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b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)
=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)
=>x*(x+20)=400*6=2400
=>x^2+20x-2400=0
=>(x+60)(x-40)=0
=>x=-60 hoặc x=40
c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)
=>(2x+1)^2-(2x-1)^2=8
=>4x^2+4x+1-4x^2+4x-1=8
=>8x=8
=>x=1(nhận)
a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)
\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow x+8-12+20x=0\)
\(\Leftrightarrow21x-4=0\)
\(\Leftrightarrow21x=4\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)
Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!
1: \(\Leftrightarrow\left(\dfrac{x+1}{85}+1\right)+\left(\dfrac{x+3}{83}+1\right)=\left(\dfrac{x+5}{81}+1\right)+\left(\dfrac{x+7}{79}+1\right)\)
=>x+86=0
=>x=-86
2: \(\Leftrightarrow\left(\dfrac{x-1}{2015}+1\right)-\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+7}{2007}+1\right)-\left(\dfrac{x+11}{2003}+1\right)\)
=>x+2014=0
=>x=-2014
3: \(\Leftrightarrow3\left(x+4\right)-2\left(x-3\right)=4x\)
=>4x=3x+12-2x+6
=>4x=x+18
=>3x=18
=>x=6
4: \(\Leftrightarrow15x-5\left(x+1\right)=3\left(2x+1\right)\)
=>15x-5x-5=6x+3
=>10x-5=6x+3
=>4x=8
=>x=2
5: \(\Leftrightarrow2\left(2x-7\right)+5\left(x+11\right)=-40\)
=>4x-14+5x+55=-40
=>9x+41=-40
=>x=-9
Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow x+8+20x-12=0\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
a: =>\(\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{4x}{2\left(x-3\right)\left(x+1\right)}=\dfrac{-x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}\)
=>x^2-3x-4x=-x^2-x
=>x^2-7x+x^2+x=0
=>2x^2-6x=0
=>x=0(nhận) hoặc x=3(loại)
b: =>\(\dfrac{2x-3-3x-15}{x+5}>=0\)
=>\(\dfrac{-x-18}{x+5}>=0\)
=>x+18/x+5<=0
=>-18<=x<-5
\(\dfrac{x}{2x+1}-\dfrac{2x}{x^2-2x-3}=\dfrac{x}{6-2x}\) (ĐKXĐ: \(x\ne3;x\ne-1\)
\(\Leftrightarrow\dfrac{x}{2x+1}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=-\dfrac{x}{2\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\dfrac{2.2x}{2\left(x-3\right)\left(x+1\right)}=-\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}\)
\(\Rightarrow x^2-3x-4x=-x^2-x\)
\(\Leftrightarrow x^2-7x=-x^2-x\)
\(\Leftrightarrow x^2+x^2-7x+x=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\)
*TM: Thỏa mãn, KTM: Ko thỏa mãn
Vậy phương trình có tập nghiệm là \(S=\left\{0\right\}\)
\(\dfrac{2x-3}{x+5}\ge3\) (ĐKXĐ: \(x\ne-5\)
\(\Leftrightarrow\dfrac{2x-3}{x+5}-3\ge0\)
\(\Leftrightarrow\dfrac{2x-3}{x+5}-\dfrac{3x+15}{x+5}\ge0\)
\(\Leftrightarrow\dfrac{2x-3-3x-15}{x+5}\ge0\)
\(\Leftrightarrow\dfrac{-x-18}{x+5}\ge0\)
\(\Leftrightarrow-18\le x\le-5\)
\(\dfrac{x+5}{4}-\dfrac{2x-5}{3}=\dfrac{6x-1}{3}+\dfrac{2x-3}{12}\)
\(\Rightarrow\dfrac{3\left(x+5\right)-4\left(2x-5\right)}{12}=\dfrac{4\left(6x-1\right)+2x-3}{12}\)
\(\Rightarrow3\left(x+5\right)-4\left(2x-5\right)=4\left(6x-1\right)+2x-3\)
\(\Rightarrow3x+15-8x+20=24x-4+2x-3\)
\(\Rightarrow3x-8x-24x-2x=-4-3-15-20\)
\(\Rightarrow-31x=-42\Rightarrow x=\dfrac{42}{31}\)
\(\dfrac{x+5}{4}-\dfrac{2x-3}{12}=\dfrac{6x-1}{3}+\dfrac{2x-5}{3}\\ \dfrac{3x+15-2x+3}{12}=\dfrac{6x-1+2x-5}{3}\\ \dfrac{x+18}{12}=\dfrac{8x-6}{3}\\\left(x+18\right).3=\left(8x-6\right).12\\ 3x+54=96x-72\\ 3x-96x+54+72=0\\ 126-93x=0\\ -93x=-126\\ x=\dfrac{126}{93}\\ x=\dfrac{42}{31}\)
Chúc bạn học tốt ^^
a: =>\(\dfrac{2x-4}{2014}+\dfrac{2x-2}{2016}< \dfrac{2x-1}{2017}+\dfrac{2x-3}{2015}\)
=>\(\dfrac{2x-2018}{2014}+\dfrac{2x-2018}{2016}< \dfrac{2x-2018}{2017}+\dfrac{2x-2018}{2015}\)
=>2x-2018<0
=>x<2019
b: \(\Leftrightarrow\left(\dfrac{3-x}{100}+\dfrac{4-x}{101}\right)>\dfrac{5-x}{102}+\dfrac{6-x}{103}\)
=>\(\dfrac{x-3}{100}+\dfrac{x-4}{101}-\dfrac{x-5}{102}-\dfrac{x-6}{103}< 0\)
=>\(x+97< 0\)
=>x<-97
1: Sửa đề: 2/x+2
\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+2}=\dfrac{3}{2-x}\)
=>\(\dfrac{2x+1+2x-4}{x^2-4}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
=>4x-3=-3x-6
=>7x=-3
=>x=-3/7(nhận)
2: \(\Leftrightarrow\dfrac{\left(3x+1\right)\left(3-x\right)+\left(3+x\right)\left(1-3x\right)}{\left(1-3x\right)\left(3-x\right)}=2\)
=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)
=>-6x^2+6=2(3x^2-10x+3)
=>-6x^2+6=6x^2-20x+6
=>-12x^2+20x=0
=>-4x(3x-5)=0
=>x=5/3(nhận) hoặc x=0(nhận)
3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)
=>x*19/6=35/12
=>x=35/38
Ta có: \(\dfrac{2x-5}{2015}+\dfrac{2x-6}{2016}=\dfrac{2x+3}{2007}+\dfrac{2x+4}{2006}\)
\(\Leftrightarrow\dfrac{2x-5}{2015}+1+\dfrac{2x-6}{2016}+1=\dfrac{2x+3}{2007}+1+\dfrac{2x+4}{2006}+1\)
\(\Leftrightarrow\dfrac{2x+2010}{2015}+\dfrac{2x+2010}{2016}-\dfrac{2x+2010}{2007}-\dfrac{2x+2010}{2006}=0\)
\(\Leftrightarrow\left(2x+2010\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
mà \(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\)
nên 2x+2010=0
\(\Leftrightarrow2x=-2010\)
hay x=-1005
Vậy: S={-1005}
\(\dfrac{2x-5}{2015}+\dfrac{2x-6}{2016}=\dfrac{2x+3}{2007}+\dfrac{2x+4}{2006}\)
= \(\left(\dfrac{2x-5}{2015}-1\right)+\left(\dfrac{2x-6}{2016}-1\right)=\left(\dfrac{2x+3}{2007}+1\right)+\left(\dfrac{2x+4}{2006}+1\right)\)
=\(\dfrac{2x-2010}{2015}+\dfrac{2x-2010}{2016}=\dfrac{2x+2010}{2007}+\dfrac{2x+2010}{2006}\)
=\(\dfrac{2x-2010}{2015}+\dfrac{2x-2010}{2016}-\dfrac{2x-2010}{2007}-\dfrac{2x-2010}{2006}\)=0
=(2x-2010)\(\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)\)=0
vì \(\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)\)≠0
⇒ 2x-2010=0
⇔2x=2010
⇔x=1005