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a) (x - 12). 105 = 0
x-12= 0 : 105
x - 12 = 0
x = 0+ 12
x = 12
b) (47.(27 - x) = 94
27-x =94 : 47
27 - x = 2
x = 27 - 2
x = 25
c) 2x + 69 .2 = 69.5
2x = 69.5 - 69.2
2x = 69 . (5 - 2)
2x = 69.3
2x = 207
x = 207 : 2
x = 103,5
a)
\(\left(x+1\right)\left(y-2\right)=5\\ \Rightarrow\left(x+1\right),\left(y-2\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
Ta có bảng:
| x+1 | 1 | -1 | 5 | -5 |
| y-2 | 5 | -5 | 1 | -1 |
| x | 0 | -2 | 4 | -6 |
| y | 7 | -3 | 3 | 1 |
Vậy \(\left(x;y\right)=\left(0;7\right),\left(-2;-3\right),\left(4;3\right),\left(-6;1\right)\)
b)
\(\left(x-5\right)\left(y+4\right)=-7\\ \Rightarrow\left(x-5\right),\left(y+4\right)\inƯ\left(-7\right)=\left\{1;-1;7;-7\right\}\)
Ta có bảng:
| x-5 | 1 | -1 | 7 | -7 |
| y+4 | -7 | 7 | -1 | 1 |
| x | 6 | 4 | 12 | -2 |
| y | -11 | 3 | -5 | -3 |
Vậy \(\left(x;y\right)=\left(6;-11\right),\left(4;3\right),\left(12;-5\right),\left(-2;-3\right)\)
a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)
\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)
1.Tính nhanh nếu có thể:
a) 22 + 23 + 89 + 77
= ( 77 + 23 ) + 22 + 89
= 100 + 22 + 89
= 122 + 89
= 211
b) 35 . 15 + 15 . 65
= 15 . ( 35 + 65 )
= 15 . 100
= 1500
c) 7^2 - 36 : 3^2
= 7^2 - 36 : 9
= 7^2 - 4
= 49 - 4
= 45
d) 476 - {5 . [409 - (8 . 3 - 21)2] - 1724}
= 476 - {5 . [409 - (24 - 21)^2] - 1724}
= 476 - {5 . [409 - (3^2)] - 1724}
= 476 - {5 . [409 - 9 ] - 1724}
= 476 - {5. 400 - 1724}
= 476 - {2000 - 1724}
= 476 - 276
= 200
a: 7x+58=100
nên 7x=42
hay x=6
c: x-56:x=16
nên x-14=16
hay x=30
c)x - 56 : 4 = 16
x - 56 = 16 : 4
x- 56 = 4
x =4 + 56
x = 60
d)101 + (36 - 4x) = 105
(36- 4x ) = 105 - 101
36 - 4x = 4
4x = 36 - 4
4x = 32
x = 32:4
x = 8
Bài 10:
a: 2x-3 là bội của x+1
=>\(2x-3⋮x+1\)
=>\(2x+2-5⋮x+1\)
=>\(-5⋮x+1\)
=>\(x+1\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{0;-2;4;-6\right\}\)
b: x-2 là ước của 3x-2
=>\(3x-2⋮x-2\)
=>\(3x-6+4⋮x-2\)
=>\(4⋮x-2\)
=>\(x-2\inƯ\left(4\right)\)
=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(x\in\left\{3;1;4;0;6;-2\right\}\)
Bài 14:
a: \(4n-5⋮2n-1\)
=>\(4n-2-3⋮2n-1\)
=>\(-3⋮2n-1\)
=>\(2n-1\inƯ\left(-3\right)\)
=>\(2n-1\in\left\{1;-1;3;-3\right\}\)
=>\(2n\in\left\{2;0;4;-2\right\}\)
=>\(n\in\left\{1;0;2;-1\right\}\)
mà n>=0
nên \(n\in\left\{1;0;2\right\}\)
b: \(n^2+3n+1⋮n+1\)
=>\(n^2+n+2n+2-1⋮n+1\)
=>\(n\left(n+1\right)+2\left(n+1\right)-1⋮n+1\)
=>\(-1⋮n+1\)
=>\(n+1\in\left\{1;-1\right\}\)
=>\(n\in\left\{0;-2\right\}\)
mà n là số tự nhiên
nên n=0
a) (x - 12).105= 0 b) 47. (27 - x) = 94 c) 2x + 69.2 = 69.5
--> x - 12 = 0: 105 --> 27 - x = 94 : 47 = 2 --> 2x + 138 = 345
--> x - 12 = 0 --> 27 - x = 2 --> 2x = 345 - 138
--> x = 0 + 12 --> x = 27 - 2 --> 2x = 207
--> x = 12 --> x = 25 --> x = 207 : 2
vậy x = 12 vậy x bằng 25 --> x = 103,5
vậy x = 103,5
d) (x - 7).(2x - 8)= 0 e) (x + 1)+(x + 2)+...+(x + 10)= 165
--> x - 7 = 0 hoặc 2x - 8 = 0 ( x + x + .... + x ) + ( 1 + 2 + 3 + ..... + 10 ) = 165
--> x = 0 + 7 hoặc 2x = 0 + 8 10x + 52 = 165
--> x = 7 hoặc x = 8 : 2 10x = 165 - 52
--> x = 7 hoặc x = 4 10x = 113
vậy x = 7 hoặc x = 4 x = 113 : 10
x = 11,3
vậy x = 11,3