F.4 Maths+M2 quick

用戶25529

2010-10-23 9:32 pm

1) Solve x^2+ 3 = 5/3x + 83/86 by using qudratic formula.

2a) Expand (3x+2/x)^4 by using Pascal's triangle
2b) Hence, find the constant term

3a) Expand [ (x^2)/3 - (y/x) ] ^6 by using Pascal's triangle
3b) If the constant term after its expansion is 36 , find all possible values of y

4) Find the coefficient of x^2y^6 in the expansion of (5x-y)^8

5) Find the constant term in the expansion of [ (y/2) - (2/y) ]^9

回答 (1)

用戶19657

2010-10-23 10:27 pm

✔ 最佳答案

x^2+ 3 = 5/3x + 83/86
x^2-5/3x-175/86=0
by quadratic formula,
x=[(5/3)+- square root of (25/9 - 4(1)(-175/86)]/2
=( 5/3 +- 3.304)/2
=2.4854 or -0.8187

2a) Expand (3x+2/x)^4 by using Pascal's triangle
=(3x)^4 + 4(3x)^3(2/x) +6(3x)^2(2/x)^2 +4(3x)(2/x)^3 + (2/x)^4
2b)=81x^4 +216x^2 +216 +96/(x^2) + 16/(x^4)
.'. constant term216

3a) Expand [ (x^2)/3 - (y/x) ] ^6 by using Pascal's triangle
=[(x^2)/3)]^6 +6{[(x^2)/3]^5 (-y/x)}+15{(x^2)/3]^4 (y/x)^2} +20{[(x^2)/3]^3 (-y/x)^3} +15{[(x^2)/3]^2 (-y/x)^4} +6 [(x^2)/3 (-y/x)^5] +(-y/x)^6
3b) If the constant term after its expansion is 36 , find all possible values of y

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