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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. 2x+\(\dfrac{4}{5}\)=0 hoặc 3x-\(\dfrac{1}{2}\)=0
2x=- 4/5 hoặc 3x=1/2
x=-2/5 hoặc x=\(\dfrac{1}{6}\)
b. x-\(\dfrac{2}{5}\)=0 hoặc x+\(\dfrac{4}{7}\)=0
x=2/5 hoặc x=-\(\dfrac{4}{7}\)
d. x(1+5/8-12/16)=1
\(\dfrac{7}{8}\)x=1=> x=8/7
b: \(\Leftrightarrow x+8\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-7;-9;-3;-13\right\}\)
1
a, 4x - 3x + 1 = 5
x =5-1
x =4
Vậy x=4
b, (2x - 4 ) . 3x =0
=> 2x - 4 =0 hoặc 3x = 0
=> 2x =4 hoặc x=0
=> x =2 hoặc x=0
vậy x= 2 hoặc x=0
c, x . ( x -1 ) - ( x-1 )=0
(x-1) . (x-1 ) =0
(x-1)2 =02
x-1 =0
x =1
vậy x=1
2/ a, 7 . (x - 1 ) = 6x + 3
7x -7 = 6x +3
7x - 6x =7+3
x =10
vậy x=10
b, 8 . ( 2x - 3 ) -15x =4
16x - 24 -15x =4
16x - 15x =4+24
x =28
vậy x=28
c, 7 . 10 + ( x-1 ) .2 =100
70 + 2x -2 =100
2x -2 =100-70
2x -2 =30
2x =30+2
2x =32
x =16
vậy x=16
chúc bn học tốt
a: \(x+7⋮x+2\)
=>\(x+2+5⋮x+2\)
=>\(5⋮x+2\)
=>\(x+2\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-1;-3;3;-7\right\}\)
b: \(2x+5⋮x+1\)
=>\(2x+2+3⋮x+1\)
=>\(3⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;-2;2;-4\right\}\)
c: \(3x-2⋮x+3\)
=>\(3x+9-11⋮x+3\)
=>\(-11⋮x+3\)
=>\(x+3\in\left\{1;-1;11;-11\right\}\)
=>\(x\in\left\{-2;-4;8;-14\right\}\)
d: \(12x+1⋮3x+2\)
=>\(12x+8-7⋮3x+2\)
=>\(-7⋮3x+2\)
=>\(3x+2\in\left\{1;-1;7;-7\right\}\)
=>\(3x\in\left\{-1;-3;5;-9\right\}\)
=>\(x\in\left\{-\dfrac{1}{3};-1;\dfrac{5}{3};-3\right\}\)
e: \(x^2+3x+5⋮x+3\)
=>\(x\left(x+3\right)+5⋮x+3\)
=>\(5⋮x+3\)
=>\(x+3\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-2;-4;2;-8\right\}\)
f: \(x^2-2x+3⋮x+2\)
=>\(x^2+2x-4x-8+11⋮x+2\)
=>\(11⋮x+2\)
=>\(x+2\in\left\{1;-1;11;-11\right\}\)
=>\(x\in\left\{-1;-3;9;-13\right\}\)