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\(\Leftrightarrow\frac{1}{x^2+7x+12}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+3x+2}-\frac{1}{10}=0\)
\(\Rightarrow-\frac{x^2+5x-26}{10\left(x+1\right)\left(x+4\right)}=0\)
\(\Rightarrow x^2+5x-26=0\)
\(\Rightarrow5^2-\left(-4\left(1.26\right)\right)=129\)(cái này là D)
\(\Rightarrow x_{1,2}=\frac{-b+-\sqrt{D}}{2a}=\frac{-5+-\sqrt{129}}{2}\)
\(x=+-\frac{\sqrt{129}}{2}-2\frac{1}{2}\)
b/ (x + 5)(x + 2) - 3(4x - 3) = (5 - x)2
=> x2 + 7x + 10 - 12x + 9 = 25 - 10x + x2
=> x2 + 7x + 10 - 12x + 9 - 25 + 10x - x2 = 0
=> 5x - 6 = 0
=> x = 6/5
\(a,\dfrac{1+x}{2017}+\dfrac{2+x}{2016}+\dfrac{3+x}{2015}=-3\)
\(\Leftrightarrow\dfrac{1+x}{2017}+1+\dfrac{2+x}{2016}+1+\dfrac{3+x}{2015}+1=-3+3\)
\(\Leftrightarrow\dfrac{1+x+2017}{2017}+\dfrac{2+x+2016}{2016}+\dfrac{3+x+2015}{2015}=0\)
\(\Leftrightarrow\dfrac{x+2018}{2017}+\dfrac{x+2018}{2016}+\dfrac{x+2018}{2015}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\right)=0\)
\(\Leftrightarrow x+2018=0\)
\(\Leftrightarrow x=-2018\)
b,\(\dfrac{x-\dfrac{3x-4}{5}}{15}=\dfrac{5x-\dfrac{3-x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{\dfrac{5x-3x+4}{5}}{15}=\dfrac{\dfrac{10x-3+x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{\dfrac{2x+4}{5}}{15}=\dfrac{\dfrac{11x-3}{2}}{5}-\dfrac{5x-5}{5}\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{11x-3}{10}-\dfrac{10x-10}{10}\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{11x-3-10x+10}{10}\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{x+7}{10}\)
\(\Leftrightarrow10\left(2x+4\right)=75\left(x+7\right)\)
\(\Leftrightarrow20x+40=75x+525\)
\(\Leftrightarrow20x-75x=525-40\)
\(\Leftrightarrow-55x=485\)
\(\Leftrightarrow x=-\dfrac{97}{11}\)
a) \(\dfrac{1+x}{2017}+\dfrac{2+x}{2016}+\dfrac{3+x}{2015}=-3\)
\(\Leftrightarrow\dfrac{1+x}{2017}+1+\dfrac{2+x}{2016}+1+\dfrac{3+x}{2015}+1=0\)
\(\Leftrightarrow\dfrac{x+2018}{2017}+\dfrac{x+2018}{2016}+\dfrac{x+2018}{2015}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\right)=0\)
\(\Rightarrow x+2018=0\)
\(\Leftrightarrow x=-2018\)
b) \(\dfrac{x-\dfrac{3x-4}{5}}{15}=\dfrac{5x-\dfrac{3-x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{\dfrac{5x-3x+4}{5}}{15}=\dfrac{\dfrac{10x-3+x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{11x-3}{10}-x+1\)
\(\Leftrightarrow\dfrac{4x+8}{150}=\dfrac{165x-45}{150}-\dfrac{150x-150}{150}\)
\(\Leftrightarrow4x+8=165x-45-150x+150\)
\(\Leftrightarrow4x-165x+150x=-45+150-8\)
\(\Leftrightarrow-11x=97\)
\(\Leftrightarrow x=-\dfrac{97}{11}\)
\(S=\left\{-\dfrac{97}{11}\right\}\)
\(x^3-x^2+x^2-x+6x-6=0\Leftrightarrow\left(x-1\right)\left(x^2-x+6\right)=0\Leftrightarrow\left(x-1\right)=0\Leftrightarrow x=2;x^2-x+6>0\)
\(4x^2-12x+9=9-5\Leftrightarrow\left(2x-3\right)^2-4=0\Leftrightarrow\left(2x-1\right)\left(2x-5\right)=0\Leftrightarrow x=\frac{1}{2};x=\frac{5}{2}\)
khó ( x =2040)
a/ x.(x + 1)(x2 + x + 1) = 42
=> (x2 + x)(x2 + x + 1) = 42
Đặt a = x2 + x ta đc:
a.(a + 1) = 42
=> a2 + a - 42 = 0
=> (a - 6)(a + 7) = 0
=> a = 6 hoặc a = -7
Với a = 6 => x2 + x = 6 => x2 + x - 6 = 0 => (x - 2)(x + 3) = 0 => x = 2 hoặc x = -3
Với a = -7 => x2 + x = -7 => x2 + x + 7 = 0 , mà x2 + x + 7 > 0 => pt vô nghiệm
Vậy x = 2 , x = -3
b/ (3x - 1)2 - 5(2x + 1)2 + (6x - 3)(2x + 1) = (x - 1)2
=> 9x2 - 6x + 1 - 5.(4x2 + 4x + 1) + (12x2 - 3) = x2 - 2x + 1
=> 9x2 - 6x + 1 - 20x2 - 20x - 5 + 12x2 - 3 - x2 + 2x - 1 = 0
=> - 24x - 8 = 0
=> -24x = 8
=> x = -1/3
Vậy x = -1/3
pt <=> (x/2012 - 1) + (x+1/2013 - 1) + (x+2/2014 - 1) + (x+3/2015 - 1) + (x+4/2016 - 1) = 0
<=> x-2012/2012 + x-2012/2013 + x-2012/2014 + x-2012/2015 + x-2012/2016 = 0
<=> (x-2012).(1/2012+1/2013+1/2014+1/2015+1/2016) = 0
<=> x-2012 = 0 ( vì 1/2012+1/2013+1/2014+1/2015+1/2016 > 0 )
<=> x=2012
Vậy x=2012
Tk mk nha
Ta có :
\(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)
\(\Leftrightarrow\)\(\left(\frac{x}{2012}-1\right)+\left(\frac{x+1}{2013}-1\right)+\left(\frac{x+2}{2014}-1\right)+\left(\frac{x+3}{2015}-1\right)+\left(\frac{x+4}{2016}-1\right)=5-5\)
\(\Leftrightarrow\)\(\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)
\(\Leftrightarrow\)\(\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\ne0\)
\(\Rightarrow\)\(x-2012=0\)
\(\Rightarrow\)\(x=2012\)
Vậy \(x=2012\)
Chúc bạn học tốt ~
câu b nè
\(\frac{3x+1}{\left(x-1\right)^2}-\frac{1}{x+1}-\frac{x+3}{x^2-1}\)
=\(\frac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
=\(\frac{\left(3x^2+x+3x+1\right)-\left(x^2-2x+1\right)-\left(x^2-x-3+3x\right)}{\left(x-1\right)^2\left(x+1\right)}\)
=\(\frac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}=\frac{x^2+4x+3}{\left(x+1\right)\left(x-1^2\right)}\)
=\(\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x-1\right)^2}=\frac{x+3}{\left(x-1\right)^2}\)