Yahoo 知識+ 將於 2021 年 5 月 4 日 (美國東岸時間) 停止服務。從 2021 年 4 月 20 日 (美國東岸時間) 起,Yahoo 知識+ 網站將轉為僅限瀏覽模式。其他 Yahoo 資產或服務,或你的 Yahoo 帳戶將不會有任何變更。你可以在此服務中心網頁進一步了解 Yahoo 知識+ 停止服務的事宜,以及了解如何下載你的資料。

YuenKi 發問於 科學及數學數學 · 1 十年前

IDENTITIES

1. If x(x-4)(2x+3) is an identity of Px^3 + Qx^2 +Rx + S for all values of x, find the values of P, Q, R and S.

2. If 3x^2 - 7x + 6 is an identity of Ax(x-1) + Bx(x-2) + C(x-1)(x-2), what are the values of A, b and C?

3. If Px^2 + Q(x+2) - R(x-3) is an identity of (x+3)(3x-2), find the values of P, Q and R.

2 個解答

評分
  • sunny
    Lv 5
    1 十年前
    最愛解答

    1. If x(x-4)(2x+3) is an identity of Px^3 + Qx^2 +Rx + S for all values of x, find the values of P, Q, R and S.

    x(x-4)(2x+3) ≡ Px³ + Qx² +Rx + S

    L.H.S. = (x² - 4x)(2x + 3)

    ..........= 2x³ + 3x² - 8x² - 12x

    R.H.S. = Px³ + Qx² +Rx + S

    2x³ + 3x² - 8x² - 12x ≡ Px³ + Qx² +Rx + S

    By comparing the coefficient,

    P = 2 , Q = 3, R = -8, S = 0

    2. If 3x^2 - 7x + 6 is an identity of Ax(x-1) + Bx(x-2) + C(x-1)(x-2), what are the values of A, b and C?

    3x² - 7x + 6 ≡ Ax(x-1) + Bx(x-2) + C(x-1)(x-2)

    L.H.S. = 3x² - 7x + 6

    R.H.S. = Ax(x-1) + Bx(x-2) + C(x-1)(x-2)

    ...........= Ax² - Ax + Bx² - 2Bx + C(x² - 3x + 2)

    ...........= Ax² - Ax + Bx² - 2Bx + Cx² - 3Cx + 2C

    ...........= Ax² + Bx² + Cx² - Ax - 2Bx - 3Cx + 2C

    ...........= ( A+B+C)x² - (A + 2B + 3C)x + 2C

    3x² - 7x + 6 ≡ ( A+B+C)x² - (A + 2B + 3C)x + 2C

    By comparing the constant term ,

    2C = 6

    C = 3 //

    By comparing the coefficient of x²

    A + B + 3 = 3

    A + B = 0......................(1)

    By comparing the coefficient of x

    A + 2B + 9 = 7

    A + 2B = -2.....................(2)

    (2) - (1):

    A + 2B - A - B = -2 - 0

    B = -2 //

    Substitute B = -2 into (1),

    A - 2 = 0

    A = 2 //

    A = 2, B = -2, C = 3

    3. If Px^2 + Q(x+2) - R(x-3) is an identity of (x+3)(3x-2), find the values of P, Q and R

    Px² + Q(x+2) - R(x-3) ≡ (x+3)(3x-2)

    L.H.S. = Px² + Qx + 2Q - Rx + 3R

    ..........= Px² + (Q-R)x + 2Q + 3R

    R.H.S. = (x+3)(3x-2)

    ...........= 3x² - 2x + 9x - 6

    ...........= 3x² + 7x - 6

    Px² + (Q-R)x + 2Q + 3R ≡ 3x² + 7x - 6

    By comparing the coefficient of x²

    P = 3 //

    By comparing the coefficient of x

    Q - R = 7.............(1)

    By comparing the constant term,

    2Q + 3R = -6........(2)

    From (1),

    Q = R + 7.............(3)

    Substitute (3) into (2)

    2( R+7 ) + 3R = -6

    2R + 14 + 3R = -6

    5R = -20

    R = -4 //

    Substitute R = -4 into (3)

    Q = -4 + 7

    Q = 3 //

    P = 3, Q = 3, R = -4

    2007-02-11 13:02:57 補充:

    小提示: 有時 D identity 的題目, 計完之後, 唔可以直接找出答案, 所以我地要將整條式 expand 左佢, 再 factorize D like terms, 然後仲要用二元一次方程式解 ~( simultaneous equations in two unknowns )第二, 三題就係一個gd. example, 由於需要寫好多 steps , 所以需要好小心, 答完最好都係驗算多幾次, 咁就萬無一失喇~

    資料來源: me~
  • 1 十年前

    1.If x(x-4)(2x+3) is an identity of Px^3 + Qx^2 +Rx + S for all values of x, find the values of P, Q, R and S.

    LHS=x(x-4)(2x+3)

    ------=(x^2-4x)(2x+3)

    ------=2x^3-8x^2+3x^2-12x

    ------=2x^3-5x^2-12x

    RHS=Px^3+Qx^2+Rx+S

    Compare the coefficient of x^3 , x^2 , x^1 and the constant P,Q,R,S =

    2 , -5 , -12 & 0 respectively.

    2.If 3x^2 - 7x + 6 is an identity of Ax(x-1) + Bx(x-2) + C(x-1)(x-2), what are the values of A, b and C?

    LHS=3x^2 - 7x + 6

    RHS=Ax(x-1) + Bx(x-2) + C(x-1)(x-2)

    ------=Ax^2-Ax + Bx^2-2Bx + (Cx-C)(x-2)

    ------=(A+B)x^2 - (A+2B)x + Cx^2 - Cx - 2Cx + 2C

    ------=(A+B+C)x^2 - (A+2B+3C)x + 2C

    Compare the coefficient of x^2 , x^1 and the constant

    2C = 6

    C=3

    (A+B+C)x^2 = 3x^2

    (A+B+3)x^2 = 3x^2

    (A+B+3) = 3

    A+B=0------(1)

    (A+2B+3C)x =7x

    (A+2B+9)=7

    A + 2B = -2-----(2)

    (2) - (1)

    (A + 2B) - (A+B) = -2 - 0

    B = -2

    Sub B = -2 into (1)

    A - 2 =0

    A = 2

    ∴A = 2 , B = -2 , C = 3

    3. If Px^2 + Q(x+2) - R(x-3) is an identity of (x+3)(3x-2), find the values of P, Q and R.

    LHS= Px^2 + Q(x+2) - R(x-3)

    ------= Px^2 + Qx + 2Q – Rx + 3R

    ------= Px^2 + x(Q-R) + 2Q + 3R

    RHS= (x+3)(3x-2)

    ------=3x^2 + 9x – 2x -6

    ----- =3x^2 + 7x -6

    Compare the coefficient of x^2 , x^1 and the constant

    Px^2 = 3x^2

    P=3

    x(Q-R) = 7x

    Q – R = 7

    Q = 7 +R ------(1)

    2Q + 3R = -6 -----(2)

    Sub (1) into (2)

    2 (7+R) + 3R = -6

    14 + 5R = -6

    R = -4

    Sub R = 4 into (1)

    Q = 7 - 4

    Q =3

    ∴P=3 , R = -4 , Q =3

還有問題嗎?立即提問即可得到解答。