设b²-2ab>0,关于x的不等式a²x² b²(1-a)≥[ax+b(1-x)]²的解集为对不起,原不等式为a²x² +b²(1-x)≥[ax+b(1-x)]²
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![设b²-2ab>0,关于x的不等式a²x² b²(1-a)≥[ax+b(1-x)]²的解集为对不起,原不等式为a²x² +b²(1-x)≥[ax+b(1-x)]²](/uploads/image/z/959258-2-8.jpg?t=%E8%AE%BEb%26%23178%3B-2ab%3E0%2C%E5%85%B3%E4%BA%8Ex%E7%9A%84%E4%B8%8D%E7%AD%89%E5%BC%8Fa%26%23178%3Bx%26%23178%3B+b%26%23178%3B%281-a%29%E2%89%A5%5Bax%2Bb%281-x%29%5D%26%23178%3B%E7%9A%84%E8%A7%A3%E9%9B%86%E4%B8%BA%E5%AF%B9%E4%B8%8D%E8%B5%B7%EF%BC%8C%E5%8E%9F%E4%B8%8D%E7%AD%89%E5%BC%8F%E4%B8%BAa%26%23178%3Bx%26%23178%3B+%2Bb%26%23178%3B%281-x%29%E2%89%A5%5Bax%2Bb%281-x%29%5D%26%23178%3B)
设b²-2ab>0,关于x的不等式a²x² b²(1-a)≥[ax+b(1-x)]²的解集为对不起,原不等式为a²x² +b²(1-x)≥[ax+b(1-x)]²
设b²-2ab>0,关于x的不等式a²x² b²(1-a)≥[ax+b(1-x)]²的解集为
对不起,原不等式为a²x² +b²(1-x)≥[ax+b(1-x)]²
设b²-2ab>0,关于x的不等式a²x² b²(1-a)≥[ax+b(1-x)]²的解集为对不起,原不等式为a²x² +b²(1-x)≥[ax+b(1-x)]²
a^2x^2-b^2x+b^2≥[(a-b)x+b]^2
a^2x^2-b^2x+b^2≥(a-b)^2x^2+2b(a-b)x+b^2
[(a-b)^2-a^2]x+[2b(a-b)+b^2]x≤0
(b^2-2ab)x^2+(2ab-b^2)x≤0
(b^2-2ab)x^2-(b^2-2ab)x≤0
因为b^-2ab>0,所以
x^2-x≤0
故得 0≤x≤1.