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a: x/2=-5/y
=>xy=-10
=>\(\left(x,y\right)\in\left\{\left(1;-10\right);\left(-10;1\right);\left(-1;10\right);\left(10;-1\right);\left(2;-5\right);\left(-5;2\right);\left(-2;5\right);\left(5;-2\right)\right\}\)
b: =>xy=12
mà x>y>0
nên \(\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)
c: =>(x-1)(y+1)=3
=>\(\left(x-1;y+1\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(2;2\right);\left(4;0\right);\left(0;-4\right);\left(-2;-2\right)\right\}\)
d: =>y(x+2)=5
=>\(\left(x+2;y\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-1;5\right);\left(3;1\right);\left(-3;-5\right);\left(-7;-1\right)\right\}\)
a: =>x-3=9
=>x=12
b: =>10-x=-26
=>x=36
c: =>x:4-1=2
=>x:4=3
=>x=12
d: =>x^2=4
=>x=2 hoặc x=-2
e: =>(x-2)^2=100
=>x-2=10 hoặc x-2=-10
=>x=12 hoặc x=-8
a) 3(x - 1) - 1 = 26
=> 3x - 3 = 26 + 1
=> 3x - 3 = 27
=> 3x = 27 + 3
=> 3x = 30
=> x = 30 : 3
=> x = 10
b) |x + 4| - 9 = (-2)3
=> |x + 4| - 9 = -8
=> |x + 4| = -8 + 9
=> |x + 4| = 1
=> \(\orbr{\begin{cases}x+4=1\\x+4=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=-3\\x=-5\end{cases}}\)
Vậy ...
c) 2x - 5 \(⋮\)x - 1
<=> 2(x - 1) - 3 \(⋮\)x - 1
<=> 3 \(⋮\)x - 1
<=> x - 1 \(\in\)Ư(3) = {1; -1; 3; -3}
Lập bảng :
| x - 1 | 1 | -1 | 3 | -3 |
| x | 2 | 0 | 4 | -2 |
Vậy ...
`@` ` \text {Ans}`
`\downarrow`
`a,`
`1/4+3/4*x=3/2-x`
`=> 1/4 + 3/4x - 3/2 + x = 0`
`=> (1/4 - 3/2) + (3/4x + x) = 0`
`=> -5/4 + 7/4x = 0`
`=> 7/4x = 5/4`
`=> x = 5/4 \div 7/4`
`=> x = 5/7`
Vậy, `x=5/7`
`b,`
`3/5*x-1/4=1/10*x-1/2`
`=> 3/5x - 1/4 - 1/10x + 1/2 = 0`
`=> (3/5x - 1/10x) + (-1/4 + 1/2)=0`
`=> 1/2x + 1/4 = 0`
`=> 1/2x = -1/4`
`=> x = -1/4 \div 1/2`
`=> x = -1/2`
Vậy, `x=-1/2`
`c,`
`3x-3/5=x-1/4`
`=> 3x - 3/5 - x + 1/4 = 0`
`=> (3x - x) - (3/5 - 1/4) = 0`
`=> 2x - 7/20 = 0`
`=> 2x = 0,35`
`=> x = 0,35 \div 2`
`=> x = 7/40`
Vậy, `x=7/40`
`d,`
`3/2*x-2/5=1/3*x-1/4`
`=> 3/2x - 2/5 - 1/3x + 1/4 = 0`
`=> (3/2x - 1/3x) - (2/5 - 1/4) = 0`
`=> 7/6x - 3/20 = 0`
`=> 7/6x = 3/20`
`=> x = 3/20 \div 7/6`
`=> x = 9/70`
Vậy, `x=9/70`
`@` `\text {Kaizuu lv uuu}`
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 = 3
b) x = 3
c) x = 1
d) x = 23