Tìm x,y nguyên :
x2 + y2 + 5x2y2 + 60 = 37xy
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xem như pt bậc 2 ẩn x
x^2 + y^2 + 5(xy)^2 + 60 =37xy
<>(1+5y^2).x^2 -37xy + 60 + y^2 =0
denta = 37^2*y^2 - 4*(60+y^2)*(1+5y^2)
= -20y^4+165y^2- 240 >=0
=> 1 < y^2 <7 => y= +-2
với y= 2 => x = 2 thỏa mãn
với y =-2 => x =- 2 thỏa mãn
d: \(x\left(x^2-1\right)+3\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
e: \(x^2-10x+25=\left(x-5\right)^2\)
g: \(x^2-64=\left(x-8\right)\left(x+8\right)\)
h: \(\left(x+y\right)^2-\left(x^2-y^2\right)\)
\(=\left(x+y\right)\left(x+y-x+y\right)\)
\(=2y\left(x+y\right)\)
i: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
k: \(x^2+2xy+y^2-25=\left(x+y-5\right)\left(x+y+5\right)\)
l: \(2xy-x^2-y^2+16\)
\(=-\left(x^2-2xy+y^2-16\right)\)
\(=-\left(x-y-4\right)\left(x-y+4\right)\)
a: \(5x-15y=5\left(x-3y\right)\)
b: \(5x^2y^2+15x^2y+30xy^2=5xy\left(xy+3x+6y\right)\)
c: \(x^3-2x^2y+xy^2-9x\)
\(=x\left(x^2-9-2xy+y^2\right)\)
\(=x\left(x-y-3\right)\left(x-y+3\right)\)
\(PT\Leftrightarrow x^2-2xy+y^2=35xy-5x^2y^2-60\)
\(\Leftrightarrow\left(x-y\right)^2=5\left(3-xy\right)\left(xy-4\right)\)
Mà \(\left(x-y\right)^2\ge0\forall x;y\) nên \(5\left(3-xy\right)\left(xy-4\right)\ge0\Leftrightarrow3\le xy\le4\)
\(\Rightarrow\hept{\begin{cases}x;y\in\left\{3;4\right\}\\x=y\end{cases}}\) \(\Rightarrow\left(x;y\right)\in\left\{\left(2;2\right);\left(-2;-2\right)\right\}\)
a: \(=3x^4+3x^2y^2+2x^2y^2+2y^4+y^2\)
\(=\left(x^2+y^2\right)\left(3x^2+2y^2\right)+y^2\)
\(=3x^2+3y^2=3\)
b: \(=7\left(x-y\right)+4a\left(x-y\right)-5=-5\)
c: \(=\left(x-y\right)\left(x^2+xy+y^2\right)+xy\left(y-x\right)+3=3\)
d: \(=\left(x+y\right)^2-4\left(x+y\right)+1\)
=9-12+1
=-2
Ta có
PT <=> (1 + 5y2)x2 - 37yx + y2 + 60 = 0
Xét pt theo ẩn x ta có để pt có nghiệm thì
∆\(\ge0\)
<=> (37y)2 - 4(1 + 5y2)(y2 + 60) \(\ge0\)
<=> - 20y4 + 165y2 - 240\(\ge0\)
<=> 1 < y2 < 7
=> y2 = 4
=> y = (2;-2)
=> x = (2;-2)
xem như pt bậc 2 ẩn x
x^2 + y^2 + 5(xy)^2 + 60 =37xy
<>(1+5y^2).x^2 -37xy + 60 + y^2 =0
denta = 37^2*y^2 - 4*(60+y^2)*(1+5y^2)
= -20y^4+165y^2- 240 >=0
=> 1 < y^2 <7 => y= +-2
với y= 2 => x = 2 thỏa mãn
với y =-2 => x =- 2 thỏa mãn
xong nha
a: \(=5x\left(xy^2+3x+6y^2\right)\)
b: \(=\left(x-2\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(x+3-x-2\right)=\left(x-2\right)\)
c: \(=\left(x-3\right)\left(x-4\right)\)
d: \(=x\left(x^2-2xy+y^2-9\right)\)
=x(x-y-3)(x-y+3)
e: \(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)
f: \(=\left(x-4\right)\left(x+3\right)\)
Trả lời
⇔x2+y2−2xy+60=35xy−5x2y2=5(7xy−x2y2)⇔x2+y2−2xy+60=35xy−5x2y2=5(7xy−x2y2)
⇔(x−y)2+60=5.494−54(2xy−7)2⇔(x−y)2+60=5.494−54(2xy−7)2
⇔[2(x−y)]2+5(2xy−7)2=5.49−60.4=5⇔[2(x−y)]2+5(2xy−7)2=5.49−60.4=5
x;y∈Z;2xy−7≠0;5(2xy−7)2≥5⇒[2(x−y)]2=0→x=yx;y∈Z;2xy−7≠0;5(2xy−7)2≥5⇒[2(x−y)]2=0→x=y
|(2xy−7)|=1|(2xy−7)|=1 [2x2−7=−1;x2=3(l)2x2−7=1;x2=4(n)[2x2−7=−1;x2=3(l)2x2−7=1;x2=4(n)⇔(x;y)=(±2;±2)