Bài 1: Chứng minh rằng a) (a + b + c)^3 = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c + a) b) (x + y)^3 – (x – y)^3 = 2y(y^2 + 3x^2) Bài 2: Tính nhanh a) 99^3

Question

Bài 1: Chứng minh rằng
a) (a + b + c)^3 = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c + a)
b) (x + y)^3 – (x – y)^3 = 2y(y^2 + 3x^2)
Bài 2: Tính nhanh
a) 99^3
b) 91^3 + 3*91^2*9 + 3*91*9^2
c) 1001^3
d) 102^3 – 6*102^2 + 12*102 – 8
Bài 3: Tính giá trị biểu thức
a) A=x^3 – 3x^2 + 3x – 1 tại x=1001
b) B=8x^3 + 12x^2 + 6x+1 tại x=24,5
c) C=x^3/27 – 1 – x^2/3 + x tại x=303
d) D=(x+y)^3 – 9(x+y)^2 + 27(x+y) – 27 tại x=2;y=6

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Kylie 1 năm 2021-07-05T18:44:06+00:00 1 Answers 13 views 0

Answers ( )

    0
    2021-07-05T18:45:31+00:00

    Đáp án:

    $\begin{array}{l}
    1)a){\left( {a + b + c} \right)^3}\\
     = {a^3} + 3{a^2}\left( {b + c} \right) + 3a{\left( {b + c} \right)^2} + {\left( {b + c} \right)^3}\\
     = {a^3} + 3\left( {b + c} \right)\left( {{a^2} + a\left( {b + c} \right)} \right) + {b^3} + 3{b^2}c + 3b{c^2} + {c^3}\\
     = {a^3} + {b^3} + {c^3} + 3\left( {b + c} \right)\left( {{a^2} + ab + ac} \right) + 3bc\left( {b + c} \right)\\
     = {a^3} + {b^3} + {c^3} + 3\left( {b + c} \right)\left( {{a^2} + ab + ac + bc} \right)\\
     = {a^3} + {b^3} + {c^3} + 3\left( {b + c} \right)\left( {a + b} \right)\left( {a + c} \right)\\
    b){\left( {x + y} \right)^3} – {\left( {x – y} \right)^3}\\
     = {x^3} + 3{x^2}y + 3x{y^2} + {y^3}\\
     – \left( {{x^3} – 3{x^2}y + 3x{y^2} – {y^3}} \right)\\
     = 6{x^2}y + 2{y^3}\\
     = 2y\left( {{y^2} + 3{x^2}} \right)\\
    B2)a){99^3}\\
     = {\left( {100 – 1} \right)^3}\\
     = {100^3} – {3.100^2} + 3.100 – 1\\
     = 1000000 – 30000 + 300 – 1\\
     = 970000 + 299\\
     = 970299\\
    b){91^3} + {3.91^2}.9 + {3.91.9^2}\\
     = {91^3} + {3.91^2}.9 + {3.91.9^2} + {9^3} – {9^3}\\
     = {\left( {91 + 9} \right)^3} – 81.9\\
     = {100^3} – 729\\
     = 1000000 – 729\\
     = 999271\\
    c){1001^3}\\
     = {\left( {1000 + 1} \right)^3}\\
     = {1000^3} + {3.1000^2} + 3.1000 + 1\\
     = 1000000000 + 3000000 + 3000 + 1\\
     = 1003003001\\
    d){102^3} – {6.102^2} + 12.102 – 8\\
     = {\left( {102 – 2} \right)^3}\\
     = {100^3}\\
     = 1000000\\
    B3)a)A = {x^3} – 3{x^2} + 3x – 1\\
     = {\left( {x – 1} \right)^3}\\
     = {\left( {1001 – 1} \right)^3}\\
     = {1000^3}\\
     = 1000000000\\
    b)B = 8{x^3} + 12{x^2} + 6x + 1\\
     = {\left( {2x + 1} \right)^3}\\
     = {\left( {2.24,5 + 1} \right)^3}\\
     = {50^3}\\
     = 125000\\
    c)C = \dfrac{{{x^3}}}{{27}} – 1 – \dfrac{{{x^2}}}{3} + x\\
     = \dfrac{{{x^3}}}{{27}} – \dfrac{{{x^2}}}{3} + x – 1\\
     = {\left( {\dfrac{x}{3} – 1} \right)^3}\\
     = {\left( {\dfrac{{303}}{3} – 1} \right)^3}\\
     = {\left( {101 – 1} \right)^3}\\
     = {100^3}\\
     = 1000000\\
    d)D = {\left( {x + y} \right)^3} – 9{\left( {x + y} \right)^2} + 27\left( {x + y} \right) – 27\\
     = {\left( {x + y – 3} \right)^3}\\
     = {\left( {2 + 6 – 3} \right)^3}\\
     = {5^3}\\
     = 125
    \end{array}$

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