Question

Tìm x:

a) \(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)

b) \(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}\)

Answer 1

a)\(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)

\(\left(\frac{x-1}{11}+3\right)+\left(\frac{x-1}{12}+2\right)=\left(\frac{x-1}{13}+3\right)+\left(\frac{x-1}{14}+2\right)\)

\(\left(\frac{x-1}{11}+\frac{x-1}{12}\right)+\left(3+2\right)=\left(\frac{x-1}{13}+\frac{x-1}{14}\right)+\left(3+2\right)\)

\(\frac{x-1}{11}+\frac{x-1}{12}=\frac{x-1}{13}+\frac{x-1}{14}\)

\(\frac{x-1}{11}+\frac{x-1}{12}-\frac{x-1}{13}+\frac{x-1}{14}=0\)

\(\left(x-1\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)=0\)

Vì \(\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\)\(\Rightarrow\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\ne0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

Answer 2

cái này là lớp 7 mà

phần a ra x=1

Answer 3

kiến thức nâng cao à,

ax=1

b)x=2

Answer 4

a)\(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)

\(\frac{x-1}{11}+3+\frac{x-1}{12}+2=\frac{x-1}{13}+3+\frac{x-1}{14}+2\)

\(\left(\frac{x-1}{11}+\frac{x-1}{12}\right)+\left(2+3\right)=\left(\frac{x-1}{13}+\frac{x-1}{14}\right)+\left(2+3\right)\)

\(\frac{x-1}{11}+\frac{x-1}{12}=\frac{x-1}{13}+\frac{x-1}{14}\)

\(\frac{12\left(x-1\right)}{11.12}+\frac{11\left(x-1\right)}{11.12}=\frac{14\left(x-1\right)}{13.14}+\frac{13\left(x-1\right)}{13.14}\)

\(\frac{23\left(x-1\right)}{121}=\frac{27\left(x-1\right)}{182}\)

\(\Rightarrow23\left(x-1\right).182=27\left(x-1\right)121\)

4168(x-1)=3267(x-1)

4168(x-1)-3267(x-1)=0

901(x-1)=0

x-1=0

x=0+1

x=1

Vậy x=1

Answer 5

A)x=1

Bx=3