Tìm x
\(\frac{x}{200}+\frac{x+1}{199}+\frac{x+2}{198}+3=0\)
\(\frac{x}{200}+1+\frac{x+1}{199}+1+\frac{x+2}{198}+1=0\)
\(\frac{x}{200}+\frac{200}{200}+\frac{x+1}{199}+\frac{199}{199}+\frac{x+2}{198}+\frac{198}{198}=0\)
\(\frac{x+200}{200}+\frac{x+200}{199}+\frac{x+200}{198}=0\)
\(\left(x+200\right)\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}\right)=0\)
Vì \(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}\) khác 0
Suy ra: x + 200 = 0
x = 0 - 200
x = -200
Vậy x = -200.
Hình như lớp 5 chưa học số âm :v
Bài giải
\(\frac{x}{200}+\frac{x+1}{199}+\frac{x+2}{198}+3=0\)
\(\left(\frac{x}{200}+1\right)+\left(\frac{x+1}{199}+1\right)+\left(\frac{x+2}{198}+1\right)=0\)
\(\frac{x+200}{200}+\frac{x+200}{199}+\frac{x+198}{198}=0\)
\(\left(x+200\right)\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}\right)=0\)
Do \(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}\ne0\) nên \(x+200=0\)
\(\Rightarrow\text{ }x=-200\)
\(\frac{x}{200}+\frac{x+1}{199}+\frac{x+2}{198}+3=0\)
\(\Leftrightarrow\frac{x}{200}+1+\frac{x+1}{199}+1+\frac{x+2}{198}+1=0\)
\(\Leftrightarrow\frac{200+x}{200}+\frac{x+200}{199}+\frac{x+200}{198}=0\)
\(\Leftrightarrow\left(x+200\right)\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}\ne0\right)=0\)
\(\Leftrightarrow x=-200\)
Bài làm :
Ta có :
\(\frac{x}{200}+\frac{x+1}{199}+\frac{x+2}{198}+3=0\)
\(\Leftrightarrow\left(\frac{x}{200}+1\right)+\left(\frac{x+1}{199}+1\right)+\left(\frac{x+2}{198}+1\right)=0\)
\(\Leftrightarrow\frac{200+x}{200}+\frac{x+200}{199}+\frac{x+200}{198}=0\)
\(\Leftrightarrow\left(x+200\right)\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}\right)=0\)
\(\text{Vì : }\frac{1}{200}+\frac{1}{199}+\frac{1}{198}>0\Rightarrow x+200=0\)
\(\Leftrightarrow x=-200\)
Vậy x=-200