maths問一問

2007-09-12 1:08 am
solve x
Q1 (x+3)(2x+1)+3=0

Q2 15(x^2+1)=34x

Q3 (2x-5)^2=(x-3)(2x-5)

請用cross method,並寫清楚步驟!thz

回答 (2)

2007-09-12 1:23 am
Q1 (x+3)(2x+1)+3=0
2x^2+7x+3+3=0
2x^2+7x+6=0
(2x+3)(x+2)=0
x= -2 or -3/2 ~~~
Q2 15(x^2+1)=34x
15x^2-34x+15=0
(5x-3)(3x-5)=0
x= 5/3 or 3/5 ~~~
Q3 (2x-5)^2=(x-3)(2x-5)
4x^2-20x+25=2x^2-11x+15
2x^2-9x+10=0
(2x-5)(x-2)=0
x=2 or 5/2 ~~~
2007-09-12 1:21 am
Q1
2x^2+x6x+3+3=0
2x^2+7x+6=0
(2x+3)(x+2)=0
x=-3/2 or x=-2
Q2
15x^2+15-34x=0
(3x-5)(5x-3)=0
x=5/3 or x=3/5
Q3
(2x-5)=(x-3)
(2x-x)=(-3+5)
x=2
參考: 個人算出


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