1楼:我不是他舅
^原式=(4x2-y2)(x2-y2+3x2)=(4x2-y2)2
则bai(4x2-k2x2)2=x^du4(4-k2)2*x^4=x^4
所以zhi
dao(4-k2)2=1
4-k2=±1
k2=3或5
所以k=-√
专5,属k=√5,k=-√3,k=√3
设y=kx,是否存在实数k,使得代数式(x2-y2)(4x2-y2)+3x2(4x2-y2)能化简为x4?若能,请求出所有满足
2楼:匿名用户
能;(x2-y2)(4x2-y2)+3x2(4x2-y2)=(4x2-y2)(x2-y2+3x2)
=(4x2-y2)2,
当y=kx,原式=(4x2-k2x2)2=(4-k2)2x4,令(4-k2)2=1,解得k=±3或±
5,即当k=±3或±
5时,原代数式可化简为x4.
设y=kx,是否存在实数k,使得代数式(x2-y2)(4x2-y2)+3x2(4x2-y2)能化简
3楼:过分丶正义
^把袭y=kx代入(x^bai2-y^du2)(4x^2-y^2)+3x^2(4x^2-y^2)=(4x^2-y^2)^2得
[(4-k^2)x^2]^2=x^4,
∴zhi(4-k^2)^2=1,
∴k^2-4=土1,
∴k^2=5,或3,
∴k=土√
dao5或土√3.
设y=kx,是否存在实数k,使代数式(x的平方减y的平方)(4x的平方减y的平方)+3x的平方乖(
4楼:vicky繁
能;(baix2-y2)(4x2-y2)+3x2(du4x2-y2)=(4x2-y2)(x2-y2+3x2)
=(4x2-y2)2,zhi
当y=kx,原式=(4x2-k2x2)2=(4-k2)2x4,令(4-k2)2=1,
解得k=±3或±5,
即当daok=±3或±5时,
原代数式可化简为x4.
设y=kx,是否存在有理数k,使得代数式(x-y)(4x-y)+3x(4x-y)能化简成x2?若能
5楼:匿名用户
^^y=kx
(x-y)(4x-y)+3x(4x-y)
=4x^bai2-5xy+y^2+12x^2-3xy=16x^2-8xy+y^2
=16x^2-8x(kx)+(kx)^2
=16x^2-8kx^2+k^2x^2
=(16-8k+k^2)x^2
要使(x-y)(4x-y)+3x(4x-y)= x^2即:du(16-8k+k^2)x^2= x^216-8k+k^2=1
k^2-8k+15=0
(k-3)(k-5)=0
k=3或者k=5
当k=3或者k=5时zhi,代数式(x-y)(4x-y)+3x(4x-y)能化
dao简成x^2。
6楼:赚大钱
(x-y)(4x-y)+3x(4x-y)=(4x-y)(x-y
+3x)=(4x-y)2,把y=kx代入,得(4-k)2x2,所以4-k=1,k=3
设y=kx,是否存在实数k,使得(x^2-y^2)(4x^2-y^2)+3x^2(4x^2-y^2
7楼:琉璃易碎
^^^把y=kx代入(x^2-y^2)(4x^2-y^2)+3x^2(4x^2-y^2)=(4x^2-y^2)^2得
[(4-k^2)x^2]^2=x^4,
∴(4-k^2)^2=1,
∴k^2-4=土内1,
∴k^2=5,或3,
∴k=土√
容5或土√3.
8楼:我非柠檬
^把y=kx代入(x^bai2-y^2)(4x^du2-y^zhi2)+3x^2(4x^2-y^2)=(4x^2-y^2)^2得:[(4-k^2)x^2]^2=x^4,
∴(4-k^2)^2=1,
∴k^2-4=土1,dao
∴k^2=5,或3,
∴k=土√5或土√3