用换元法因式分解 (3x²+24x+7)(2x²+16x+15)+14
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设x^2+8x=y
则原式=(3y+7)(2y+15)+14
=6y^2+59y+119
=(6y+17)(y+7)
将y替换为x得原式=(6x^2+48x+17)(x^2+8x+7)
=(6x^2+48x+17)(x+1)(x+7)
哪里不懂再问。求采纳。
则原式=(3y+7)(2y+15)+14
=6y^2+59y+119
=(6y+17)(y+7)
将y替换为x得原式=(6x^2+48x+17)(x^2+8x+7)
=(6x^2+48x+17)(x+1)(x+7)
哪里不懂再问。求采纳。
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设x²+8x+3=m
∵(3x²+24x+7)(2x²+16x+15)+14
=6m²+23m-4
=﹙6m+1﹚﹙m-4﹚
①6m+1=6﹙x²+8x+3﹚+1=﹙6x﹢24-√462﹚﹙x﹢﹤24+√462﹥/6﹚
②m-4=x²+8x-1=﹙x+4-√17﹚﹙x+4+√17﹚
∴ (3x²+24x+7)(2x²+16x+15)+14
=﹙6x﹢24-√462﹚﹙x﹢﹤24+√462﹥/6﹚﹙x+4-√17﹚﹙x+4+√17﹚
∵(3x²+24x+7)(2x²+16x+15)+14
=6m²+23m-4
=﹙6m+1﹚﹙m-4﹚
①6m+1=6﹙x²+8x+3﹚+1=﹙6x﹢24-√462﹚﹙x﹢﹤24+√462﹥/6﹚
②m-4=x²+8x-1=﹙x+4-√17﹚﹙x+4+√17﹚
∴ (3x²+24x+7)(2x²+16x+15)+14
=﹙6x﹢24-√462﹚﹙x﹢﹤24+√462﹥/6﹚﹙x+4-√17﹚﹙x+4+√17﹚
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f(x)= (3x²+24x+7)(2x²+16x+15)+14
f(-1) = 0
f(-7)= 0
f(x) = (x+1)(x+7)(6x^2+ax+b)
coef .of const.
7b= 119, =>b=17
put x= 1
1136= 16(6+a+17)
a= 48
f(x) = (x+1)(x+7)(6x^2+48x+17)
= -1952/5
f(x) = (x-1)(x-7)(6x^2-(1952/5)x+17)
f(-1) = 0
f(-7)= 0
f(x) = (x+1)(x+7)(6x^2+ax+b)
coef .of const.
7b= 119, =>b=17
put x= 1
1136= 16(6+a+17)
a= 48
f(x) = (x+1)(x+7)(6x^2+48x+17)
= -1952/5
f(x) = (x-1)(x-7)(6x^2-(1952/5)x+17)
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(3(x²+8x+5)-8)(2(x²+8x+5)+5)+14
=(3t-8)(2t+5)+14
=6t^2-t-40+14
=6t^2-t+26
=(t+2)(6t-13)
=(x^2+8x+5+2)(6x^2+48x+30-13)
=(x^2+8x+7)(6x^2+48x+17)
=(x+1)(x+7)(6x^2+48x+17)
=(3t-8)(2t+5)+14
=6t^2-t-40+14
=6t^2-t+26
=(t+2)(6t-13)
=(x^2+8x+5+2)(6x^2+48x+30-13)
=(x^2+8x+7)(6x^2+48x+17)
=(x+1)(x+7)(6x^2+48x+17)
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不会。
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