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Clearnote
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Junior High
Mathematics
11的2跟挑戰1
Mathematics
Junior High
6 monthsago
Chloe
11的2跟挑戰1
最 m-n= 3 7.若√(x+y-3)²+v (4x-3y-5)2=0,求: .2x+y=5 Sx+y=3 {4x+4y=12 (4x-3y = 5 14x-3y=5 y = 1 X=2 8.已知: : (1) - 3是4a-3的平方根 49-3=9 5 40=17 (2)b為正整數,且~1806為整數 9=3 由(1)、(2)可得知a+b的最小值為 8 9.設/15-a為整數,則正整數a的值可能為6.11.14.15 0.1.4.9 19 ×4 x6 ⑨⑩.若(10x2-2x²-3x)除以(2x-1)所得的商式為 (5x+2x-2),餘式為(-1),則(40x-8x²-12-19) 除以(12x-6)所得的餘式為 22 11.(1)求x4+2x3-4x²-5x+5除以x²+x-3的商式為何? 2 X+X-2 (2)承(1),若x²+x=3,則x++2x²-4x²-5x+5之值為 四、挑戰題:(2分) 多項式x²+ax²+bx+c能被多項式x²+3x-4整除,求 2a-2b-c 的值
Answers
Lillian🌿🍁🐚
6 monthsago
因該是這樣🧠🧠🧠
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