单选题If (x-1) 2-(x-1) = 0, then, ______.Ax = 0 or x = 2Bx = 0 or x = 1Cx =-1 or x = 2Dx = 1 or x = 2Ex = 0 or x =-1
单选题
If (x-1) 2-(x-1) = 0, then, ______.
A
x = 0 or x = 2
B
x = 0 or x = 1
C
x =-1 or x = 2
D
x = 1 or x = 2
E
x = 0 or x =-1
参考解析
解析:
Solution 1: If(x-1) 2 -(x-1) = 0, then factoring the left side of the equation gives (x- 1) [(x-1)-1] = 0 which simplifies to (x-1)(x-2) = 0. Hence, x may be equal to 1 or 2. Solution 2: Substitute the pair of values in each of the answer choices into the given equation, (x-1) 2 -(x -1) = 0, until you find a pair, x = 1 or x = 2, that works.
Solution 1: If(x-1) 2 -(x-1) = 0, then factoring the left side of the equation gives (x- 1) [(x-1)-1] = 0 which simplifies to (x-1)(x-2) = 0. Hence, x may be equal to 1 or 2. Solution 2: Substitute the pair of values in each of the answer choices into the given equation, (x-1) 2 -(x -1) = 0, until you find a pair, x = 1 or x = 2, that works.
相关考题:
单选题曲面z-ez+2xy=3在点(1,2,0)处的切平面方程为( )。A4(x-1)+2(y-2)=0B-4(x-1)+2(y-2)=0C4(x-1)-2(y-2)=0D-4(x-1)-2(y-2)=0
单选题曲面z-ez+2xy=3在点(1,2,0)处的切平面方程为( )。A3(x-1)+2(y-2)=0B4(x-1)+2(y-2)=0C3(x-1)+(y-2)=0D4(x-1)+(y-2)=0
单选题球面x2+y2+z2=14在点(1,2,3)处的切平面方程是().A(x-1)+2(y-2)-(z-3)=0B(x+1)+2(y+2)+3(z+3)=0C(x-1)+2(y-2)+3(z-3)=0D(x+1)+2(y+2)-(z+3)=0
单选题曲面z-ez+2xy=3在点(1,2,0)处的切平面方程为( )。A4(x+1)+2(y-2)=0B4(x-1)+2(y-2)=0C4(x-1)-2(y-2)=0D4(x-1)+2(y+2)=0
单选题曲面x2+2y2+3z2=21在点(1,-2,2)的法线方程为( )。A(x-1)/1=(y+2)/4=(z-2)/6B(x-1)/1=(y+2)/(-4)=(z-2)/6C(x-1)/6=(y+2)/(-4)=(z-2)/1D(x-1)/1=(y+2)/6=(z-2)/(-4)
单选题实现集合运算求补集~A运算的对应表达式是()A~ABA==0CA(~(AB))D1(x-1)A==1(x-1)