1.某正數與它的平方數之和是72,求此數
2.三個連續的正整數中,最小的一個是k,另外兩個數的乘積等於k+26.求這三個數
3.兩個連續正奇數的平方和為202,求此兩數
4.一正方形的邊長為10cm.假設每邊增長為x cm,則正方形的面積增加了96 cm的2次方.
求x的值.
各位高手,請於今天內help me.please~~
thx~~
F.4 Maths please help me
2007-09-27 1:11 am
回答 (2)
2007-09-27 1:31 am
✔ 最佳答案
1)設該正數為 x
x + x^2 = 72
x^2 + x - 72 = 0
(x+9)(x-8) = 0
x = -9 (rej) or x = 8
該正數為 8
2)
(k+1)(k+2) = k + 26
k^2 + 3k + 2 = k + 26
k^2 + 2k - 24 = 0
(k+6)(k-4) = 0
k = -6 (Rej) or k = 4
因此, 該三個正整數為 4, 5, 6
3)
設該正奇數為 x
x^2 + (x+2)^2 = 202
x^2 + x^2 + 4x + 4 = 202
2x^2 + 4x + 4 = 202
x^2 + 2x + 2 = 101
x^2 + 2x - 99 = 0
(x+11)(x-9) = 0
x = -9 (rej) or x = 11
因此該兩個奇數為 11 和 13
4)
(10+x)^2 - (10)^2 = 96
x^2 + 20x + 100 - 100 = 96
x^2 + 20x - 96 = 0
(x+24)(x-4) = 0
x = -24(rej) or x = 4
參考: Me
2007-09-27 1:36 am
1.假設某數是x
x+x^2=72
x^2+x -72=0
(x+9)*(x-8)=0
(x+9)=0 or (x - 8)=0
x= - 9 or 8
2. (k+1)*(k+2)=k+26
k^2+3k+2=k+26
k^2+3k+2 - k -26=0
k^2+2k - 24=0
(k+6)*(k - 4)=0
(k+6)=0 or (k - 4)=0
k= -6(rej.) or 4
那三個數是4.5和6
3. 假設較小的數是x
x^2+(x+2)^2=202
x^2+(x^2+4x+4)=202
2(x^2)+4x+4=202
2(x^2)+4x+4-202=0
2(x^2)+4x -198=0
(x+11)*(x-9)=0
(x+11)=0 or (x-9)=0
x= -11(rej) or 9
此兩數為 9 .11
4. (10+x)^2 -10^2=96
(100+20x+x^2) - 100 = 96
x^2+20x+100 -100 - 96 = 0
x^2+20x - 96 = 0
(x+24)*(x-4) = 0
(x+24) = 0 or (x-4) = 0
x= -24 (rej.) or 4
x = 4
x+x^2=72
x^2+x -72=0
(x+9)*(x-8)=0
(x+9)=0 or (x - 8)=0
x= - 9 or 8
2. (k+1)*(k+2)=k+26
k^2+3k+2=k+26
k^2+3k+2 - k -26=0
k^2+2k - 24=0
(k+6)*(k - 4)=0
(k+6)=0 or (k - 4)=0
k= -6(rej.) or 4
那三個數是4.5和6
3. 假設較小的數是x
x^2+(x+2)^2=202
x^2+(x^2+4x+4)=202
2(x^2)+4x+4=202
2(x^2)+4x+4-202=0
2(x^2)+4x -198=0
(x+11)*(x-9)=0
(x+11)=0 or (x-9)=0
x= -11(rej) or 9
此兩數為 9 .11
4. (10+x)^2 -10^2=96
(100+20x+x^2) - 100 = 96
x^2+20x+100 -100 - 96 = 0
x^2+20x - 96 = 0
(x+24)*(x-4) = 0
(x+24) = 0 or (x-4) = 0
x= -24 (rej.) or 4
x = 4
參考: me
收錄日期: 2021-04-13 19:02:57
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