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\(a,\) Vì \(x,y\in Z\) nên \(\left(3x+2\right):3R2;R1\)
Mà \(\left(3x+2\right)\left(y-8\right)=12\) nên \(3x+2\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
Do đó \(3x+2\in\left\{-4;-1;2\right\}\)
\(\Rightarrow x\in\left\{-2;-1;0\right\}\)
Với \(x=-2\Rightarrow\left(-4\right)\left(y-8\right)=12\Rightarrow y-8=-3\Rightarrow y=5\)
Với \(x=-1\Rightarrow\left(-3\right)\left(y-8\right)=12\Rightarrow y-8=-4\Rightarrow y=4\)
Với \(x=0\Rightarrow2\left(y-8\right)=12\Rightarrow y-8=6\Rightarrow y=14\)
Vậy PT có nghiệm \(\left(x;y\right)\) là \(\left(-2;5\right);\left(-1;4\right);\left(0;14\right)\)
\(b,\) Vì \(x,y\in Z\) nên \(\left(5x-4\right):5R1;R4\)
Mà \(\left(5x-4\right)\left(y+3\right)=-18\)
\(\Rightarrow5x-4\inƯ\left(-18\right)=\left\{-18;-9;-6;-3;-2;-1;1;2;3;6;9;18\right\}\\ \Rightarrow5x-4\in\left\{-9;1;6\right\}\\ \Rightarrow x\in\left\{-1;1;2\right\}\)
Với \(x=-1\Rightarrow-9\left(y+3\right)=-18\Rightarrow y+3=2\Rightarrow y=-1\)
Với \(x=1\Rightarrow y+3=18\Rightarrow y=15\)
Với \(x=2\Rightarrow6\left(y+3\right)=18\Rightarrow y+3=3\Rightarrow y=0\)
Vậy PT có nghiệm \(\left(x;y\right)\) là \(\left(-1;-1\right);\left(1;15\right);\left(2;0\right)\)
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a.\(2^x-2^4.2^7.32=0\)
\(2^x-2^{16}=0\)
\(=>x=16\)
b.\(3^x+3^{x+2}=270\)
\(3^x+3^x.3^2=270\)
\(3^x.10=270\)
\(3^x=27\)
\(=>x=3\)
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a: =>(2x-1)^3=4^12:4^10=4^2=8
=>2x-1=2
=>2x=3
=>x=3/2(loại)
b: 6x+5 chia hết cho 3x-1
=>6x-2+7 chia hết cho 3x-1
=>7 chia hết cho 3x-1
mà x là số tự nhiên
nên 3n-1=-1
=>n=0
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`2x-15 = 17``
`=> 2x = 17 + 15`
`=> 2x = 32`
`=> X = 32 : 2`
`=> x = 16`
`156 - (x + 61) = 82`
`=> x + 61 = 156 - 82`
`=> x + 61 = 74`
`=> x = 13`
`2x - 138 = 2^3 . 3^2`
`=>2x - 138 = 72`
`=> 2x = 210`
`=> x = 105`
bài 2:
`23-3x = 8`
`=> 3x = 23 - 8`
`=> 3x = 15`
`=> x = 5`
`(x-35) - 120 = 0`
`=>(x-35) = 120`
`=> x = 120 +35`
`=> x = 155`
`3^x + 2 = 29`
`=> 3^x = 27`
`=> 3^x = 3^3`
`=> x = 3`
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 4x -15=-75-x b,72-3x= 5x+8
4x+x=-75+15 -3x-5x=8-72
5x=-60 -8x=-64
x=-60:5 8x=64
x=-14 x=64:8
x=8
c,3Ix-7I=21 d,-7Ix+3I=-49
Ix-7I=21:3 Ix+3I=-49:-7
Ix-7I=7 Ix+3I=7
x-7=7 hoặc x-7=-7 x+3=7 hoặc x+3=-7
x=14 hoặc x=0 x=4 hoặc x=-10
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{2x+5}{2x+1}=\dfrac{2x+1+4}{2x+1}=\dfrac{2x+1}{2x+1}+\dfrac{4}{2x+1}=1+\dfrac{4}{2x+1}\)
Để \(\dfrac{2x+5}{2x+1}\in Z\) thì \(\dfrac{4}{2x+1}\in Z\)
\(\Rightarrow4\) ⋮ \(2x+1\)
\(\Rightarrow2x+1\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
\(\Rightarrow2x\in\left\{0;-2;1;-3;3;-5\right\}\)
\(\Rightarrow x\in\left\{0;-1;\dfrac{1}{2};-\dfrac{3}{2};\dfrac{3}{2};-\dfrac{5}{2}\right\}\)
Mà x nguyên \(\Rightarrow\text{x}\in\left\{0;-1\right\}\)
b) \(\dfrac{3x+5}{x+1}=\dfrac{3x+3+2}{x+1}=\dfrac{3\left(x+1\right)+2}{x+1}=\dfrac{3\left(x+1\right)}{x+1}+\dfrac{2}{x+1}=3+\dfrac{2}{x+1}\)
Để \(\dfrac{3x+5}{x+1}\in Z\) thì \(\dfrac{2}{x+1}\in Z\)
\(\Rightarrow2\) ⋮ \(x+1\)
\(\Rightarrow x+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
\(\Rightarrow x\in\left\{0;-2;1;-3\right\}\)
c) \(\dfrac{3x+8}{x-1}=\dfrac{3x-3+11}{x-1}=\dfrac{3\left(x-1\right)+11}{x-1}=\dfrac{3\left(x-1\right)}{x-1}+\dfrac{11}{x-1}=3+\dfrac{11}{x-1}\)
Để: \(\dfrac{3x+8}{x-1}\in Z\) thì \(\dfrac{11}{x-1}\in Z\)
\(\Rightarrow11\) ⋮ \(x-1\)
\(\Rightarrow x-1\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)
\(\Rightarrow x\in\left\{2;0;12;-10\right\}\)
d) \(\dfrac{5x+12}{x-2}=\dfrac{5x-10+22}{x-2}=\dfrac{5\left(x-2\right)+22}{x-2}=\dfrac{5\left(x-2\right)}{x-2}+\dfrac{22}{x-2}=5+\dfrac{22}{x-2}\)
Để: \(\dfrac{5x+12}{x-2}\in Z\) thì \(\dfrac{22}{x-2}\in Z\)
\(\Rightarrow22\) ⋮ \(x-2\)
\(\Rightarrow x-2\inƯ\left(22\right)=\left\{1;-1;2;-2;11;-11;22;-22\right\}\)
\(\Rightarrow x\in\left\{3;1;4;0;13;-9;24;-20\right\}\)
e) \(\dfrac{7x-12}{x+16}=\dfrac{7x+112-124}{x+16}=\dfrac{7\left(x+16\right)-124}{x+16}=\dfrac{7\left(x+16\right)}{x+16}-\dfrac{124}{x+16}=7-\dfrac{124}{x+16}\)
Để \(\dfrac{7x-12}{x+16}\in Z\) thì \(\dfrac{124}{x+16}\in Z\)
\(\Rightarrow124\) ⋮ \(x+16\)
\(\Rightarrow x+16\inƯ\left(124\right)=\left\{1;-1;2;-2;4;-4;31;-31;62;-62;124;-124\right\}\)
\(\Rightarrow x\in\left\{-15;-17;-14;-18;-12;-20;15;-47;46;-78;108;-140\right\}\)
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a) |2x - 1| - 3 = 5
=> |2x - 1| = 8
Có 2 TH xảy ra:
TH1 : 2x - 1 = 8 => 2x = 9 => x = 9/2 (ko thỏa mãn x thuộc Z)
TH2 : -(2x - 1) = 8 => -2x + 1 = 8 => -2x = 9 => x = -9/2 (ko thỏa mãn x thuộc Z)
b) |3x - 5| = 4
Có 2 TH xảy ra :
TH1 : 3x - 5 = 4 => 3x = 9 => x = 3
TH2 : -(3x - 5) = 4 => -3x + 5 = 4 => -3x = -1 => x = 1/3 (ko thỏa mãn x thuộc Z)
c) |5x - 1| = |-3 - 3x|
Có 2 TH xảy ra :
TH1 : 5x - 1 = -3 - 3x => 5x + 3x = -3 + 1 => 8x = -2 => x = -1/4 (ko thỏa mãn x thuộc Z)
TH2 : 5x - 1 = -(-3 - 3x) => 5x - 1 = 3 + 3x => 5x - 3x = 3 +1 => 2x = 4 => x = 2
d) |4x - 8| = |x + 1|
Có 2 TH xảy ra :
TH1 : 4x - 8 = x + 1 => 4x - x = 1 + 8 => 3x = 9 => x = 3
TH2 : 4x - 8 = -(x + 10) => 4x - 8 = -x - 10 => 4x + x = -10 + 8 => 5x = -2 => x = -2/5 (ko thỏa mãn x thuộc Z)
e) |3x - 5| - |4x + 9| = 0
=> |3x - 5| = |4x + 9|
Có 2 TH xảy ra :
TH1 : 3x - 5 = 4x + 9 => 3x - 4x = 9 + 5 => -x = 14 => x = -14
TH2 : 3x - 5 = -(4x + 9) => 3x - 5 = -4x - 9 => 3x + 4x = -9 + 5 => 7x = -4 => x = -4/7 (ko thỏa mãn x thuộc Z)
a ) \(\left[x+8\right]=-3x\)
\(\Leftrightarrow\left(-3\right)x=-3x\)
\(\Leftrightarrow x+8=-3x\)
\(\Leftrightarrow4x=-8\)
\(\Leftrightarrow x=-2\)
b ) \(\left[5x-3\right]=2x+1\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\)
\(\Leftrightarrow x=1\frac{1}{3}\)
Mình làm hơi tắt , thông cảm nhé ! ( Mình sắp phải đi học )