Bài tập Chuyên đề Nhị thức Newton có đáp án

Khai triển các biểu thức sau: a) (2x + y)^6; b) (x – 3y)^6; c) (x – 1)^n

10/23

Khai triển các biểu thức sau:

a) (2x + y)6;

b) (x – 3y)6;

c) (x – 1)n;

d) (x + 2)n;

e) (x + y)2n;

g) (x – y)2n;

trong đó n lả số nguyên dương.

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Giải thích

a) (2x + y)6

=C602x6+C612x5y+C622x4y2+C632x3y3+C622x2y2+C612xy5+C66y6

=26x6+C6125x5y+C6224x4y2+C6323x3y3+C6422x2y4+C652xy5+y6.

b) (x – 3y)6

= [x + (–3y)]6

=C60x6+C61x5−3y+C62x4−3y2+C63x3−3y3+C64x2−3y4+C65x−3y5+C66−3y6

=x6−C613x5y+C6232x4y2−C6333x3y3+C6434x2y4−C6535xy5+36y6.

c) (x – 1)n

= [(x + (–1)]n

=Cn0xn+Cn1xn−1−1+Cn2xn−2−12+...+Cnn−1x−1n−1+Cnn−1n

=xn+Cn1−1xn−1+Cn2−12xn−2+...+Cnn−1−1n−1x+−1n.

d) (x + 2)n

=Cn0xn+Cn1xn−12+Cn2xn−222+...+Cnn−1x2n−1+Cnn2n

=xn+Cn12xn−1+Cn222xn−2+...+Cnn−12n−1x+2n.

e) (x + y)2n

=C2n0x2n+C2n1x2n−1y+C2n2x2n−2y2+...+C2n2n−1xy2n−1+C2n2ny2n

=x2n+C2n1x2n−1y+C2n2x2n−2y2+...+C2n2n−1xy2n−1+y2n.

g) (x – y)2n

=C2n0x2n+C2n1x2n−1−y+C2n2x2n−2−y2+...+C2n2n−1x−y2n−1+C2n2n−y2n

=C2n0x2n−C2n1x2n−1y+C2n2x2n−2y2−...−C2n2n−1xy2n−1+C2n2ny2n

=x2n−C2n1x2n−1y+C2n2x2n−2y2−...−C2n2n−1xy2n−1+y2n.