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a: \(\Leftrightarrow\left(x+5;y-3\right)\in\left\{\left(1;11\right);\left(11;1\right);\left(-1;-11\right);\left(-11;-1\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(-4;14\right);\left(6;4\right);\left(-6;-8\right);\left(-16;2\right)\right\}\)
b: =>x(y-2)=-19
=>\(\left(x,y-2\right)\in\left\{\left(1;-19\right);\left(-19;1\right);\left(-1;19\right);\left(19;-1\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(1;-17\right);\left(-19;3\right);\left(-1;21\right);\left(19;1\right)\right\}\)
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(x+1)+(x+3)+...+(x+99)=0
Tổng các số hạng là: (99+1):2=50 (số hạng)
=> (x+1)+(x+3)+...+(x+99)=0 <=> 50.x+(1+3+5+...+99) = 0
<=> 50.x+=0 <=> 50.x+2500=0 => x=-2500/50=-50
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b) A=\(\frac{5x-2}{x-3}=\frac{5x-15+13}{x-3}=\frac{5x-15}{x-3}+\frac{13}{x-3}=\frac{5\left(x-3\right)}{x-3}+\frac{13}{x-3}=5+\frac{13}{x-3}\)
Để A thuộc Z thì \(5+\frac{13}{x-3}\in Z\)
=>13 chia hết cho x-3
=>x-3 \(\in\)Ư(13)={-1;1;-13;13}
x-3=-1 x-3=1 x-3 =-13 x-3=13
x =-1+3 x =1+3 x =-13+3 x =13+3
x=2 x =4 x=-10 x=16
Vậy x=2;4;-10;16 thì A thuộc Z
c)B=\(\frac{6x-1}{3x+2}=\frac{6x+4-5}{3x+2}=\frac{6x+4}{3x+2}+\frac{-5}{3x+2}=\frac{2\left(3x+2\right)}{3x+2}+\frac{-5}{3x+2}=2+\frac{-5}{3x+2}\)
Để B thuộc Z thì \(2+\frac{-5}{3x+2}\in Z\)
=>-5 chia hết cho 3x+2
=>3x+2\(\in\)Ư(-5)={-1;1;-5;5}
3x+2=-1 3x+2=1 3x+2=-5 3x+2=5
3x =-3 3x =-1 3x =-7 3x =3
x =-1 x =-1/3 x =-7/3 x =1
Vậy x=-1;-1/3;-7/3;1 thì B thuộc Z
d) C=\(\frac{10x}{5x-2}=\frac{10x-4+4}{5x-2}=\frac{10-4}{5x-2}+\frac{4}{5x-2}=\frac{2\left(5x-2\right)}{5x-2}+\frac{4}{5x-2}=2+\frac{4}{5x-2}\)
Để C thuộc Z thì \(2+\frac{4}{5x-2}\in Z\)
=> 4 chia hết cho 5x-2
=>5x-2\(\in\)Ư(4)={-1;1;-2;2;-4;4}
5x-2=-1 5x-2=1 5x-2=2 5x-2=-2 5x-2=4 5x-2=-4
bạn tự giải tìm x như các bài trên nhé
d) bạn ghi đề mjk ko hjeu
e)E=\(\frac{4x+5}{x-3}=\frac{4x-12+17}{x-3}=\frac{4x-12}{x-3}+\frac{17}{x-3}=\frac{4\left(x-3\right)}{x-3}+\frac{17}{x-3}=4+\frac{17}{x-3}\)
Để E thuộc Z thì\(4+\frac{17}{x-3}\in Z\)
=>17 chia hết cho x-3
=>x-3 \(\in\)Ư(17)={1;-1;17;-17}
x-3=1 x-3=-1 x-3=17 x-3=-17
bạn tự giải tìm x nhé
điều cuối cùng cho mjk ****
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a) \(xy+x+2y=5\\ \Rightarrow y\left(x+2\right)+x+2=5+2\\ \Rightarrow\left(x+2\right)\left(y+1\right)=7\)
Ta xét bảng:
x+2 | 1 | 7 | -1 | -7 |
x | -1 | 5 | -3 | -9 |
y+1 | 7 | 1 | -7 | -1 |
y | 6 | 0 | -8 | -2 |
Vậy \(\left(x;y\right)\in\left\{\left(-1;6\right);\left(5;0\right);\left(-3;-8\right);\left(-9;-2\right)\right\}\)
b) \(xy-3x-y=0\\ \Rightarrow x\left(y-3\right)-y+3=3\\ \Rightarrow\left(y-3\right)\left(x-1\right)=3\)
Ta xét bảng:
x-1 | 1 | 3 | -1 | -3 |
x | 2 | 4 | 0 | -2 |
y-3 | 3 | 1 | -3 | -1 |
y | 6 | 4 | 0 | 2 |
Vậy \(\left(x;y\right)\in\left\{\left(2;6\right);\left(4;4\right);\left(0;0\right);\left(-2;2\right)\right\}\)
c) \(xy+2x+2y=-16\\ \Rightarrow x\left(y+2\right)+2y+4=-12\\ \Rightarrow\left(y+2\right)\left(x+2\right)=-12\)
Ta xét bảng:
x+2 | 1 | 2 | 3 | 4 | 6 | 12 | -1 | -2 | -3 | -4 | -6 | -12 |
x | -1 | 0 | 1 | 2 | 4 | 10 | -3 | -4 | -5 | -6 | -8 | -14 |
y+2 | -12 | -6 | -4 | -3 | -2 | -1 | 12 | 6 | 4 | 3 | 2 | 1 |
y | -14 | -8 | -6 | -5 | -4 | -3 | 10 | 4 | 2 | 1 | 0 | -1 |
Vậy \(\left(x;y\right)\in\left\{\left(-1;-14\right);\left(0;-8\right);\left(1;-6\right);\left(2;-5\right);\left(4;-4\right);\left(10;-3\right);\left(-3;10\right);\left(-4;4\right);\left(-5;2\right);\left(-6;1\right);\left(-8;0\right);\left(-14;-1\right)\right\}\)
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b) xy - 2y = -19
=> y(x - 2) = -19
=> x - 2 \(\in\) Ư(-19) = Ư(19) = {-1;1;-19;19}
Ta có bảng sau:
Vậy: (x;y) \(\in\){(1;19);(3;-19);(-17;1);(21;-1)}
a) x + 5 \(\in\)Ư(11) = {-1;1;-11;11}
Ta có bảng sau:
Vậy: (x; y) \(\in\) {(-6;-8);(-4;14);(-16;2);(6;4)}