Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1 :
A ) 3 < x < 5
=> x thuộc { 4 }
Vậy x = 4
Câu b và câu c cứ theo vậy mà làm .
Bài 2 :
| x + 7 | = 0
x = 0 - 7
x = -7
Vậy x = -7
1: \(\Leftrightarrow\left(x-1\right)^x\cdot\left(x-1\right)^2-\left(x-1\right)^x=0\)
=>\(\left(x-1\right)^x\cdot\left[\left(x-1\right)^2-1\right]=0\)
=>\(x\left(x-1-1\right)\cdot\left(x-1\right)^x=0\)
=>x(x-2)(x-1)^x=0
=>x=0;x=2;x=1
2: \(\Leftrightarrow\left(6-x\right)^{2003}\left(x-1\right)=0\)
=>6-x=0 hoặc x-1=0
=>x=6;x=1
3: =>(7x-11)^3=32*25+200=1000
=>7x-11=10
=>7x=21
=>x=3
4: =>x^2-1=-3 hoặc x^2-1=3
=>x^2=-2(loại) hoặc x^2=4
=>x=2 hoặc x=-2
Bài 2:
a)|x| < 3
x\(\in\){-2;-1;0;1;2}
b)|x - 4 | < 3
x\(\in\){ 6 ; 5 ; 4 ; 3 ; 2 }
c) | x + 10 | < 2
x\(\in\){ -2 ; -10 }
Bài 1:
A = 1 + 2 - 3 + 4 + 5 - 6 +...+98 - 99
A = (1 + 4 + 7 +...+97) + [(2-3)+(5-6)+...+(98-99)]
A = 1617 + [(-1)+(-1)+...+(-1)]
A = 1617 + (-49)
A = +(1617-49) = A = 1568
B = - 2 - 4 + 6 - 8 + 10 + 12 - .... + 60
B =
2)
a) \(x\in\left\{2;1;0;-1;-2\right\}\)
b) \(x\in\left\{6;-6;5;-5;4\right\}\)
c) \(x\in\left\{-9;-11;-10\right\}\)
3)
\(\left(a;b\right)\in\left\{\left(0;1\right);\left(0;-1\right);\left(1;0\right);\left(-1;0\right)\right\}\)
Bài 1 :
a) x={2,4}
b) x-1={-3,-2,-1,0,1,2,3,4}
=> x={-2,-1,0,1,2,3,4,5}
c) x+2={-7,-6,-5,-4}
=> x={-9,-8,-7,-6}
Bài 2 :
(x-3)(x+2)=0
=> x-3=0 => x=3
=> x+2=0 => x=-2
Vậy x=-2 hoặc x=3
BÀI 1
A) 3<X<5
=>X=4
B) -4<X+2<5
=>X-1\(\in\left(-3;-2;-1;0;1;2;3;4\right)\)
=> X-1=-3 => X-1=-2 =>X-1=-1 =>X-1=0 => X-1=1
X=-2 X=-1 X= 0 X=1 X=2
=>X-1=2 => X-1=3 =>X-1=4
X=3 X=4 X=5
C) -8<X+2<-3
=> X+2\(\in\left(-7;-6;-5;-4\right)\)
=> X+2=-7 =>X+2=-6 =>X+2=-5 =>X+2=-4
X=-9 X=-8 X=-7 X=-6
BÀI 2
\(\left(X-3\right).\left(X+2\right)=0\)
\(\Rightarrow X-3=X+2=O\)
\(TH1:X-3=0\)
X=3
TH2: X+2=0
X=-2
VẬY X=3 HOẶC X=-2
a, 2\(xy\) - 2\(x\) + 3\(y\) = -9
(2\(xy\) - 2\(x\)) + 3\(y\) - 3 = -12
2\(x\)(\(y-1\)) + 3(\(y-1\)) = -12
(\(y-1\))(2\(x\) + 3) = -12
Ư(12) = {-12; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6; 12}
Lập bảng ta có:
| \(y\)-1 | -12 | -6 | -4 | -3 | -2 | -1 | 1 | 2 | 3 | 4 | 6 | 12 |
| \(y\) | -11 | -5 | -3 | -2 | -1 | 0 | 2 | 3 | 4 | 5 | 7 | 13 |
| 2\(x\)+3 | 1 | 2 | 3 | 4 | 6 | 12 | -12 | -6 | -4 | -3 | -2 | -1 |
| \(x\) | -1 | -\(\dfrac{1}{2}\) | 0 | \(\dfrac{1}{2}\) | \(\dfrac{3}{2}\) | \(\dfrac{9}{2}\) | \(-\dfrac{15}{2}\) | \(-\dfrac{9}{2}\) | -\(\dfrac{7}{2}\) | -3 | \(-\dfrac{5}{2}\) | -2 |
Theo bảng trên ta có: Các cặp \(x\);\(y\) nguyên thỏa mãn đề bài là:
(\(x;y\)) = (-1; -11); (0; -3); (-3; 5); ( -2; 13)
b, (\(x+1\))2(\(y\) - 3) = -4
Ư(4) = {-4; -2; -1; 1; 2; 4}
Lập bảng ta có:
| \(\left(x+1\right)^2\) | - 4(loại) | -2(loại) | -1(loại) | 1 | 2 | 4 |
| \(x\) | 0 | \(\pm\)\(\sqrt{2}\)(loại) | 1; -3 | |||
| \(y-3\) | 1 | 2 | 4 | -4 | -2 | -1 |
| \(y\) | -1 | 2 |
Theo bảng trên ta có: các cặp \(x;y\) nguyên thỏa mãn đề bài là:
(\(x;y\)) = (0; -1); (-3; 2); (1; 2)
a) \(20\cdot2^x+1=10\cdot4^2+1\)
\(\Leftrightarrow2\cdot10\cdot2^x=10\cdot4^2\)
\(\Leftrightarrow10\cdot2^{x+1}=10\cdot2^4\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=3\)
b) \(\left(4-\frac{x}{2}\right)^3-1=2\cdot\left(2^3-\frac{5}{2^0}\right)+1\)
\(\Leftrightarrow\left(4-\frac{x}{2}\right)^3=2\cdot3+1+1\)
\(\Leftrightarrow\left(4-\frac{x}{2}\right)^3=8=2^3\)
\(\Rightarrow4-\frac{x}{2}=2\)
\(\Leftrightarrow\frac{x}{2}=2\)
\(\Rightarrow x=4\)
1. Tìm x
a) 1+2+3+...+x = 210
=> \(\frac{x\left(x+1\right)}{2}=210\)
=> x = 20
b) \(32.3^x=9.3^{10}+5.27^3\)
=>\(32.3^x=9.3^{10}+5.3^9\)(\(27^3=\left(3^3\right)^3=3^9\))
=>\(32.3^x=9.3.3^9+5.3^9\)
=>\(32.3^x=3^9\left(9.3+5\right)\)
=>\(32.3^x=3^9.32\)
=>x = 9
2.
Ta có 2A = 3A - A
=> 2A = \(3\left(1+3+3^2+3^3+....+3^{10}\right)\)\(-\)\(1-3-3^2-3^3-....-3^{10}\)
=> 2A = \(3+3^2+3^3+.....+3^{11}-\)\(1-3-3^2-3^3-...-3^{10}\)
=> 2A = \(3^{11}-1\)
=> 2A+1 = \(3^{11}-1+1\)=\(3^{11}\)
=> n = 11
Ta có : a)1 + 2 + 3 + ... + x = 210
=> \(\frac{x\left(x+1\right)}{2}=210\)
=> x(x + 1) = 420
=> x(x + 1) = 20.21
=> x = 20
`@` `\text {Ans}`
`\downarrow`
`2+(x+3)=7`
`\Rightarrow x+3=7-2`
`\Rightarrow x+3=5`
`\Rightarrow x=5-3`
`\Rightarrow x=2`
`5+(3+x)=10`
`\Rightarrow 3+x=10-5`
`\Rightarrow 3+x=5`
`\Rightarrow x=5-3`
`\Rightarrow x=2`
`(4+x)+1=7`
`\Rightarrow 4+x=7-1`
`\Rightarrow 4+x=6`
`\Rightarrow x=6-4`
`\Rightarrow x=2`
`(x+5)+3=9`
`\Rightarrow x+5=9-3`
`\Rightarrow x+5=6`
`\Rightarrow x=6-5`
`\Rightarrow x=1`
`(x-1)-4=7`
`\Rightarrow x-1=7+4`
`\Rightarrow x-1=11`
`\Rightarrow x=11+1`
`\Rightarrow x=12`
`4-(6-x)=1`
`\Rightarrow 6-x=4-1`
`\Rightarrow 6-x=3`
`\Rightarrow x=6-3`
`\Rightarrow x=3`
\(2+\left(x+3\right)=7\)
\(\Rightarrow2+x+3=7\)
\(\Rightarrow x+5=7\)
\(\Rightarrow x=2\)
\(5+\left(3+x\right)=10\)
\(\Rightarrow5+3+x=10\)
\(\Rightarrow x+8=10\)
\(\Rightarrow x=2\)
\(\left(4+x\right)+1=7\)
\(\Rightarrow4+x+1=7\)
\(\Rightarrow x+5=7\)
\(\Rightarrow x=2\)
\(\left(x+5\right)+3=9\)
\(=x+5+3=9\)
\(\Rightarrow x+8=9\)
\(\Rightarrow x=1\)
\(\left(x-1\right)-4=7\)
\(\Rightarrow x-1-4=7\)
\(\Rightarrow x-5=7\)
\(\Rightarrow x=12\)
\(4-\left(6-x\right)=1\)
\(\Rightarrow4-6-x=1\)
\(\Rightarrow-2-x=1\)
\(\Rightarrow x=-3\)
a) \(2^0+2^1+2^2+...+2^{2017}=4^x-1\)
\(\Rightarrow2^1+2^2+2^3+...+2^{2018}=2\left(4^x-1\right)\)
\(\Rightarrow2^{2018}-1=4^x-1\)
\(\Rightarrow2^{2018}=4^x\\ \Rightarrow2^{2018}=2^{2x}\\ \Rightarrow2x=2018\\ \Rightarrow x=1009\)
b)
\(3^x+3^{x+1}+3^{x+2}+3^{x+3}=360\\ \Rightarrow3^{x+1}+3^{x+2}+3^{x+3}+3^{x+4}=1080\\ \Rightarrow3^{x+4}-3^x=720\\ \Rightarrow3^x\left(3^4-1\right)=720\\ \Rightarrow3^x.80=720\\ \Rightarrow3^x=9\\ \Rightarrow x=2\)
a) Gọi \(2^0+2^1+...+2^{2017}\) là A
\(A=2^0+2^1+2^2+...+2^{2017}\\ 2A=2+2^2+2^3+....+2^{2018}\\ 2A-A=\left(2+2^2+2^3+....+2^{2018}\right)-\left(2^0+2^1+2^2+...+2^{2017}\right)\\ A=2^{2018}-1=4^x-1\\ =>2^{2018}=4^x=>4^{1009}=4^x=>x=1009\)
\(b,3^x+3^{x+1}+3^{x+2}+3^{x+3}=360\\ 3^x+3^x\cdot3+3^x\cdot3^2+3^x\cdot3^3=360\\ 3^x\left(1+3+3^2+3^3\right)=360\\ 3^x\cdot40=360\\ 3^x=9\\ 3^x=3^2\\ =>x=2\)