解
原式
=2√2+2-4×(√2/2)-2×2×2——最后面是2÷1/2再乘以2吗
=2√2+2-2√2-8
=-6
x²+2x-15=0
x²+2x=15
(x-2)/(x²-x)÷[(x+1)-3/(x-1)]
=(x-2)/x(x-1)÷[(x²-1)-3]/(x-1)
=(x-2)/x(x-1)×(x-1)/(x-2)(x+2)
=1/x(x+2)
=1/(x²+2x)
=1/15
解
原式
=2√2+2-4×(√2/2)-2×2×2——最后面是2÷1/2再乘以2吗
=2√2+2-2√2-8
=-6
x²+2x-15=0
x²+2x=15
(x-2)/(x²-x)÷[(x+1)-3/(x-1)]
=(x-2)/x(x-1)÷[(x²-1)-3]/(x-1)
=(x-2)/x(x-1)×(x-1)/(x-2)(x+2)
=1/x(x+2)
=1/(x²+2x)
=1/15