Question

Tìm x biết:

\(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}=\dfrac{x+1}{5}+\dfrac{x+1}{6}\)

CÒN NỮA NÈ, HELP ME!!!!!!!!!!

Answer 1

\(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}=\dfrac{x+1}{5}+\dfrac{x+1}{6}\)

\(\Leftrightarrow\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}-\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)

Mà \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\ne0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy ..

Answer 2

\(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}=\dfrac{x+1}{5}+\dfrac{x+1}{6}\)

=> \(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}-\dfrac{x+1}{5}-\dfrac{x+1}{6}\)= 0

(x + 1).(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\)) = 0

Ta thấy \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\) > 0

=> x + 1 = 0

x = 0 - 1

x = -1

Answer 3

sai rui ahihi

Answer 4

Answer 5

Ta có: \(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}=\dfrac{x+1}{5}+\dfrac{x+1}{6}\)

\(\Rightarrow\left(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}\right)-\left(\dfrac{x+1}{5}+\dfrac{x+1}{6}\right)=0\)\(\Rightarrow\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}-\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)

\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)

Vì \(\dfrac{1}{2}>\dfrac{1}{3}>\dfrac{1}{4}>\dfrac{1}{5}>\dfrac{1}{6}\) nên \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\ne0\)

Do đó, \(x+1=0\)

\(\Rightarrow x=0-1=-1\)

Vậy \(x=-1\)

Answer 6

Ta có :

\(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}=\dfrac{x+1}{5}+\dfrac{x+1}{6}\)

\(\Rightarrow\left(x+1\right).\dfrac{1}{2}+\left(x+1\right).\dfrac{1}{3}+\left(x+1\right).\dfrac{1}{4}=\left(x+1\right).\dfrac{1}{5}+\left(x+1\right).\dfrac{1}{6}\)

\(\Rightarrow\left(x+1\right).\dfrac{1}{2}+\left(x+1\right).\dfrac{1}{3}+\left(x+1\right).\dfrac{1}{4}-\left(x+1\right).\dfrac{1}{5}-\left(x+1\right).\dfrac{1}{6}=0\)

\(\Rightarrow\left(x+1\right).\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)

\(\Rightarrow\left(x+1\right).\dfrac{43}{60}=0\)

Mà \(\dfrac{43}{60}\ne0\)

\(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

Vậy \(x=-1\)