设A=[5,2,0,0;2,1,0,0;0,0,5,2;0,0,2,1],则丨2A丨=?这个可以用分块矩阵去算吗?还是先算A*和A^(-1)呢?

设a>1,0 设a>1,0 (1/2)设a>0,|x| 设a>b>0,证明a^2+1/ab+1/a(a-b)>=4 设A={-2≤x ≤a,0 设集合A={0 设a>0,0 设a>0,0 设a>0,0 设a>0,0 数列极限：设2a^2-5a+2=0,计算lim a^n/[1+a^(n+1)] 设0＜a 设0〈a 设0≤a 设a>0,b 设0≤a 设0≤a 设0≤a