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a: (x-2)(y-3)=5
=>\(\left(x-2\right)\cdot\left(y-3\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)
=>\(\left(x-2;y-3\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(3;8\right);\left(7;4\right);\left(1;-2\right);\left(-3;2\right)\right\}\)
b: (2x-1)*(y-4)=-11
=>\(\left(2x-1\right)\cdot\left(y-4\right)=1\cdot\left(-11\right)=\left(-11\right)\cdot1=\left(-1\right)\cdot11=11\cdot\left(-1\right)\)
=>\(\left(2x-1;y-4\right)\in\left\{\left(1;-11\right);\left(-11;1\right);\left(-1;11\right);\left(11;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(-5;5\right);\left(0;15\right);\left(6;3\right)\right\}\)
c: xy-2x+y=3
=>\(x\left(y-2\right)+y-2=1\)
=>\(\left(x+1\right)\left(y-2\right)=1\)
=>\(\left(x+1\right)\cdot\left(y-2\right)=1\cdot1=\left(-1\right)\cdot\left(-1\right)\)
=>\(\left(x+1;y-2\right)\in\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;3\right);\left(-2;1\right)\right\}\)
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a, \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x-3,2y-6\in Z\\x-3,2y-6\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\end{matrix}\right.\)
Ta có bảng:
x-3 | -1 | -5 | 1 | 5 |
2y-6 | -5 | -1 | 5 | 1 |
x | 2 | -2 | 4 | 8 |
y | \(\dfrac{1}{2}\left(loại\right)\) | \(\dfrac{5}{2}\left(loại\right)\) | \(\dfrac{11}{2}\left(loại\right)\) | \(\dfrac{7}{2}\left(loại\right)\) |
Vậy không có x,y thỏa mãn đề bài
b, tương tự câu a
\(c,xy-5x+2y=7\\ \Rightarrow x\left(y-5\right)+2y-10=-3\\ \Rightarrow x\left(y-5\right)+2\left(y-5\right)=-3\\ \Rightarrow\left(x+2\right)\left(y-5\right)=-3\)
Rồi làm tương tự câu a
\(d,xy-3x-4y=5\\ \Rightarrow x\left(y-3\right)-4y+12=17\\ \Rightarrow x\left(y-3\right)-4\left(y-3\right)=17\\ \Rightarrow\left(x-4\right)\left(y-3\right)=17\)
Rồi làm tương tự câu a
![](https://rs.olm.vn/images/avt/0.png?1311)
a: x/3-1/6=1/5
=>x/3=11/30
hay x=11/90
b: =>1/2x=2
hay x=4
c: =>2/3:x=-7-1/3=-22/3
=>x=-1/11
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 70 - 5(x - 3 ) = 45
5( x - 3 ) = 70 - 45 = 25
x - 3 = 25 : 5 = 5
x = 5 + 3 = 8
b) (2x - 1 )4 = 3 . 62 - 27
(2x - 1 )4 = 3 . 36 - 27
(2x - 1 )4 = 81
Ta thấy 81 = 34 vậy suy ra (2x - 1)4 = 34
Để vế trong ngoặc tròn (2x - 1 ) = 3 thì x cần bằng 2
Thử lại : 2 . 2 - 1 = 4 - 1 = 3
Vậy x = 2
c) 3x3 + 43 = 102 - 33
3x3 + 43 = 100 - 33 = 67
3x3 = 67 + 43 = 110 ( Đoạn này đề bài sai hay tao sai z :)?)
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a: \(\Leftrightarrow2x+\dfrac{7}{2}=\dfrac{16}{3}:\dfrac{8}{3}=2\)
=>2x=-3/2
hay x=-3/4
b: 2x+3=5
=>2x=2
hay x=1
c: =>3(x-2)=4(5+x)
=>4x+20=3x-6
=>x=-26
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>1/3x-2/5x=5
=>-1/15x=5
=>x=-75
b: =>4x=4
=>x=1
c: =>6*3^x-5*3^x=243
=>3^x=243
=>x=5
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>x/-3=3
hay x=-9
b: =>x/9=-1/9
hay x=-1
c: =>x+1/5=-1/3
hay x=-8/15
d: =>-7/x=-7/9
hay x=9
a, \(\dfrac{x}{-3}=3\Leftrightarrow x=-9\)
b, \(\dfrac{x}{9}=-\dfrac{1}{9}\Rightarrow x=-1\)
c, \(\dfrac{x+3}{15}=-\dfrac{6}{15}\Rightarrow x=-9\)
d, \(\dfrac{42}{-54}=-\dfrac{42}{6x}\Rightarrow6x=54\Leftrightarrow x=9\)
a, \(2x-1=5\)
\(2x=6\)
\(\Leftrightarrow x=3\)
\(\left(5^x-1\right).3-2=70\)
\(\left(5^x-1\right).3=72\)
\(5^x-1=24\)
\(5^x=25=5^2\)
\(\Leftrightarrow x=2\)