已知实数a,b满足(a+b)^2=1,(a-b)^2=25,求a^2+b^2+ab的值
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![已知实数a,b满足(a+b)^2=1,(a-b)^2=25,求a^2+b^2+ab的值](/uploads/image/z/1812221-53-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AE%9E%E6%95%B0a%2Cb%E6%BB%A1%E8%B6%B3%28a%2Bb%29%5E2%3D1%2C%28a-b%29%5E2%3D25%2C%E6%B1%82a%5E2%2Bb%5E2%2Bab%E7%9A%84%E5%80%BC)
已知实数a,b满足(a+b)^2=1,(a-b)^2=25,求a^2+b^2+ab的值
已知实数a,b满足(a+b)^2=1,(a-b)^2=25,求a^2+b^2+ab的值
已知实数a,b满足(a+b)^2=1,(a-b)^2=25,求a^2+b^2+ab的值
(a+b)^2=a^2+2ab+b^2=1
(a-b)^2=a^2-2ab+b^2=25
两式相加2a^2+2b^2=26
a^2+b^2=13
两式相减4ab=-24
ab=-6
a^2+b^2+ab=7
7
∵(a+b)2=1,(a-b)2=25,
∴a2+b2+2ab=1①,a2+b2-2ab=25②,
①+②得:a2+b2=13,
①-②得:ab=-6,
∴a2+b2+ab=13-6=7.
故答案为:7.
a^2+b^2
=[(a+b)的平方+(a-b)的平方]÷2
=(1+25)÷2
=13
ab
==[(a+b)的平方-(a-b)的平方]÷4
=(1-25)÷4
=-6
a的平方+b的平方+ab
=13-6
=7
(a+b)^2-(a-b)^2=25-1=24
4ab=24
ab=6
a^2+b^2+ab=a^2+b^2+2ab-ab
=(a+b)^2-ab
=1-6
=-5
(a+b)^2=a^2+2ab+b^2=1
(a-b)^2=a^2-2ab+b^2=25
两式相加2a^2+2b^2=26
a^2+b^2=13
两式相减4ab=-24
ab=-6
a^2+b^2+ab=7
(a+b)^2=a^2+2ab+b^2=1
(a-b)^2=a^2-2ab+b^2=25
两式相加2a^2+2b^2=26
a^2+b^2=13
两式相减4ab=-24
ab=-6
a^2+b^2+ab=7