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14 tháng 7 2018

400/x = 100/x + 300/x + 10 + 1
(=) 400/x = 100/x + 300/x + 10x/x + x/x = 0
(=) 400/x - 100/x - 300/x - 10x/x - x/x = 0
(=) (400 - 100 - 300 - 10x - x )/x = 0
(=) -11x/x = 0
(=) 11x/x = 0
=) 11x = 0
(=) x=0
 

14 tháng 7 2018

x phải khác 0 thì mới thỏa măn ĐKXĐ của phương trình.

Vậy phương trình trên vô nghiệm

14 tháng 7 2018

\(\frac{400}{x}=\frac{100}{x}+\frac{300}{x}+10+\)\(1\)

<=> \(\frac{400-100-300}{x}=11\)

<=> \(\frac{0}{x}=11\)

AH
Akai Haruma
Giáo viên
19 tháng 4 2021

Lời giải:
PT $\Leftrightarrow \frac{x-342}{15}-1+\frac{x-323}{17}-2+\frac{x-300}{19}-3+\frac{x-273}{21}-4=0$

$\Leftrightarrow \frac{x-357}{15}+\frac{x-357}{17}+\frac{x-357}{19}+\frac{x-357}{21}=0$

$(x-357)\left(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\right)=0$

Dễ thấy: $\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\neq 0$

$\Rightarrow x-357=0$

$\Rightarrow x=357$

 

4 tháng 2 2020

\(\left(x^2-x+1\right)+\left(x^2-2x+3\right)+...+\left(x^2-100x+199\right)=300\)

\(\Leftrightarrow100x^2-100x+\frac{\left[\left(199-1\right):2+1\right]\left(199+1\right)}{2}=300\)

\(\Leftrightarrow100x^2-100x+10000=300\)

\(\Leftrightarrow100x^2-100x+9700=0\)

\(\Leftrightarrow100\left(x^2-x+97\right)=0\)

\(\Leftrightarrow x^2-x+97=0\)

\(\Leftrightarrow x^2-2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+97=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{387}{4}=0\left(1\right)\)

Vì \(\left(x-\frac{1}{2}\right)^2\ge0;\forall x\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{387}{4}\ge\frac{387}{4}>0;\forall x\)

\(\Rightarrow\)pt\(\left(1\right)\)vô nghiệm

Vậy pt trên vô nghiệm

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

\(\dfrac{x-130}{20}\)+\(\dfrac{x-100}{25}\)+\(\dfrac{x-60}{30}\)+\(\dfrac{x-10}{35}\)=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: =>\(\dfrac{2x-4}{2014}+\dfrac{2x-2}{2016}< \dfrac{2x-1}{2017}+\dfrac{2x-3}{2015}\)

=>\(\dfrac{2x-2018}{2014}+\dfrac{2x-2018}{2016}< \dfrac{2x-2018}{2017}+\dfrac{2x-2018}{2015}\)

=>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