tim x,y thuoc z
xy-3x+19=0
giup minh voi !
LL
Những câu hỏi liên quan
tim x, y thuoc N biet
3x2 + 5y2=380
giup minh voi
tim x,y thuoc Z biet
a,xy-3x+2y=17
b,xy+3x-4y=19
tich minh cho minh len thu 8 tren bang sep hang cai
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tim x,y thuoc N,biet x<y:1/x+1/y=1/8
giup minh voi!
4(x-3)-8x(x-3)=0
5x(x-7)-10(7-x)=0
2x-8=3x(x-4)
3x(x-5)=10-2x
6x(x-3)-3(3-x)=0
x^2(x+4)+9(-x-4)=0
giup voi dang can gap a
\(4\left(x-3\right)-8x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(4-8x\right)=0\\ \Leftrightarrow2\left(1-2x\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ 5x\left(x-7\right)-10\left(7-x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(5x+10\right)=0\\ \Leftrightarrow5\left(x+2\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\\ 2x-8=3x\left(x-4\right)\\ \Leftrightarrow2\left(x-4\right)-3x\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(2-3x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\\ 3x\left(x-5\right)=10-2x\\ \Leftrightarrow3x\left(x-5\right)+2\left(x-5\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=5\end{matrix}\right.\\ 6x\left(x-3\right)-3\left(3-x\right)=0\\ \Leftrightarrow\left(6x+3\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
\(x^2\left(x+4\right)+9\left(-x-4\right)=0\\ \Leftrightarrow\left(x^2-9\right)\left(x+4\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=-4\end{matrix}\right.\)
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\(\left(4-8x\right)\left(x-3\right)=0\)
\(\left[{}\begin{matrix}4-8x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\3\end{matrix}\right.\)
\(2\left(x-4\right)-3x\left(x-4\right)=0\)
\(\left(2-3x\right)\left(x-4\right)=0\)
\(\left[{}\begin{matrix}2-3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.\)
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cach lam bai : giup minh voi
tim x thuoc Z biet:
a, /x/+/y/=0
b, /x/ + /y/= 1
a. x=y=0
b.x=0;y=-1 hoac 1
y=0;x=-1 hoac 1
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tim x thuoc z ; gia tri tuyet doi cua - 3x = 18 voi x < o
|-3x| = 18
Có 2 TH xảy ra :
TH1 : -3x = 18 => x = -6 (thỏa mãn x < 0)
TH2 : -(-3x) = 18 => 3x = 18 => x = 6 (ko thỏa mãn x < 0)
Vậy x = -6 thì thỏa mãn đề bài
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cac ban oi giup minh voi
1.tim a,b thuoc Z,biet:a.(2b-3)=-6
2.cho x,y thuoc Z thoa man x mu 2 +y mu 2 chia het cho 3.chung to x va y chia het cho 3.
tim x biet :
a) ( 6x - 84 ) : 2 -72 = 201 voi x thuoc N
b) ( 3x -34 ) . 63 = 65voi x thuoc N
a) 6x / 2 - 84 / 2 - 72 =201
3x - 42 -72 = 201
3x - 114 =201
3x = 315
x = 105
b)3x -3^4 =6^5 / 6^3
3x - 3^4 = 36
3x - 3 * 3^3 = 36
3*(x-27) = 36
x - 27 = 36 / 3
x - 27 = 12
x = 39
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Tim x va y, biet :
xy + 3y = 66
( x,y thuoc N )
Giup minh voi !
<=> y(x+3) =66
hay y ; x+3 thuộc ước của 66
Ư(66) = { 1;2;3;6;11 ;22 ;33;66}
Ta có bảng sau
| y | 1 | 2 | 3 | 6 | 11 | 22 | 33 | 66 |
| x+3 | 66 | 33 | 22 | 11 | 6 | 3 | 2 | 1 |
| y | 1 | 2 | 3 | 6 | 11 | 22 | 33 | 66 |
| x | 63 | 30 | 19 | 8 | 3 | 0 | / | / |
Vậy \hept{y=1x=63;