cho a,b thuoc Z.Tim x thuoc Z sao cho:
x-a=2
x+b=4
a-x=21
14-x=b+9
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cho a,b thuoc Z.Tim x thuoc Z sao cho
a, x-a=2 b, x+b=4
c, a-x=21 d, 14-x=b+9
Cho a,b thuoc z.Tim x sao cho:
a,x+a=10 b,a-x=5 c,x+a=b d,a-x=b
Cho:x=a/m, y=b/m, z=a+b/2m(a,b,m thuoc Z, m khac 0)Chung to :x<z<y
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Tim x thuoc Z de A thuoc Z va tim gia tri do .
a/ A= x+3/x-2 .
b/ A= 1-2x/x+3 .
cho A= { 8; 45}
B={ 15; 4}
a) tim tap hop C cac so tu nhien x= a+b sao cho a thuoc A va b thuoc B
b) tim tap hpo D cac so tu nhien x = a-b sao cho a thuoc A va b thuoc B
c) tim tap hop E cac so tu nhien x = a.b sao cho a thuoc A va b thuoc B
d) tim tap hop G cac so tu nhien x sao cho a=b.x va a thuoc A; b thuoc B
làm đầy đủ nha
bai 2
a)chung to rang P=4a2+4a chia het cho 8 voi moi a thuoc z
b)tim a thuoc z sao cho Q=a+7 chia het cho a (a khac 0)
c)tim a thuoc z sao cho M=a+1 chia het a -2 (a khac 2)
Tim x thuoc Z, sao cho:
a) 2x-5 chia het cho x-1.
b) x+1 la uoc cua x2+7.
Bài giải
\(a,\text{ }\frac{2x-5}{x-1}=\frac{2\left(x-1\right)-1}{x-1}=\frac{2\left(x-1\right)}{x-1}-\frac{1}{x-1}=2-\frac{1}{x-1}\)
\(2x-5\text{ }⋮\text{ }x-1\text{ khi }1⋮\text{ }x-1\)\(\Leftrightarrow\text{ }x\inƯ\left(1\right)\)
\(\Rightarrow\orbr{\begin{cases}x-1=-1\\x-1=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{0\text{ ; }2\right\}\)
\(b,\text{ }x+1\text{ là ước của }x^2+7\text{ }\Rightarrow\text{ }x^2+7\text{ }⋮\text{ }x+1\)
Ta có : \(\frac{x^2+7}{x+1}=\frac{x\left(x+1\right)-\left(x+1\right)+8}{x+1}=\frac{\left(x-1\right)\left(x+1\right)+8}{x+1}=\frac{\left(x-1\right)\left(x+1\right)}{x+1}+\frac{8}{x+1}\)
\(=x-1+\frac{8}{x+1}\)
\(\text{ }x^2+7\text{ }⋮\text{ }x+1\text{ khi }8\text{ }⋮\text{ }x+1\text{ }\Rightarrow\text{ }x+1\inƯ\left(8\right)\)
Ta có bảng :
| x + 1 | - 1 | 1 | - 2 | 2 | - 4 | 4 | - 8 | 8 |
| x | - 2 | 0 | - 3 | 1 | - 5 | 3 | - 9 | 7 |
\(\Rightarrow\text{ }x\in\left\{-2\text{ ; }0\text{ ; }-3\text{ ; }1\text{ ; }-5\text{ ; }3\text{ ; }-9\text{ ; }7\right\}\)
cho a;b thuoc Z tim x sao ch x+a=10
Cho he phuong trinh
\(\left(a+1\right)x-ay=5\) (1)
\(x+ay=a^2+4a\) (2)
Tim gia tri cua a thuoc Z sao cho he phuong trinh co nghiem (x;y) voi x, y thuoc Z
Lấy (1) cộng (2), ta có:
\(\left(2a+1\right)x=a^2+4a+5\)\(\Rightarrow x=\dfrac{a^2+4a+5}{2a+1}\)
Thay vào (1): \(\dfrac{\left(a^2+4a+5\right)\left(a+1\right)-10a-5}{2a+1}.\dfrac{1}{a}\)\(=\dfrac{a^3+5a^2-a}{2a+1}.\dfrac{1}{a}=\dfrac{a^2+5a-1}{2a+1}\)
Để x,y nguyên thì \(\left\{{}\begin{matrix}a^2+4a+5⋮2a+1\\a^2+5a-1⋮2a+1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\left(a+2\right)+2a+5⋮2a+1\\a^2+2a+3a-1⋮2a+1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}4⋮2a+1\\a+2⋮2a+1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}4⋮2a+1\\3⋮2a+1\end{matrix}\right.\)\(\Rightarrow2a+1\in\left\{\pm1\right\}\)\(\Rightarrow a\in\left\{-1;0\right\}\)
Vậy với a=-1;0 thì hpt có nghiệm (x;y) với x,y thuộc Z.
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