Tìm a và b biết $a^{2}+b^{2}=61$ và $ab=30$

Đáp án:

     `a^2+b^2 = 61`

`⇔ (a^2+2ab+b^2)-2ab = 61`

`⇔ (a+b)^2 = 61+2ab`

`⇔ (a+b)^2 = 61+2.30`

`⇔ (a+b)^2 = 61+60`

`⇔ (a+b)^2 = 121`

`⇔ a+b = ±11`

⇔ \(\left[ \begin{array}{l}a=11-b\\a=-11-b\end{array} \right.\) 

      `a.b = 30`

`⇒ TH1 : (11-b).b = 30`

`⇔ 11b-b^2-30 = 0`

`⇔ (11b-55)-(b^2-25)=0`

`⇔ 11(b-5)-(b-5)(b+5) = 0`

`⇔ (b-5)(11-b-5) = 0`

`⇔ (b-5)(6-b) = 0`

⇔ \(\left[ \begin{array}{l}b-5=0\\6-b=0\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}b=5\\b=6\end{array} \right.\) 

⇒ \(\left[ \begin{array}{l}a=6\\a=5\end{array} \right.\) 

`TH2 : (-11-b).b = 30`

`⇔ -11b-b^2-30 = 0`

`⇔(-11b-55)-(b^2-25) = 0`

`⇔ -11(b+5)-(b-5)(b+5) = 0`

`⇔ (b+5)(-11-b+5) = 0`

`⇔ (b+5)(-6-b) = 0`

⇔ \(\left[ \begin{array}{l}b+5=0\\-6-b=0\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}b=-5\\b=-6\end{array} \right.\) 

⇒ \(\left[ \begin{array}{l}a=-6\\a=-5\end{array} \right.\) 

+Vậy `(a;b) = (5;6) ; (6;5) ; (-5;-6) ; (-6;-5)`