Đề kiểm tra Nhị thức Newton lớp 10 (có lời giải) - Đề 1

Ta có \({\left( {6x + 1} \right)^4} = C_4^0{\left( {6x} \right)^4} + C_4^1{\left( {6x} \right)^3} + C_4^2{\left( {6x} \right)^2} + C_4^3{\left( {6x} \right)^1} + C_4^4.\)

Ta có:

\[\begin{array}{l}x{\left( {2x - y} \right)^4} = x\left( {C_4^0{{\left( {2x} \right)}^4} + C_4^1{{\left( {2x} \right)}^3}\left( { - y} \right) + C_4^2{{\left( {2x} \right)}^2}{{\left( { - y} \right)}^2} + C_4^3\left( {2x} \right){{\left( { - y} \right)}^3} + C_4^4{{\left( { - y} \right)}^4}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = x\left( {16{x^4} - 32{x^3}y + 24{x^2}{y^2} - 8x{y^3} + {y^4}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 16{x^5} - 32{x^4}y + 24{x^3}{y^2} - 8{x^2}{y^3} + x{y^4}\end{array}\].

Ta có \(P\left( x \right) = {\left( {2x + 1} \right)^4} - {\left( {2x - 1} \right)^4}\)

\( = \left[ {C_4^0{{(2x)}^4} + C_4^1{{(2x)}^3} + C_4^2{{(2x)}^2} + C_4^3(2x) + C_4^4} \right] - \left[ {C_4^0{{(2x)}^4} - C_4^1{{(2x)}^3} + C_4^2{{(2x)}^2} - C_4^3(2x) + C_4^4} \right]\)\( = 16x + 64{x^3}\).

Vậy hệ số \({a_3} = 64\).

Xét khai triển nhị thức \[P\left( x \right) = {\left( {{x^2} + \frac{1}{2}} \right)^5} = \sum\limits_{k = 0}^5 {C_5^k{{\left( {{x^2}} \right)}^{5 - k}}{{\left( {\frac{1}{2}} \right)}^k}}  = \sum\limits_{k = 0}^5 {C_5^k{x^{10 - k}}{{\left( {\frac{1}{2}} \right)}^k}} \]

\[ = C_5^0.{x^{10}} + C_5^1.{x^8}.\frac{1}{2} + C_5^2.{x^6}.{\left( {\frac{1}{2}} \right)^2} + C_5^3.{x^4}.{\left( {\frac{1}{2}} \right)^3} + C_5^4.{x^2}.{\left( {\frac{1}{2}} \right)^4} + C_5^5.{\left( {\frac{1}{2}} \right)^5}\].

Ta có: \({\left( {3x - 2} \right)^4} = {\left( { - 2 + 3x} \right)^4}.\)

Số hạng tổng quát trong khai triển là:

\({T_{k + 1}} = C_4^k{\left( { - 2} \right)^{4 - k}}.{\left( {3x} \right)^k} = C_4^k.{\left( { - 2} \right)^{4 - k}}{3^k}{x^k}\), (\(0 \le k \le 4,\,k \in \mathbb{N}\)).

Hệ số của \({x^2}\) là: \(C_4^2.{\left( { - 2} \right)^2}{.3^2} = 216\).