设集合A={x|4x–2x+2+a=0,x∈R}.(1)若A中仅有一个元素,求实数a的取值集合B设集合A={x|4x–2x+2+a=0,x∈R}.(1)若A中仅有一个元素,求实数a的取值集合B;(2)若对于任意a∈B,不等式x2–6x<a(x–2)
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![设集合A={x|4x–2x+2+a=0,x∈R}.(1)若A中仅有一个元素,求实数a的取值集合B设集合A={x|4x–2x+2+a=0,x∈R}.(1)若A中仅有一个元素,求实数a的取值集合B;(2)若对于任意a∈B,不等式x2–6x<a(x–2)](/uploads/image/z/2724196-4-6.jpg?t=%E8%AE%BE%E9%9B%86%E5%90%88A%3D%7Bx%EF%BD%9C4x%E2%80%932x%2B2%2Ba%3D0%2Cx%E2%88%88R%7D.%EF%BC%881%EF%BC%89%E8%8B%A5A%E4%B8%AD%E4%BB%85%E6%9C%89%E4%B8%80%E4%B8%AA%E5%85%83%E7%B4%A0%2C%E6%B1%82%E5%AE%9E%E6%95%B0a%E7%9A%84%E5%8F%96%E5%80%BC%E9%9B%86%E5%90%88B%E8%AE%BE%E9%9B%86%E5%90%88A%3D%7Bx%EF%BD%9C4x%E2%80%932x%2B2%2Ba%3D0%2Cx%E2%88%88R%7D.%EF%BC%881%EF%BC%89%E8%8B%A5A%E4%B8%AD%E4%BB%85%E6%9C%89%E4%B8%80%E4%B8%AA%E5%85%83%E7%B4%A0%2C%E6%B1%82%E5%AE%9E%E6%95%B0a%E7%9A%84%E5%8F%96%E5%80%BC%E9%9B%86%E5%90%88B%EF%BC%9B%EF%BC%882%EF%BC%89%E8%8B%A5%E5%AF%B9%E4%BA%8E%E4%BB%BB%E6%84%8Fa%E2%88%88B%2C%E4%B8%8D%E7%AD%89%E5%BC%8Fx2%E2%80%936x%EF%BC%9Ca%28x%E2%80%932%29)
设集合A={x|4x–2x+2+a=0,x∈R}.(1)若A中仅有一个元素,求实数a的取值集合B设集合A={x|4x–2x+2+a=0,x∈R}.(1)若A中仅有一个元素,求实数a的取值集合B;(2)若对于任意a∈B,不等式x2–6x<a(x–2)
设集合A={x|4x–2x+2+a=0,x∈R}.(1)若A中仅有一个元素,求实数a的取值集合B
设集合A={x|4x–2x+2+a=0,x∈R}.
(1)若A中仅有一个元素,求实数a的取值集合B;
(2)若对于任意a∈B,不等式x2–6x<a(x–2)恒成立,求x的取值范围.
设集合A={x|4x–2x+2+a=0,x∈R}.(1)若A中仅有一个元素,求实数a的取值集合B设集合A={x|4x–2x+2+a=0,x∈R}.(1)若A中仅有一个元素,求实数a的取值集合B;(2)若对于任意a∈B,不等式x2–6x<a(x–2)
题目打错了吧,是不是4^x-2^(x+2)+a=0
(1)令2x=t(t>0),设f(t)=t^2-4t+a,由f(t)=0在(0,+∞)上仅有一根或两相等实根、有
①f(t)=0有两等根时,△=0 ∴16-4a=0 解得a=4.
验证:t^2-4t+4=0 t=2∈(0,+∞)这时x=1.
②f(t)=0有一正根和一负根时,f(0)