對于0≤m≤4的m,不等式x^2+mx>4x+m-3恒成立,則x的取值范圍是?
熱心網(wǎng)友
對于0≤m≤4的m,不等式x^2+mx4x+m-3恒成立,則x的取值范圍是?x^2-4x+3+mx-m0(x-1)(x-3+m)0當 x1 x-3+m0 === m3-x === 03-x === x3 x m 4 x3 或 x<-1即： x∈(-OO,-1)U(3,+00)
熱心網(wǎng)友
x^2+mx4x+m-3m(x-1)-x^2+4x-3if x1m(-x^2+4x-3)/(x-1)=(x-3)(1-x)/(x-1)=3-x===xx-1-1 本題目有問題,請改正后再上傳熱心網(wǎng)友