\)
\(\Leftrightarrow\frac{12x+1}{2\left(3x-1\right)}-\frac{9x-5}{3x+1}=\frac{9\left(12x-4x^2-1\right)}{4\left(9x^2-1\right)}\)
\(\Leftrightarrow\frac{2\left(12x+1\right)\left(3x+1\right)}{4\left(3x-1\right)\left(3x+1\right)}-\frac{4\left(9x-5\right)\left(3x-1\right)}{4\left(3x+1\right)\left(3x-1\right)}=\frac{9\left(12x-4x^2-1\right)}{4\left(3x+1\right)\left(3x-1\right)}\)
\(\Leftrightarrow72x^2+30x+2-\left(108x^2-96x+20\right)=108x-36x^2-9\)
\(\Leftrightarrow72x^2+30x+2-108x^2+96x-20-108x+36x^2+9=0\)
\(\Leftrightarrow18x-9=0\)
\(\Leftrightarrow9\left(2x-1\right)=0\)
mà 9≠0
nên 2x-1=0
⇔2x=1
hay \(x=\frac{1}{2}\)(tm)
Vậy: \(x=\frac{1}{2}\)
b)ĐKXĐ: x≠0
Ta có: \(x+\frac{1}{x}=x^2+\frac{1}{x^2}\)
\(\Leftrightarrow x+\frac{1}{x}-x^2-\frac{1}{x^2}=0\)
\(\Leftrightarrow\frac{x^3}{x^2}+\frac{x}{x^2}-\frac{x^4}{x^2}-\frac{1}{x^2}=0\)
\(\Leftrightarrow x^3+x-x^4-1=0\)
\(\Leftrightarrow x^3\left(1-x\right)+\left(x-1\right)=0\)
\(\Leftrightarrow x^3\left(1-x\right)-\left(1-x\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(x-1\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow-\left(x-1\right)^2\cdot\left(x^2+x+1\right)=0\)(1)
Ta có: \(x^2+x+1=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)(2)
Từ (1) và (2) suy ra x-1=0
hay x=1(tm)
Vậy: x=1
c) ĐKXĐ: x≠0
Ta có: \(\frac{1}{x}+2=\left(\frac{1}{x}+2\right)\left(x^2+2\right)\)
\(\Leftrightarrow\frac{1}{x}+2-\left(\frac{1}{x}+2\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}+2\right)\left(2-x^2-2\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}+2\right)\cdot\left(-x^2\right)=0\)(3)
Ta có: 1≠0
x≠0
Do đó: \(\frac{1}{x}\ne0\)
\(\Leftrightarrow\frac{1}{x}+2\ne0\)(4)
Từ (3) và (4) suy ra x=0(ktm)
Vậy: x∈∅
d) ĐKXĐ: x≠0
Ta có: \(\left(x+1+\frac{1}{x}\right)^2=\left(x-1-\frac{1}{x}\right)^2\)
\(\Leftrightarrow\left(x+1+\frac{1}{x}\right)^2-\left(x-1-\frac{1}{x}\right)^2=0\)
\(\Leftrightarrow\left(x+1+\frac{1}{x}+x-1-\frac{1}{x}\right)\left(x+1+\frac{1}{x}-x+1+\frac{1}{x}\right)=0\)
\(\Leftrightarrow2x\cdot\left(2+\frac{2}{x}\right)=0\)
\(\Leftrightarrow4x\left(1+\frac{1}{x}\right)=0\)
mà 4≠0
và x≠0
nên \(1+\frac{1}{x}=0\)
\(\Leftrightarrow\frac{1}{x}=-1\)
hay x=-1(tm)
Vậy: x=-1
21 tháng 7 2019
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
= \(\frac{3x\left(x-y\right)}{5.2.\left(x+y\right)\left(x-y\right)}-\frac{x\left(x+y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x^2-3xy-x^2-xy}{10\left(x^2-y^2\right)}\)
= \(\frac{3x\left(x-y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x}{10\left(x+y\right)}\)
A
2
DE
17 tháng 3 2020
\(A=\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right)\times\frac{x^2-36}{12x^2+12}\)
\(A=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right]\times\frac{\left(x+6\right)\left(x-6\right)}{12\left(x^2+1\right)}\)
\(A=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x}\times\frac{1}{12\left(x^2+1\right)}\)
\(A=\frac{12\left(x^2+1\right)}{x}\times\frac{1}{12\left(x^2+1\right)}=\frac{1}{x}\)