(1/2)三题:1)a(ac+bd)2 2)已知ad≠bc证明(a2+b2)(c2+d2)>(ac+bd)23)比较b/a和(b+m)/(a+m) (a,b,m∈R+)
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![(1/2)三题:1)a(ac+bd)2 2)已知ad≠bc证明(a2+b2)(c2+d2)>(ac+bd)23)比较b/a和(b+m)/(a+m) (a,b,m∈R+)](/uploads/image/z/3963356-44-6.jpg?t=%281%2F2%29%E4%B8%89%E9%A2%98%3A1%29a%28ac%2Bbd%292+2%29%E5%B7%B2%E7%9F%A5ad%E2%89%A0bc%E8%AF%81%E6%98%8E%28a2%EF%BC%8Bb2%29%28c2%EF%BC%8Bd2%29%3E%28ac%2Bbd%2923%29%E6%AF%94%E8%BE%83b%2Fa%E5%92%8C%28b%2Bm%29%2F%28a%2Bm%29+%28a%2Cb%2Cm%E2%88%88R%2B%29)
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(1/2)三题:1)a(ac+bd)2 2)已知ad≠bc证明(a2+b2)(c2+d2)>(ac+bd)23)比较b/a和(b+m)/(a+m) (a,b,m∈R+)
(1/2)三题:1)a(ac+bd)2
2)已知ad≠bc证明(a2+b2)(c2+d2)>(ac+bd)2
3)比较b/a和(b+m)/(a+m) (a,b,m∈R+)
(1/2)三题:1)a(ac+bd)2 2)已知ad≠bc证明(a2+b2)(c2+d2)>(ac+bd)23)比较b/a和(b+m)/(a+m) (a,b,m∈R+)
(1)构造方程ax^2+bx+c=0,x=1时有a+b+c=0,所以x=1是该方程的解.△=b^2-4ac≥0
(2)(a^2+b^2)(c^2+d^2)-(ac+bd)^2=a^2d^2+b^2c^2-2abcd=(ab-bc)^2≥0,原不等式成立.
(3)当a>b时,b/a<(b+m)/(a+m)
(a^2+b^2)(c^2+d^2) (a,b,c,d∈R)
=a^2·c^2 +b^2·d^2+a^2·d^2+b^2·c^2
=a^2·c^2 +2abcd+b^2·d^2+a^2·d^2-2abcd+b^2·c^2
=(ac+bd)^2+(ad-bc)^2
≥(ac+bd)^2,
写不过啊!