Question

x∧2-x-7/4=0

Answer 1

x²-x-7/4=0,
4x²-4x-7=0，
判别式△=(-4)²-4×4×(-7)=128=(8√2)²,
x1=(4+8√2)/(4×2)=(1+2√2)/2,
x2=(4-8√2)/(4×2)=(1-2√2)/2,

Answer 2

解： x=2代入曲线方程，得 22-2y+y2=7 y2-2y-3=0 (y+1)(y-3)=0 y=-1或y=3 x2-xy+y2=7，等式两边同时求导 2x-(y+xy')+2yy'=7 y'=(2x-y-7)/(x-2y) x=2，y=-1时，y'=[2·2-(-1)-7]/[2-2·(-1)]=-? y-(-1)=-?(x-2)，整理，得x+2y=0 x=2，y=3时，y'=(2·2-3-7)/(2-2·3)=3/2 y-3=(3/2)(x-2)，整理，得3x-2y=0 曲线在x=2相应点处切线有两条，在(2，-1)点处的切线方程为x+2y=0；在(2，3)处的切线方程为3x-2y=0

Answer 3

x^2-x-7/4=0
x^2-x + 1/4 = 2
(x-1/2)^2=2
x-1/2 = √2 or -√2
x= (1+2√2)/2 or (1-2√2)/2