已知(x-1)²=4,求x的值。
已知(x-1)²=4,求x的值。
相关考题:
求下列代数式的值:1)-3x²+5x-0.5x²+x-1,其中x=2;2)1/4×(-4x2+2x-8)-(x/2-1),其中1/2;3)(5a²-3b²)+(a²+b²)-(5a²+3b²),其中a=-1,b=1;4)2(a²b+ab²)-2(a²b -1)-2ab²-2,其中a=-2,b=2.
2、求f(x)=x sin(2x-1)在0附近的最小值,相应的命令是()。A.[x,fval] = fminbnd(@(x) x*sin(2*x-1),0,0.5)B.[x,fval] = fminsearch(@(x) x*sin(2*x-1),0)C.[x,fval] = fminsearch(@(x) x*sin(2*x-1),0.5)D.[x,fval] = fminunc(@(x) x*sin(2*x-1),0)
求f(x)=x sin(2x-1)在0附近的最小值,相应的命令是()。A.[x,fval] = fminbnd(@(x) x*sin(2*x-1),0,0.5)B.[x,fval] = fminsearch(@(x) x*sin(2*x-1),0)C.[x,fval] = fminsearch(@(x) x*sin(2*x-1),0.5)D.[x,fval] = fminunc(@(x) x*sin(2*x-1),0)
求f(x)=x sin(2x-1)在0附近的最小值,相应的命令是()。A.[x,fval]=fminbnd(@(x) x*sin(2*x-1),0,0.5)B.[x,fval]=fminbnd(@(x) x*sin(2*x-1),0)C.[x,fval]=fminsearch(@(x) x*sin(2*x-1),[0,0.5])D.[x,fval]=fminunc(@(x) x*sin(2*x-1),[0,0.5])