tim x va y sao cho :2x+2xy=100
Tim cac cap so thuc (x,y) sao cho x va y thoa man dong thoi 2 dieu kien:x=x2+y2 va y=2xy
tim x dua vao quan he uoc boi:
tim so tu nhien x sao cho x-1 la uoc cua 12
tim so tu nhien x sao cho 2x+1 la uoc cua 28
tim so tu nhien x sao cho x+15 la boi cua x+3
tim cac so nguyen x,y sao cho (x+1)(y-2)=3
tim so nguyen x sao cho(x+2).(y-1)=2
tim so nguyen to x vua la uoc cua 275 vua la uoc cua 180
tim so nguyen to x,y biet x+y=12 va UCLL (x:y)=5
tim so tu nhien x,y biet x+y=32 va UCLL (x:y)=8
tim so tu nhien x biet x chia het cho10; xchia het cho12; x chia het cho15 va 100<x<150
tim so x nho nhat khac 0b biet x chia het cho 24 va 30
40 chia het cho x . 56 chia het cho x va x>6
Tim cap x,y biet x,y thuoc Z va
2xy- 4+2x+y=0
tim x va y sao cho (2x+1)(y-5)=12
tim x va y biet rang
a) x2+2y2+2xy-2y +1=0
b) x2+2y2+2xy -2x+2=0
......................?
mik ko biết
mong bn thông cảm
nha ................
a) x2+2y2+2xy-2y+1=0
\(\Leftrightarrow\)(x2+2xy+y2)+(y2-2y+1)=0
\(\Leftrightarrow\)(x+y)2+(y-1)2=0
\(\Leftrightarrow\hept{\begin{cases}x+y=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)
Vậy x=-1, y=1
a/ \(x^2+2y^2+2xy-2y+1=0\)
<=> \(\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)=0\)
<=> \(\left(x+y\right)^2+\left(y-1\right)^2=0\)
<=> \(\hept{\begin{cases}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\)
<=> \(\hept{\begin{cases}x+y=0\\y-1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=-y\\y=1\end{cases}}\)
<=> \(\hept{\begin{cases}x=-1\\y=1\end{cases}}\)
b/ \(x^2+2y^2+2xy-2x+2=0\)
<=> \(\left(x^2+2xy+y^2\right)+\left(2y-2x+2\right)=0\)
<=> \(\left(x+y\right)^2+2\left(y-x+1\right)=0\)
<=> \(\hept{\begin{cases}\left(x+y\right)^2=0\\2\left(y-x+1\right)=0\end{cases}}\)
<=> \(\hept{\begin{cases}x+y=0\\y-x+1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x+y=0\\y-x=-1\end{cases}}\)
<=> \(\hept{\begin{cases}x+y=0\left(1\right)\\x-y=1\left(2\right)\end{cases}}\)
Trừ (1) và (2)
=> \(2y=-1\)
<=> \(y=-\frac{1}{2}\)
<=> \(x=\frac{1}{2}\)(vì \(x+y=0\)<=> \(x=-y\))
tim snt x va y sao cho : x2 - 2x +1 = 6y2-2x + 2
cho (x+2y)(x^2-2xy+4y^2)=0 vs (x-2y)(x^2+2xy+4y^2)=16 tim x va y
giai cach lam jup minh nha
Tim cac so tu nhien x va y sao cho:
a) (2x-3)(3y-2)=1
Ta có :
\(\left(2x-3\right)\left(3y-2\right)=1\)
Vì \(x,y\in N\Leftrightarrow2x-3;3y-2\in N,2x-3;3y-2\inƯ\left(1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3=1\\3y-2=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy \(\left(x,y\right)=\left(2,1\right)\)
Tìm các số tự nhiên x và y sao cho:
a)(2x-3)(3y-2)=1
2x-3=1
2x =1+3
2x =4
x =4:2
x =2
Hoặc :3y-2=1
3y =1+2
3y =3
y =3:3
y =1
Vậy:x=2 hoặc y =1
Chúc bạn học tốt!
a) Tim so nguyen a de a2 + a + 3 / a+1 la so nguyen.o cho x-2xy+y=0
b)Tim so nguyen x,y sao cho x-2xy+y=0.
Oái gặp bn trùng tên nè!
a) Để phân số \(\dfrac{a^2+a+3}{a+1}\) là số nguyên thì :
\(a^2+a+3⋮a+1\)
Mà \(a+1⋮a+1\)
\(\Rightarrow\left\{{}\begin{matrix}a^2+a+3⋮a+1\\a^2+a⋮a+1\end{matrix}\right.\)
\(\Rightarrow3⋮a+1\)
Vì \(a\in Z\Rightarrow a+1\in Z;a+1\inƯ\left(3\right)\)
Ta có bảng :
\(a+1\) | \(1\) | \(3\) | \(-1\) | \(-3\) |
\(a\) | \(0\) | \(2\) | \(-2\) | \(-4\) |
\(Đk\) \(a\in Z\) | TM | TM | TM | TM |
Vậy \(a\in\left\{0;2;-2;-4\right\}\) là giá trị cần tìm
b) Ta có :
\(x-2xy+y=0\)
\(\Rightarrow2x-4xy-2y=0\)
\(\Rightarrow\left(2x-4xy\right)+2y-1=0-1\)
\(\Rightarrow\left(2x-4xy\right)-\left(1-2y\right)=-1\)
\(\Rightarrow2x\left(1-2y\right)-\left(1-2y\right)=-1\)
\(\Rightarrow\left(1-2y\right)\left(2x-1\right)=-1\)
Vì \(x,y\in Z\Rightarrow1-2y;2x-1\in Z,1-2y;2x-1\inƯ\left(-1\right)\)
Ta có bảng :
\(x\) | \(2x-1\) | \(1-2y\) | \(y\) | \(Đk\) \(x,y\in Z\) |
\(0\) | \(-1\) | \(1\) | \(0\) | TM |
\(1\) | \(1\) | \(-1\) | \(1\) | TM |
Vậy cặp giá trị \(\left(x,y\right)\) cần tìm là :
\(\left(0,0\right);\left(1,1\right)\)
b) \(x-2xy+y=0\)
\(\Rightarrow x-\left(2xy-y\right)=0\)
\(\Rightarrow x-y\left(2x-1\right)=0\)
\(\Rightarrow2x-2y\left(2x-1\right)=0\)
\(\Rightarrow\left(2x-1\right)-2y\left(2x-1\right)=0-1=-1\)
\(\Rightarrow\left(2x-1\right)\left(1-2y\right)=-1\)
Ta có:
TH1: \(\left\{{}\begin{matrix}2x-1=1\\1-2y=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
TH2:\(\left\{{}\begin{matrix}2x-1=-1\\1-2y=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy...................