Question

Tìm x : 

\(\frac{x+29}{31}+\frac{x+27}{33}=\frac{x+17}{43}+\frac{x+15}{45}\)

Answer 1

\(\frac{x+29}{31}+\frac{x+27}{33}=\frac{x+17}{43}+\frac{x+15}{45}\)

\(\frac{x+29}{31}+1+\frac{x+27}{33}+1=\frac{x+17}{43}+1+\frac{x+15}{45}+1\)

\(\frac{x+60}{31}+\frac{x+60}{33}=\frac{x+60}{43}+\frac{x+60}{45}\)

\(\left(x+60\right)\left(\frac{1}{31}+\frac{1}{33}-\frac{1}{43}-\frac{1}{45}\right)=0\)

VÌ \(\frac{1}{31}+\frac{1}{33}-\frac{1}{43}-\frac{1}{45}\ne0\)

\(\Rightarrow x+60=0\)

\(\Rightarrow x=-60\)

Answer 2

Câu hỏi của honoka sonoka - Toán lớp 6 - Học toán với OnlineMath

Answer 3

ket qua x=-60

Answer 4

\(\frac{x+29}{31}+\frac{x+27}{33}=\frac{x+17}{43}+\frac{x+15}{45}\)

\(=>\frac{x+29+31}{31}+\frac{x+27+33}{33}=\frac{x+17+43}{43}+\frac{x+15+45}{45}\)

\(=>\frac{x+60}{31}+\frac{x+60}{33}=\frac{x+60}{43}+\frac{x+60}{45}\)

\(=>\frac{x+60}{31}+\frac{x+60}{33}-\frac{x+60}{43}-\frac{x+60}{45}=0\)

\(=>\left(x+60\right).\left(\frac{1}{31}+\frac{1}{33}-\frac{1}{43}-\frac{1}{45}\right)=0\)

\(Do:\frac{1}{31}+\frac{1}{33}-\frac{1}{43}-\frac{1}{45}\ne0\)

\(=>x+60=0\)

\(=>x=-60\)

Answer 5

\(\frac{x+29}{31}+\frac{x+27}{33}=\frac{x+17}{43}+\frac{x+15}{45}\)

\(\frac{x+29}{31}+1+\frac{x+27}{33}+1=\frac{x+17}{43}+1+\frac{x+15}{45}+1\)

\(\frac{x+29}{31}+\frac{31}{31}+\frac{x+27}{33}+\frac{33}{33}=\frac{x+17}{43}+\frac{43}{43}+\frac{x+15}{45}+\frac{45}{45}\)

\(\frac{x+29+31}{31}+\frac{x+27+33}{33}=\frac{x+17+43}{43}+\frac{x+15+45}{45}\)

\(\frac{x+60}{31}+\frac{x+60}{33}=\frac{x+60}{43}+\frac{x+60}{45}\)

\(\frac{x+60}{31}+\frac{x+60}{33}-\frac{x+60}{43}-\frac{x+60}{45}=0\)

\(\left(x+60\right)\left(\frac{1}{31}+\frac{1}{33}-\frac{1}{43}-\frac{1}{45}\right)=0\)

\(Th1:x+60=0\Leftrightarrow x=-60\)

\(Th2:\frac{1}{31}+\frac{1}{33}-\frac{1}{43}-\frac{1}{43}\ne0\)

Vậy pt cs nghiệm là -60 

Answer 6

PT trên \(< =>\frac{x+29}{31}+1+\frac{x+27}{33}+1=\frac{x+17}{43}+1-\frac{x+15}{45}+1\)

\(< =>\frac{x+60}{31}+\frac{x+60}{33}-\frac{x+60}{43}-\frac{x+60}{45}=0\)

\(< =>\left(x+60\right).\left(\frac{1}{31}+\frac{1}{33}-\frac{1}{43}-\frac{1}{45}\right)=0\)

Do : \(\frac{1}{31}+\frac{1}{33}-\frac{1}{43}-\frac{1}{45}\ne0\)

\(< =>x+60=0\)

\(< =>x=-60\)

Answer 7

Nếu \(\frac{x+29}{31}\)+\(\frac{x+27}{33}\)=\(\frac{x+17}{43}\)+\(\frac{x+15}{45}\)

Thì \(\frac{x+29+31}{31}\)+\(\frac{x+27+33}{33}\)=\(\frac{x+17+43}{43}\)+\(\frac{x+15+45}{45}\)

Suy ra: \(\frac{x+60}{31}\)+\(\frac{x+60}{33}\)- (\(\frac{x+60}{43}\)+\(\frac{\left(x+60\right)}{45}\)) = 0

(x + 60) * (1/31+1/33-1/43-1/45) = 0

Vì vế 2 khác 0

Vậy x +  60 = 0

x = -60