F2 Maths Help!!!!(40POINTS)!!－快遞百科

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F2 Maths Help!!!!(40POINTS)!!

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列式/步驟/解釋1) Determine x(y-2)+y(z-x)+z(x-y)=02) Find thw constants A&B A(x-5)+B(x+3)三4x-43) Prove that the following expressions are indentities (x2+1)(x2-x+1)三 x4+x2+14) Prove that the following expressions are indentities x(x2+9)(x+3)(x-3)三x5-81x5) Expand 1... 顯示更多列式/步驟/解釋 1) Determine x(y-2)+y(z-x)+z(x-y)=0 2) Find thw constants A&B A(x-5)+B(x+3)三4x-4 3) Prove that the following expressions are indentities (x2+1)(x2-x+1)三 x4+x2+1 4) Prove that the following expressions are indentities x(x2+9)(x+3)(x-3)三x5-81x 5) Expand 1 -- (5b-2a)(5b+2a) 5 ( 3/4 a3 - 2/3 b3)2 ** 英文字母後的數字為次方

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1) Determine x(y-2)+y(z-x)+z(x-y) =0 xy-2x+yz-xy+zx-zy=0 zx – 2x = 0 x(z – 2) = 0 則 x = 0 或 z = 2，y可以是任何數字。 2) Find the constants A&B A(x-5)+B(x+3) ≡ 4x-4 Ax – 5A + Bx + 3B ≡ 4x – 4 (A+B)x – 5A + 3B ≡ 4x – 4 A + B = 4 及 -5A + 3B = -4 因 A = 4 - B -5(4 – B) + 3B = -4 -20 + 5B + 3B = -4 8B = 16 B = 2 A + B = 4 A + 2 = 4 B = 2 3) Prove that the following expressions are indentities (x2+1)(x2-x+1) ≡ x4+x2+1 LHS = (x2+1)(x2-x+1) = x2(x2-x+1) + (x2-x+1) = x4 – x3 + x2 + x2 – x + 1 = x4 – x3 + 2x2 – x + 1 不等於 RHS 所以這式是非全等的。 4) Prove that the following expressions are indentities x(x2+9)(x+3)(x-3) ≡x5-81x RHS = x5-81x = x(x4 – 34) = x(x2 + 32)(x2 – 32) = x(x2 + 9)(x – 3)(x + 3) = LHS 所以這式是全等的 5) Expand 1 -- (5b-2a)(5b+2a) 5 = (5b-2a)(5b+2a)/5 = (25b2-4a2) /5 = 5b2 – 4a2/5 (( 3/4) a3 – (2/3) b3)2 = ( 3/4)2 a6 – 2(3/4)(2/3)a3b3 + (2/3) 2 b6 = 9a6/16 - a3b3 + 4b6/9

其他解答:

1. x(y-2)+y(z-x)+z(x-y) = 0 xy - 2x + yz - xy + xz - yz = 0 - 2x + xz = 0 x( z -2 ) = 0 x = 0 or z =2 2. A(x-5)+B(x+3) 三 4x - 4 Ax - 5A + Bx + 3B = 4x - 4 (A+B)x - (5A-3B) = 4x - 4 then, you have two equation: A + B = 4 ... (1) -5A + 3B = - 4 ... (2) then put A = 4 - B into equation (2) By substitution -5(4 - B) + 3B = -4 -20 + 5B +3B = -4 B = 2 then, A = 4 -B = 4 - 2 = 2 3. (x^2+1)(x^2-x+1)三 x^4+x^2+1 LHS = (x^2+1)(x^2-x+1) = x^4 - x^3 + x + x^2 - x +1 =x^4 - x^3 +x^2 +1 RHS = x^4+x^2+1 Thus, LHS is not equal to RHS. They are not indentities 4. x(x^2+9) (x+3)(x-3)三x^5-81x LHS = (x^3 + 9x) (x^2 - 9) = x^5 - 9x^3 + 9x^3 - 81x = x^5 - 81x RHS = x^5 - 81x LHS is equals to RHS, they are identities 5a. 1/5(5b-2a)(5b+2a) =1/5 (25b^2 - 4a^2) = 5b^2 - 4/5a^2 5b. (3/4 a3 - 2/3 b3 )^2 = (1/12)^2(9a^3 - 8b^3)^2 =1/144 (81a^6 - 144a^3b^3 - 64b^6) = 9/16a^6 - a^3b^3 - 4/9b^6