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(1)8(x²+y²)²-x²-y²=1求x²-y²的值
(2)(x-3)(x-4)(x-5)=(x-4)(x-5)(x-6)(x-7)
(3)(y+√2+√3)(y-1)-4-4y
人气:398 ℃ 时间:2020-09-29 02:43:05
解答
(1)令x²+y²=t,则原方程化为:8t²-t=1,解得:t1=(1+根号33)/16,t2=(1-根号33)/16,即:x²+y²=(1+根号33)/16或x²+y²=(1-根号33)/16(2))(x-3)(x-4)(x-5)=(x-4)(x-5)(x-6)(x-7)...(y+√2+√3)(y-1)=4-4y(y+√2+√3)(y-1)=4-4y,(y+√2+√3)(y-1)-4(y-1)=0,(y-1)(y+√2+√3-4)=0,∴。y1=1,y2=4-√2-√3
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