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\(\dfrac{x-130}{20}\)+\(\dfrac{x-100}{25}\)+\(\dfrac{x-60}{30}\)+\(\dfrac{x-10}{35}\)=10
⇔\(\dfrac{2625\left(x-130\right)}{52500}\)+\(\dfrac{2100\left(x-100\right)}{52500}\)+\(\dfrac{1750\left(x-60\right)}{52500}\)+\(\dfrac{1500\left(x-10\right)}{52500}\)=\(\dfrac{525000}{52500}\)
⇔2625\(x\)-341250+2100\(x\)-210000+1750\(x\)-105000+1500\(x\)-15000=525000
⇔ 7975\(x\) = 1196250
⇔ \(x\) = \(\dfrac{1196250}{7975}\)
⇔\(x \) = 150
a: \(\Leftrightarrow\left(\dfrac{x+2001}{5}+1\right)+\left(\dfrac{x+1999}{7}+1\right)+\left(\dfrac{x+1997}{9}+1\right)+\left(\dfrac{x+1995}{11}+1\right)=0\)
=>x+2006=0
=>x=-2006
b: \(\Leftrightarrow\left(\dfrac{x-15}{100}-1\right)+\left(\dfrac{x-10}{105}-1\right)+\left(\dfrac{x-100}{5}-1\right)=\left(\dfrac{x-100}{15}-1\right)+\left(\dfrac{x-105}{10}-1\right)+\left(\dfrac{x-110}{5}-1\right)\)
=>x-105=0
=>x=105
\(\dfrac{1.2}{1^2}.\dfrac{2.3}{2^2}.\dfrac{3.4}{3^2}...\dfrac{9.10}{9^2}.\dfrac{10.11}{10^2}\left(x-2\right)=-20\left(x+1\right)+60\)
\(\Leftrightarrow\dfrac{1.2^2.3^2.4^2...10^2.11}{1^2.2^2.3^2....10^2}\left(x-2\right)+20\left(x+1\right)=60\)
\(\Leftrightarrow11\left(x-2\right)+20\left(x+1\right)=60\)
\(\Leftrightarrow31x=62\)
\(\Rightarrow x=2\)
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
$ĐKXĐ:x \neq -4;-5;-6;-7$
$pt⇔\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}$
$⇔\dfrac{1}{(x+4)(x+5)}+\dfrac{1}{(x+5)(x+6)}+\dfrac{1}{(x+6)(x+7)}=\dfrac{1}{18}$
$⇔\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}$
$⇔\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}$
$⇔\dfrac{3}{(x+4)(x+7)}=\dfrac{1}{18}$
$⇔x^2+11x+28=54$
$⇔x^2+11x-26=0$
$⇔x^2-2x+13x-26=0$
$⇔(x-2)(x+13)=0$
$⇔$ \(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)(t/m)
Vậy phương trình đã cho có tập nghiệm $S=(2;-13)$
\(\dfrac{200x}{100}+\dfrac{300\left(x-20\right)}{100}=\dfrac{33.500}{100}\)
=> 200x + 300(x - 20) = 16500
<=> 200x + 300x - 6000 = 16500
<=> 500x = 22500
<=> x = 45
S = {45}
Ta có: \(\dfrac{x\cdot200}{100}+\dfrac{\left(x-20\right)\cdot300}{100}=\dfrac{33\cdot500}{100}\)
\(\Leftrightarrow200x+300x-6000=16500\)
\(\Leftrightarrow500x=22500\)
hay x=45
Vậy: S={45}
\(\dfrac{x}{30}=\dfrac{x}{40}+\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{4x}{120}=\dfrac{3x}{120}+\dfrac{90}{120}\)
\(\Leftrightarrow4x=3x+90\)
\(\Leftrightarrow4x-3x-90=0\)
\(\Leftrightarrow x-90=0\)
\(\Leftrightarrow x=90\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{90\right\}\)
a) \(5x - 30 = 0\)
\(5x = 0 + 30\)
\(5x = 30\)
\(x = 30:5\)
\(x = 6\)
Vậy phương trình có nghiệm \(x = 6\).
b) \(4 - 3x = 11\)
\( - 3x = 11 - 4\)
\( - 3x = 7\)
\(x = \left( { 7} \right):\left( { - 3} \right)\)
\(x = \dfrac{-7}{3}\)
Vậy phương trình có nghiệm \(x = \dfrac{7}{3}\).
c) \(3x + x + 20 = 0\)
\(4x + 20 = 0\)
\(4x = 0 - 20\)
\(4x = - 20\)
\(x = \left( { - 20} \right):4\)
\(x = - 5\)
Vậy phương trình có nghiệm \(x = - 5\).
d) \(\dfrac{1}{3}x + \dfrac{1}{2} = x + 2\)
\(\dfrac{1}{3}x - x = 2 - \dfrac{1}{2}\)
\(\dfrac{{ - 2}}{3}x = \dfrac{3}{2}\)
\(x = \dfrac{3}{2}:\left( {\dfrac{{ - 2}}{3}} \right)\)
\(x = \dfrac{{ - 9}}{4}\)
Vậy phương trình có nghiệm \(x = \dfrac{{ - 9}}{4}\).
xem lại câu b nha, tại vì trên là 7 dưới -7
a, đk : x khác 5;-6
\(x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
\(\Leftrightarrow2x+61=23x+61\Leftrightarrow21x=0\Leftrightarrow x=0\)(tm)
b, đk : x khác 1;3
\(x^2+2x-15=x^2-1-8\Leftrightarrow2x-15=-9\Leftrightarrow x=3\left(ktmđk\right)\)
pt vô nghiệm
a, đk : x khác 5;-6
x2+12x+36+x2−10x+25=2x2+23x+61x2+12x+36+x2−10x+25=2x2+23x+61
⇔2x+61=23x+61⇔21x=0⇔x=0⇔2x+61=23x+61⇔21x=0⇔x=0(tm)
b, đk : x khác 1;3
x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)
pt vô nghiệm
-90
\(\dfrac{10-x}{100}\) + \(\dfrac{20-x}{110}\)+\(\dfrac{30-x}{120}\)=3
<=> \(\dfrac{10-x}{100}\)-1+\(\dfrac{20-x}{110}\)-1+\(\dfrac{30-x}{120}\)-1 = 0
<=> \(\dfrac{-x-90}{100}\)+\(\dfrac{-x-90}{110}\)+\(\dfrac{-x-90}{120}\)=0
<=> (-x-90) ( \(\dfrac{1}{100}\)+\(\dfrac{1}{110}\)+\(\dfrac{1}{120}\))=0
<=> (-x-90) = 0 ( do 1/100 +1/110+1/120 khác 0)
<=> -x-90 = 0
<=> -x = 90
<=> x =-90
Vậy nghiệm của pt là x=-90