13-x(2-12)=2021-x(3+18)
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1. Giải:
Do \(5x+13B\in\left(2x+1\right)\Rightarrow5x+13⋮2x+1.\)
\(\Rightarrow2\left(5x+13\right)⋮2x+1\Rightarrow10x+26⋮2x+1.\)
\(\Rightarrow5\left(2x+1\right)+21⋮2x+1.\)
Do 5(2x+1)⋮2x+1⇒ Ta cần 21⋮2x+1.
⇒ 2x+1 ϵ B(21)=\(\left\{1;3;7;21\right\}.\)
Ta có bảng:
2x+1 | 1 | 3 | 7 | 21 |
x | 0 | 1 | 3 | 10 |
TM | TM | TM | TM |
Vậy xϵ\(\left\{0;1;3;10\right\}.\)
2. Giải:
Do (2x-18).(3x+12)=0.
⇒ 2x-18=0 hoặc 3x+12=0.
⇒ 2x =18 3x =-12.
⇒ x =9 x =-4.
Vậy xϵ\(\left\{-4;9\right\}.\)
3. S= 1-2-3+4+5-6-7+8+...+2021-2022-2023+2024+2025.
S= (1-2-3+4)+(5-6-7+8)+...+(2021-2022-2023+2024)+2025 Có 506 cặp.
S= 0 + 0 + ... + 0 + 2025.
⇒S= 2025.
a: \(\dfrac{-11}{18}+\dfrac{12}{29}+\dfrac{-7}{18}+\dfrac{2020}{2021}+\dfrac{17}{29}\)
=(-11/18-7/18)+(12/29+17/29)+2020/2021
=2020/2021
b: \(\dfrac{-2}{3}\cdot\dfrac{4}{9}+\dfrac{5}{9}\cdot\dfrac{-2}{3}+\dfrac{4}{3}\)
=-2/3+4/3=2/3
1) = 35 . 18 - 35 . 18 = 0
2) = 45 - 5 . 12 + 5 . 9 = 45 - 60 + 45 = 30
3) = 24 . 16 - 24 . 5 - 16 . 24 - 16 . 5
= (24 . 16 - 16 . 24) - 5 (24 - 16)
= 0 - 5 . 8 = 0 - 40 = -40
4) = 29 . 19 - 29 . 13 - 19 . 29 - 19 . 13
= (29 . 19 - 29 . 19) - 13 (29 - 19)
= 0 - 13 . 10 = 0 - 130 = -130
5) = 31 . (-18) + 31 . (-81) - 31 . 1
= 31 [(-18) + (-81) - 1]
= 31 . (-100) = -3100
6) = (-12) . 47 + (-12) . 52 + (-12). 1
= (-12) . (47 + 52 + 1)
= (-12) . 100 = -1200
7) = 13 . 45 - 3 . 45
= 45 (13 - 3)
= 45 . 10 = 450
8) = (-48) + 48 . (-78) + 48 . (-21)
= (-48) . 1 + 48 . (-78) + 48 . (-21)
= 48 . (-1) + 48 . (-78) + 48 . (-21)
= 48 . [(-1) + (-78) + (-21)]
= 48 . (-100) = -4800
1) 35x18-5x7x18
= 35x18-35x18
= 35x(18-18)
= 35x0=0
3) 24.(16-5)-16.(24-5)
= 24.16-24.5-16.24-16.5
=[24.16-16.24]-[24.5-16.5]
= [ 24. (16-16)]- [(24-16).5]
= [24.0]-[8.5]
= 0-40=-40
4) 29.(19-13)-19.(29-13)
=29.19-29.13-19.29-19.13
=[ 29.19-19.29]-[ 29.13-19.13]
= [ 29.(19-19)]-[(29-19).13]
=[29.0]-[10.13]
= 0-130= -130
5) 31.(-18)+31.(-81)-31
= 31.(-18)+31.(-81)-31.1
= 31.[(-18)+(-81)-1]
=31.(-100)
= -3100
6);7);8 tương tự 4);5);6)
Bài 2:
a: \(\left(6x-39\right):7=3\)
\(\Leftrightarrow6x-39=21\)
hay x=10
a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)
\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)
\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)
\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)
mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)
nên x-17=0
hay x=17
Vậy: x=17
b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)
\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)
\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)
\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)
mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)
nên x+20=0
hay x=-20
Vậy: x=-20
=>x/3-7/13-2/15+2/3=0
=>x/3=7/13+2/15-2/3=1/195
=>x=3/195=1/65
a: =>25-4x=1
=>4x=24
hay x=6
b: =>2x-4=0
hay x=2
c: =>x-35=115
hay x=150
d: =>x-2=12
hay x=14
e: =>x-36=216
hay x=252