Question

[Phạm Mai Linh]

a.\(\frac{x+13}{2006}\)+ \(\frac{x+2006}{13}\) + \(\frac{x+1}{2018}\) + 3 = 0

b.\(\frac{4}{\left(x+3\right)\cdot\left(x+7\right)}\) + \(\frac{3}{\left(x+7\right)\cdot\left(x+10\right)}\) = \(\frac{x}{\left(x+3\right)\cdot\left(x+10\right)}\)

mình đang cần rất gấp kết quả mong các bạn giải bài tìm x giúp mình

Các bạn nhớ viết cả lời giải nhé! Thank

Answer 1

[kim chi hàn quốc]

a) ta có: \(\frac{x+13}{2006}+\frac{x+2006}{13}+\frac{x+1}{2018}+3=0\)

\(\Rightarrow\frac{x+13}{2006}+1+\frac{x+2006}{13}+1+\frac{x+1}{2018}+1=0\)

\(\Rightarrow\frac{x+2019}{2006}+\frac{x+2019}{13}+\frac{x+2019}{2018}=0\)

\(\Rightarrow\left(x+2019\right)\left(\frac{1}{2006}+\frac{1}{13}+\frac{1}{2018}\right)=0\)

mà \(\frac{1}{2006}+\frac{1}{13}+\frac{1}{2018}>0\)

\(\Rightarrow x+2019=0\)

\(\Rightarrow x=-2019\)

Answer 2

[kim chi hàn quốc]

b) \(\frac{4}{\left(x+3\right)\left(x+7\right)}+\frac{3}{\left(x+7\right)\left(x+10\right)}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)

\(\Rightarrow\frac{\left(x+7\right)-\left(x+3\right)}{\left(x+3\right)\left(x+7\right)}+\frac{\left(x+10\right)-\left(x+7\right)}{\left(x+7\right)\left(x+10\right)}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)

\(\Rightarrow\frac{1}{x+3}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+10}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)

\(\Rightarrow\frac{1}{x+3}-\frac{1}{x+10}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)

\(\Rightarrow\frac{7}{\left(x+3\right)\left(x+10\right)}=\frac{x}{\left(x+3\right)\left(x+10\right)}\)

\(\Rightarrow x=7\)