K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
HL
1
KV
18 tháng 12 2023
a) 78 - x = 90
x = 78 - 90
x = -12
b) 15 + 2x = 5¹⁰ : 5⁸
15 + 2x = 5²
15 + 2x = 25
2x = 25 - 15
2x = 10
x = 10 : 2
x = 5
c) 48 : x + 17 = -33
48 : x + 17 = -33 - 17
48 : x = -50
x = 48 : (-50)
x = -24/25
d) Do x ⋮ 15 và x ⋮ 20
⇒ x ∈ BC(15; 20)
Ta có:
15 = 3.5
20 = 2².5
⇒ BCNN(15; 20) = 2².3.5 = 60
⇒ x ∈ BC(15; 20) = {0; 60; 120; ...}
Mà 50 < x < 70
⇒ x = 60
14 tháng 4 2020
Bài 1: Tính
a) Ta có: \(\left(-25\right)\cdot68+\left(-34\right)\cdot\left(-250\right)\)
\(=-25\cdot68+\left(-340\right)\cdot\left(-25\right)\)
\(=-25\cdot\left(68-340\right)\)
\(=-25\cdot\left(-272\right)\)
\(=6800\)
b) Ta có: \(1999+\left(-2000\right)+2001+\left(-2002\right)\)
\(=1999-2000+2001-2002\)
\(=-1-1=-2\)
c) Ta có: \(515+\left[72+\left(-515\right)+\left(-32\right)\right]\)
\(=515+72-515-32\)
\(=40\)
d) Ta có: \(\left(2736-75\right)-2736+175\)
\(=2736-75-2736+175\)
\(=100\)
e) Ta có: \(-2020-\left(157-2020\right)-\left(-257\right)\)
\(=-2020-157+2020+257\)
\(=100\)
Bài 2: Tìm x
a) Ta có: \(x-\left|-2\right|=\left|-18\right|\)
\(\Leftrightarrow x-2=18\)
hay x=20
Vậy: x=20
b) Ta có: \(2x-\left|+14\right|=\left|-14\right|\)
\(\Leftrightarrow2x-14=14\)
\(\Leftrightarrow2x=28\)
hay x=14
Vậy: x=14
c) Ta có: \(\left|x+4\right|+5=20-\left(-12-7\right)\)
\(\Leftrightarrow\left|x+4\right|+5=20+12+7\)
\(\Leftrightarrow\left|x+4\right|=39-5=34\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=34\\x+4=-34\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30\\x=-38\end{matrix}\right.\)
Vậy: x∈{30;-38}
d) Ta có: \(15-\left|2-x\right|=\left(-2\right)^2\)
\(\Leftrightarrow15-\left|2-x\right|=4\)
\(\Leftrightarrow\left|2-x\right|=11\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=11\\2-x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=13\end{matrix}\right.\)
Vậy: x∈{-9;13}
e) Ta có: \(\left|15-x\right|+\left|-25\right|=\left|-55\right|\)
\(\Leftrightarrow\left|15-x\right|+25=55\)
\(\Leftrightarrow\left|15-x\right|=30\)
\(\Leftrightarrow\left[{}\begin{matrix}15-x=30\\15-x=-30\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-15\\x=45\end{matrix}\right.\)
Vậy: x∈{-15;45}
g) Ta có: \(\left|17-\left(-4\right)\right|+\left|-24-\left(-5\right)\right|=\left|-x+3\right|\)
\(\Leftrightarrow\left|17+4\right|+\left|-24+5\right|=\left|3-x\right|\)
\(\Leftrightarrow\left|3-x\right|=40\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=40\\3-x=-40\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-37\\x=43\end{matrix}\right.\)
Vậy: x∈{-37;43}
18 tháng 8 2023
a: =>3^x=3^4*3=3^5
=>x=5
b: =>\(2^{x+1}=2^5\)
=>x+1=5
=>x=4
c: \(\Leftrightarrow3^{x+2-3}=3\)
=>x-1=1
=>x=2
d: \(\Leftrightarrow x^2=\dfrac{32}{2}=16\)
=>x=4 hoặc x=-4
e: (2x-1)^4=81
=>2x-1=3 hoặc 2x-1=-3
=>2x=4 hoặc 2x=-2
=>x=-1 hoặc x=2
f: (2x-6)^4=0
=>2x-6=0
=>x-3=0
=>x=3
H9
HT.Phong (9A5)
CTVHS
18 tháng 8 2023
a) \(3^x=81\cdot3\)
\(\Rightarrow3^x=3^4\cdot3\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
b) \(2^{x+1}=32\)
\(\Rightarrow2^{x+1}=2^5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=4\)
c) \(3^{x+2}:27=3\)
\(\Rightarrow3^{x+2}:3^3=3\)
\(\Rightarrow3^{x+2-3}=3\)
\(\Rightarrow3^{x-1}=3\)
\(\Rightarrow x-1=1\)
\(\Rightarrow x=2\)
d) \(2x^2=32\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x^2=4^2\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
e) \(\left(2x-1\right)^4=81\)
\(\Rightarrow\left(2x-1\right)^4=3^4\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
f) \(\left(2x-6\right)^4=0\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=6:2\)
\(\Rightarrow x=3\)
21 tháng 1
a: Sửa đề: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{2}{-z}=\dfrac{-t}{-9}\)
=>\(\dfrac{x}{5}=\dfrac{y}{-3}=\dfrac{-2}{z}=\dfrac{t}{9}=-2\)
=>\(x=-2\cdot5=-10;y=-2\cdot\left(-3\right)=6;z=\dfrac{-2}{-2}=1;t=9\cdot\left(-2\right)=-18\)
b: \(\dfrac{-24}{-6}=\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}\)
=>\(\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}=4\)
=>\(\left\{{}\begin{matrix}x=4\cdot3=12\\y^2=\dfrac{4}{4}=1\\z^3=-2\cdot4=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=12\\y\in\left\{1;-1\right\}\\z=-2\end{matrix}\right.\)
LQ
3
PN
4
21 tháng 1 2022
a: =>2x-x=-5/2-1/3
=>x=-17/6
b: =>4(x-2)2=36
=>(x-2)2=9
=>x-2=3 hoặc x-2=-3
hay x=5 hoặc x=-1
c: =>2x+1/2=5/6
=>2x=1/3
hay x=1/6
N
1
DN
23 tháng 12 2021
a.\(2^x-2^4.2^7.32=0\)
\(2^x-2^{16}=0\)
\(=>x=16\)
b.\(3^x+3^{x+2}=270\)
\(3^x+3^x.3^2=270\)
\(3^x.10=270\)
\(3^x=27\)
\(=>x=3\)
24 tháng 1 2021
a) Ta có: \(\left|-5\right|+\left|x-1\right|=\left|7\right|\)
\(\Leftrightarrow\left|x-1\right|+5=7\)
\(\Leftrightarrow\left|x-1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-1\right\}\)
b) Ta có: \(2\cdot\left|2x-4\right|-\left|-4\right|=\left|-50\right|\)
\(\Leftrightarrow4\cdot\left|x-2\right|-4=50\)
\(\Leftrightarrow4\cdot\left|x-2\right|=54\)
\(\Leftrightarrow\left|x-2\right|=\dfrac{27}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=\dfrac{27}{2}\\x-2=-\dfrac{27}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{31}{2}\left(loại\right)\\x=-\dfrac{23}{2}\left(loại\right)\end{matrix}\right.\)
Vậy: \(x\in\varnothing\)
24 tháng 1 2021
a, | -5 | + | x-1 | = | 7 |
5 + | x - 1 | = 7
| x - 1 | = 2
TH1 x -1 = 2
x = 3
TH2 x -1 = -2
x= -1
a) 29 + 2x = 59
2x = 59 - 29
2x = 30
x = 30 : 2
x = 15
b) 2020 - 6(17-x) = 2002
6(17-x) = 2020 - 2002
6(17-x) = 18
17-x = 18 : 6
17-x = 3
x = 17 - 3
x = 14