Tìm x
( – 5/2 . x + 1 ) ^ 2 – 10 = – 31/4
( – 1/2 x + 2/3 ) : -5/7 = 28 / 15
1/3 . I x – 2 I – 3/7 = 4/5
Tìm x ( – 5/2 . x + 1 ) ^ 2 – 10 = – 31/4 ( – 1/2 x + 2/3 ) : -5/7 = 28 / 15 1/3 . I x – 2 I – 3/7 = 4/5
By Jade
By Jade
Tìm x
( – 5/2 . x + 1 ) ^ 2 – 10 = – 31/4
( – 1/2 x + 2/3 ) : -5/7 = 28 / 15
1/3 . I x – 2 I – 3/7 = 4/5
Đáp án + giải thích bước giải :
$1/$ `( (-5)/2x + 1)^2 – 10 = (-31)/4`
`⇔ ( (-5)/2x + 1)^2 = 9/4`
`⇔ ( (-5)/2x + 1)^2 = (3/2)^2` hoặc `( (-5)/2x + 1)^2 = (-3/2)^2`
`⇔ (-5)/2x + 1 = 4/2` hoặc `(-5)/2x + 1 = (-3)/2`
`⇔ (-5)/2x = 1` hoặc `(-5)/2x = (-5)/2`
`⇔ x = (-2)/5` hoặc `x = 1`
$2/$ `( (-1)/2x + 2/3) : (-5)/7 = 28/15`
`⇔ (-1)/2x + 2/3 = (-4)/3`
`⇔ (-1)/2x = -2`
`⇔ x = 4`
$3/$ `1/3 |x – 2| – 3/7 = 4/5`
`⇔ 1/3 |x – 2| = 43/35`
`⇔ |x – 1| = 129/35`
`⇔ x – 1 = 129/35` hoặc `x- 1 = (-129)/35`
`⇔ x = 164/35` hoặc `x = (-94)/35`
$#Đáp án + Giải thích các bước giải:$
`1.`
`( – 5/2x + 1 ) ^ 2 – 10 = – 31/4`
`→( – 5/2x + 1 ) ^ 2= 9/4`
+ TH`1:`
`( – 5/2x + 1 ) ^ 2= 9/4`
`→( – 5/2x + 1 ) ^ 2= (3/2)^2`
`→ – 5/2x + 1 = 4/2`
`→ – 5/2 x = 1`
`→ x=-2/5`
+ TH`2:`
`( – 5/2x + 1 ) ^ 2= 9/4`
`→( – 5/2x + 1 ) ^ 2= (-3/2)^2`
`→ – 5/2x + 1 = -3/2`
`→ – 5/2 x = -5/2`
`→ x=1`
Vậy `x∈{-2/5;1}`
————————————————-
`2.`
`( – 1/2 x + 2/3 ) : -5/7 = 28 / 15`
`→- 1/2 x + 2/3 = -4/3`
`→- 1/2 x = -2`
`→x=4`
Vậy `x=4`
—————————————————-
`1/3 | x – 2 | – 3/7 = 4/5`
`→1/3 | x – 2 | = 43/35`
`→ x – 1 = 129/35`
+ TH`1:`
`x – 1 = 129/35`
`→ x = 164/35`
+ TH`2:`
`x – 1 = -129/35`
`→x = -94/35`
Vậy `x∈{164/35 ; -94/35}`
$#Cam$
$#XIN HAY NHẤT CHO NHÓM AK$