Question

x/200+x+1/199+x+2/198+3=0

Answer 1

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

Answer 2

em cảm ơn ạ

Answer 3

                                               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 \(x+200=0\)

                                                           \(\Rightarrow\text{ }x=-200\)

Answer 4

\(\frac{x}{200}+\frac{x+1}{199}+\frac{x+2}{198}+3=0\)

\(\Leftrightarrow\frac{x}{200}+1+\frac{x+1}{199}+1+\frac{x+2}{198}+1=0\)

\(\Leftrightarrow\frac{200+x}{200}+\frac{x+200}{199}+\frac{x+200}{198}=0\)

\(\Leftrightarrow\left(x+200\right)\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}\ne0\right)=0\)

\(\Leftrightarrow x=-200\)

Answer 5

               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