方程组{█(x+y=a@xy=b)的两组解为{█(x1=a1@y1=b1) {█(x2=a2@y2=b2)┤ ┤ ┤ 求a1a2-b1b2=?

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方程组{█(x+y=a@xy=b)的两组解为{█(x1=a1@y1=b1) {█(x2=a2@y2=b2)┤ ┤ ┤ 求a1a2-b1b2=?
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方程组{█(x+y=a@xy=b)的两组解为{█(x1=a1@y1=b1) {█(x2=a2@y2=b2)┤ ┤ ┤ 求a1a2-b1b2=?
方程组{█(x+y=a@xy=b)的两组解为{█(x1=a1@y1=b1) {█(x2=a2@y2=b2)┤ ┤ ┤ 求a1a2-b1b2=?

方程组{█(x+y=a@xy=b)的两组解为{█(x1=a1@y1=b1) {█(x2=a2@y2=b2)┤ ┤ ┤ 求a1a2-b1b2=?
y=a-x
x(a-x)=b
x^2-ax+b=0
由韦达定理知
x1x2=b,x1+x2=a
y1y2=(a-x1)(a-x2)=a^2-a(x1+x2)+x1x2=b
a1a2-b1b2=0

a₁+b₁=a a₂+b₂=a 则a₁+b₁=a₂+b₂即a₁-a₂=b₂-b₁
(a₁-a₂)²=(b₂-b₁)² ∴a...

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a₁+b₁=a a₂+b₂=a 则a₁+b₁=a₂+b₂即a₁-a₂=b₂-b₁
(a₁-a₂)²=(b₂-b₁)² ∴a₁²-2a₁a₂+a₂²=b₁²-2b₁b₂+b₂²
∴2(a₁a₂﹣b₁b₂)=a₁²+a₂²-b₁²-b₂²
=(a₁+b₁)(a₁-b₁)+(a₂+b₂)(a₂﹣b₂)
(a₁-b₁)²=(a₁+b₁)²-4a₁a₂=a²-4b,则 (a₁-b₁)=根号下(a²-4b)
同理,(a₂-b₂)=根号下(a²-4b)
则a₁a₂﹣b₁b₂=a²×根号下(a²-4b)

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a1a2-b1b2=0

方程组消掉x得:x(a-x)=b,即x²-ax+b=0,∴x1x2=b,即a1a2=b;
方程组消掉y得:y(a-y)=b,即y²-ay+b=0,∴y1y2=b,即b1b2=b;
所以a1a2-b1b2=b-b=0