Phương trình bậc I' 1 ẩn
`(x+1)/35 + (x+3)/33 = (x+5)/31 + (x+7)/29`
`<=> (x+1)/35 + 1 + (x+3)/33 +1 = (x+5)/31+1 + (x+7)/29+1`
`<=> (x+1)/35+ 35/35 + (x+3)/33 + 33/33 = (x+5)/31 + 31/31 + (x+7)/29 + 29/29`
`<=> (x+36)/35 + (x+36)/33 = (x+36)/31 + (x+36)/29`
`<=> (x+36)/35 + (x+36)/33 -(x+36)/31 - (x+36)/29=0`
`<=> (x+36)(1/35 + 1/33 - 1/31 - 1/29)=0`
Vì `1/35 + 1/33 - 1/31 - 1/29 ne 0`
`=> x+36 = 0`
`<=> x= -36`
Vậy tập nghiệm của phương trình là `S ={-36}`
\(\dfrac{x+1}{35}+\dfrac{x+3}{33}=\dfrac{x+5}{31}+\dfrac{x+7}{29}\)
⇔\(\dfrac{x+1}{35}+1+\dfrac{x+3}{33}+1=\dfrac{x+5}{31}+1+\dfrac{x+7}{29}+1\)
⇔\(\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)
⇔\(\dfrac{x+36}{35}+\dfrac{x+36}{33}-\dfrac{x+36}{31}-\dfrac{x+36}{29}=0\)
⇔\(\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{33}-\dfrac{1}{31}-\dfrac{1}{29}\right)=0\)
-Vì \(\dfrac{1}{35}+\dfrac{1}{33}-\dfrac{1}{31}-\dfrac{1}{29}\ne0\) nên phương trình đã cho tương đương:
\(x+36=0\)
⇔\(x=-36\).
-Vậy \(S=\left\{-36\right\}\)
; 
;2x2 + 3 = 0 ; 4– 0,2x = 0
