1.(x+14)^3-(x+12)^3=1352
2.(x-1)^4+(x+3)^4=32
3.(x+3)×(x+4)×(x+5)=x
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12 tháng 10 2021
11: Ta có: \(\left(x+3\right)^3=125\)
\(\Leftrightarrow x+3=5\)
hay x=2
12: Ta có: \(\left(2x\right)^4=16\)
\(\Leftrightarrow x^4=1\)
hay \(x\in\left\{1;-1\right\}\)
7 tháng 3 2021
b) Ta có: \(x^3+4x+5=0\)
\(\Leftrightarrow x^3-x+5x+5=0\)
\(\Leftrightarrow x\left(x^2-1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+5\right)=0\)
mà \(x^2-x+5>0\forall x\)
nên x+1=0
hay x=-1
Vậy: S={-1}
18 tháng 3 2021
a)x2-(x+3)(3x+1)=9
⇔(x-3)(x+3)-(x+3)(3x+1)=0
⇔x+3=0 hoặc 3x+1=0
1.x+3=0 ⇔x=-3
2.3x+1=0⇔x=-1/3
phương trình có 2 nghiệm x=-3 và x=-1/3
12 tháng 5 2022
Bài 1:
1: =-5/24+16/27+3/4
=-5/24+18/24+16/27
=13/24+16/27
=117/216+128/216=245/216
2: =-1/3+1/3+6/7=6/7
3: \(=\dfrac{1}{2}-\dfrac{7}{12}+\dfrac{1}{2}=1-\dfrac{7}{12}=\dfrac{5}{12}\)
4: \(=-\dfrac{5}{8}+\dfrac{14}{25}-\dfrac{6}{10}=\dfrac{-125+112-120}{200}=\dfrac{-133}{200}\)
VT
1
H
10 tháng 7 2023
`(x+2)-2=0`
`=>x+2=0+2`
`=>x+2=2`
`=>x=2-2`
`=>x=0`
__
`(x+3)+1=7`
`=>x+3=7-1`
`=>x+3=6`
`=>x=6-3`
`=>x=3`
__
`(x+3)+4=12`
`=>x+3=12-4`
`=>x+3=8`
`=>x=8-3`
`=>x=5`
__
`(5x+4)-1=13`
`=>5x+4=13+1`
`=>5x+4=14`
`=>5x=14-4`
`=>5x=10`
`=>x=10:5`
`=>x=2`
__
`(4x-8)+3=12`
`=>4x-8=12-3`
`=>4x-8=9`
`=>4x=9+8`
`=>4x=17`
`=> x=17/4`
__
`3+(x-5)=14`
`=>x-5=14-3`
`=>x-5=11`
`=>x=11+5`
`=>x=16`
8 tháng 2 2022
6: =>x=9/10+1/5=9/10+2/10=11/10
7: =>x=3/8-5/12=36/96-40/96=-1/24
8: =>x=7/6-5/4=14/12-15/12=-1/12
9: =>x=1/35+2/7=1/35+10/35=11/35
10: =>x=2/7-7/10=20/70-49/70=-29/70
H
4 tháng 7 2021
a)\(1\dfrac{3}{4}x-5=3\dfrac{1}{3}\text{⇔}\dfrac{7}{4}x-5=\dfrac{10}{3}\text{⇔}\dfrac{7}{4}x=\dfrac{25}{3}\text{⇔}x=\dfrac{100}{21}\)
b)\(\dfrac{2}{3}x+\dfrac{1}{4}=\dfrac{7}{12}\text{⇔}\dfrac{2}{3}x=\dfrac{1}{3}\text{⇔}x=\dfrac{1}{2}\)
c)\(\dfrac{1}{3}+\dfrac{2}{5}\left(x+1\right)=1\text{⇔}\dfrac{2}{5}\left(x+1\right)=\dfrac{2}{3}\text{⇔}x+1=\dfrac{5}{3}\text{⇔}x=\dfrac{2}{3}\)
d)\(\dfrac{1}{4}+\dfrac{1}{3}:3x=-5\text{⇔}\dfrac{1}{3}:3x=-\dfrac{21}{4}\text{⇔}\dfrac{1}{9x}=-\dfrac{21}{4}\text{⇔}9x=-\dfrac{4}{21}\text{⇔}x=-\dfrac{4}{189}\)
4 tháng 7 2021
a, \(1\dfrac{3}{4}x-5=3\dfrac{1}{3}\)
\(\Rightarrow\dfrac{7}{4}x=5+\dfrac{10}{3}=\dfrac{25}{3}\)
\(\Rightarrow x=\dfrac{25}{3}:\dfrac{7}{4}=\dfrac{100}{21}\)
Vậy ...
b, \(PT\Leftrightarrow\dfrac{2}{3}x=\dfrac{7}{12}-\dfrac{1}{4}=\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{1}{3}:\dfrac{2}{3}=\dfrac{1}{2}\)
Vậy ....
c, \(PT\Leftrightarrow\dfrac{2}{5}\left(x+1\right)=1-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\Rightarrow x+1=\dfrac{2}{3}:\dfrac{2}{5}=\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{5}{3}-1=\dfrac{2}{3}\)
Vậy ...
d, \(PT\Leftrightarrow\dfrac{1}{3}:3x=-5-\dfrac{1}{4}=\dfrac{1}{9}x=-\dfrac{21}{4}\)
\(\Rightarrow x=-\dfrac{189}{4}\)
Vậy ...
31 tháng 7 2023
1) 2(x + 5) + 3(x + 7) = 41
2x + 10 + 3x + 21 = 41
5x + 31 = 41
5x = 10
x = 2
6) 7(x - 1) + 5(3 - x) = 11x - 10
7x - 7 + 15 - 5x = 11x - 10
2x + 8 = 11x - 10
-9x = -18
x = 2
2) 5(x + 6) + 2(x - 3) = 38
5x + 30 + 2x - 6 = 38
7x + 24 = 38
7x = 14
x = 2
7) 4(2 + x) + 3(x - 2) = 12
8 + 4x + 3x - 6 = 12
7x + 2 = 12
7x = 10
x = 10/7
3) 7(5 + x) + 2(x - 10) = 15
35 + 7x + 2x - 20 = 15
9x + 15 = 15
9x = 0
x = 0
8) 5(2 + x) + 4(3 - x) = 10x - 15
10 + 5x + 12 - 4x = 10x - 15
x + 22 = 10x - 15
9x = 37
x = 37/9
4) 3(x + 4) + (8 - 2x) = 22
3x + 12 + 8 - 2x = 22
x + 20 = 22
x = 2
9) 7(x - 2) + 5(3 - x) = 11x - 6
7x - 14 + 15 - 5x = 11x - 6
2x + 1 = 11x - 6
-9x = -7
x = 7/9
5) 4(x + 5) + 3(7 - x) = 49
4x + 20 + 21 - 3x = 49
x + 41 = 49
x = 8
10) 5(3 - x) + 5(x + 4) = 6 + 4x
15 - 5x + 5x + 20 = 6 + 4x
35 = 6 + 4x
4x = 29
x = 29/4
Cảm ơn bạn nhé
3: Đặt x+3=a
Ta có: (x+3)(x+4)(x+5)=x
⇔a(a+1)(a+2)=a-3
⇔\(a^3+3a^2+2a-a+3=0\)
\(\Leftrightarrow a^3+3a^2+a+3=0\)
\(\Leftrightarrow a^2\left(a+3\right)+\left(a+3\right)=0\)
\(\Leftrightarrow\left(a+3\right)\left(a^2+1\right)=0\)(1)
Ta có: \(a^2\ge0\forall a\)
\(\Rightarrow a^2+1\ge1>0\forall a\)(2)
Từ (1) và (2) suy ra a+3=0
hay \(x+6=0\)
⇔x=-6
Vậy: x=-6