Toán Lớp 8: Tìm nghiệm nguyên dương của phương trình x^2y^2 − 16xy + 99 = 9x^2 + 36y^2 + 13x + 26y.

Question

Toán Lớp 8:  Tìm nghiệm nguyên dương của phương trình x^2y^2 − 16xy + 99 = 9x^2 + 36y^2 + 13x + 26y.

buianhtuan · Answer

x^2y^2 &minus; 16xy + 99 = 9x^2 + 36y^2 + 13x + 26y   ->x^2y^2+20xy-36xy+100-1=9x^2+36y^2+13x+26y   ->x^2y^2+20xy+100=9x^2+36y^2+13x+26y+36xy+1   ->(x^2y^2+10)^2=(9x^2+36xy+36y^2)+13x+26y+1   ->(xy+10)^2=(3x+6y)^2+13(x+2y)+1   ->(xy+10)^2=9(x+2y)^2+13(x+2y)+1   Đặt {(x+2y=a(a>0)),(xy+10=b(b>10)):}   ->b^2-1=9a^2+13a   &harr;9a^2+6a. 13/6+169/36-169/36=b^2-1   &harr;(3a+13/6)^2-b^2=133/36   &harr;(18a+13)^2-36b^2=133   &harr;(18a-6b+13)(18a+6b+13)=133(1)   Ta c&oacute; : a,b>0->18a+6b+13>18a-6b+13>0   Lại c&oacute; : 133=19.7   ->(1)&harr; ( left[  begin{array}{l} left  { {{18a+6b+13=133}  atop {18a-6b+13=1}}  right.    left  { {{18a+6b+13=19}  atop {18a-6b+13=7}}  right.  end{array}  right. )    &harr; ( left[  begin{array}{l} left  { {{18a+6b=120}  atop {18a-6b=-12}}  right.    left  { {{18a+6b=32}  atop {18a-6b=-6}}  right.  end{array}  right. )    &harr; ( left[  begin{array}{l} left  { {{12b=132}  atop {3a-b=-2}}  right.    left  { {{12b=38}  atop {3a-b=-1}}  right.  end{array}  right. )    &harr; ( left[  begin{array}{l} left  { {{b=11}  atop {a=3}}  right.    left  { {{b= frac{19}{6}(KTM)}  atop {b=- frac{25}{18}(KTM)}}  right.  end{array}  right. )    ->{(x+2y=3),(xy+10=11):}   &harr;{(x=3-2y),(xy=1):}   &harr;{(x=3-2y),(2y^2-3y+1=0):}   &harr;{(x=3-2y),(2y-1=0),(y-1=0):}   &harr;{(x=1),(y=1):}   Vậy x,y={1;1}