Toán Lớp 8: Giải phg trình
b, $\frac{x-1}{13}$ – $\frac{2x-13}{15}$= $\frac{3x-15}{27}$- $\frac{4x-27}{29}$
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Toán Lớp 8: Giải phg trình
b, $\frac{x-1}{13}$ – $\frac{2x-13}{15}$= $\frac{3x-15}{27}$- $\frac{4x-27}{29}$
Comments ( 2 )
(x – 1)/13 – (2x – 13)/15 = (3x – 15)/27 – (4x – 27)/29
⇔ (x – 1)/13 – 1 – (2x – 13)/15 + 1 = (3x – 15)/27 – 1 – (4x – 27)/29 + 1
⇔ ((x – 1)/13 – 1) – ((2x – 13)/15 – 1) = ((3x – 15)/27 – 1) – ((4x – 27)/29 – 1)
⇔ (x – 1 – 13)/13 – (2x – 13 – 15)/15 = (3x – 15 – 27)/27 – (4x – 27 – 29)/29
⇔ (x – 14)/13 – (2x – 28)/15 = (3x – 42)/27 – (4x – 56)/29
⇔ (x – 14)/13 – (2(x – 14))/15 = (3(x – 14))/27 – (4(x – 14))/29
⇔ (x – 14)/13 – (2(x – 14))/15 – (3(x – 14))/27 + (4(x – 14))/29 = 0
⇔ (x – 14)(1/13 – 2/15 – 3/27 + 4/29) = 0
Vì: 1/13 – 2/15 – 3/27 + 4/29 $\neq$ 0
⇒ x – 14 = 0
⇔ x = 14
(x-1)/13-(2x-13)/15=(3x-15)/27 – (4x-27)/29
<=>(x-1-13)/13 – (2x-13 -15)/15 = (3x-15-27)/27 – (4x-27 – 29)/29
<=>(x-14)/13 – (2(x-14))/15 – (3(x-14))/27 + (4(x-14))/29=0
<=> (x-14)(1/13 – 2/15 – 3/27 + 4/29)=0
Do 1/13 – 2/15-3/27 +4/29 \ne 0
<=>x-14=0
<=>x=14
Vậy S={14}