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Bài 1:
\(\left(x^{-\frac{1}{5}}+x^{\frac{1}{3}}\right)^{10}=\sum\limits^{10}_{k=0}C_{10}^k\left(x^{-\frac{1}{5}}\right)^k\left(x^{\frac{1}{3}}\right)^{10-k}=\sum\limits^{10}_{k=0}C_{10}^kx^{\frac{10}{3}-\frac{8k}{15}}\)
Trong khai triển trên có 11 số hạng nên số hạng đứng giữa có \(k=6\)
\(\Rightarrow\) Số hạng đó là \(C_{10}^6x^{\frac{10}{3}-\frac{48}{15}}=C_{10}^6x^{\frac{2}{15}}\)
Bài 2:
\(\left(1+x^2\right)^n=a_0+a_1x^2+a_2x^4+...+a_nx^{2n}\)
Cho \(x=1\Rightarrow2^n=a_0+a_1+...+a_n=1024=2^{10}\)
\(\Rightarrow n=10\)
\(\left(1+x^2\right)^{10}=\sum\limits^{10}_{k=0}C_{10}^kx^{2k}\)
Số hạng chứa \(x^{12}\Rightarrow2k=12\Rightarrow k=6\) có hệ số là \(C_{10}^6\)
Bài 3:
\(\left(x-\frac{1}{4}\right)^n=\sum\limits^n_{k=0}C_n^kx^k\left(-\frac{1}{4}\right)^{n-k}\)
Với \(k=n-2\Rightarrow\) hệ số là \(C_n^{n-2}\left(-\frac{1}{4}\right)^2=\frac{1}{16}C_n^2\)
\(\Rightarrow\frac{1}{16}C_n^2=31\Rightarrow C_n^2=496\Rightarrow n=32\)
Bài 4:
Xét khai triển:
\(\left(1+x\right)^n=C_n^0+xC_n^1+x^2C_n^2+...+x^nC_n^n\)
Cho \(x=2\) ta được:
\(\left(1+2\right)^n=C_n^0+2C_n^1+2^2C_n^2+...+2^nC_n^n\)
\(\Rightarrow S=3^n\)
Bài 5:
Xét khai triển:
\(\left(1+x\right)^n=C_n^0+xC_n^1+x^2C_n^2+...+x^{2k}C_n^{2k}+x^{2k+1}C_n^{2k+1}+...\)
Cho \(x=-1\) ta được:
\(0=C_n^0-C_n^1+C_n^2-C_n^3+...+C_n^{2k}-C_n^{2k+1}+...\)
\(\Rightarrow C_n^0+C_n^2+...+C_n^{2k}+...=C_n^1+C_n^3+...+C_n^{2k+1}+...\)
Bài 6:
\(\left(1-4x+x^2\right)^5=\sum\limits^5_{k=0}C_5^k\left(-4x+x^2\right)^k=\sum\limits^5_{k=0}\sum\limits^k_{i=0}C_5^kC_k^i\left(-4\right)^ix^{2k-i}\)
Ta có: \(\left\{{}\begin{matrix}2k-i=5\\0\le i\le k\le5\\i;k\in N\end{matrix}\right.\) \(\Rightarrow\left(i;k\right)=\left(1;3\right);\left(3;4\right);\left(5;5\right)\)
Hệ số: \(\left(-4\right)^1.C_5^3C_3^1+\left(-4\right)^3C_5^4.C_4^3+\left(-4\right)^5C_5^5.C_5^5\)
a)
=
= -4.
b)
=
=
(2-x) = 4.
c)
=
=
=
=
.
d)
=
= -2.
e)
= 0 vì
(x2 + 1) =
x2( 1 +
) = +∞.
f)
=
a: \(\left(x+1\right)^5\)
\(=x^5+C_5^1\cdot x^4\cdot1^1+C_5^2\cdot x^3\cdot1^2+C_5^3\cdot x^2\cdot1^3+C_5^4\cdot x\cdot1^4+1\)
\(=x^5+5x^4+10x^3+10x^2+5x+1\)
b: \(\left(x-2y\right)^6\)
\(=x^6-C_6^1\cdot x^5\cdot\left(2y\right)^1+C_6^2\cdot x^4\cdot\left(2y\right)^2-C_6^3\cdot x^3\cdot\left(2y\right)^3+C_6^4\cdot x^2\cdot\left(2y\right)^4-C_6^5\cdot x\cdot\left(2y\right)^5+\left(2y\right)^6\)
\(=x^6-12x^5y+60x^4y^2-160x^3y^3+960x^2y^4-192xy^5+64y^6\)
c: \(\left(x^2+\frac{1}{x}\right)^5\)
\(=\left(x^2\right)^5+C_5^1\cdot\left(x^2\right)^4\cdot\left(\frac{1}{x}\right)^1+C_5^2\cdot\left(x^2\right)^3\cdot\left(\frac{1}{x}\right)^2+C_5^3\cdot\left(x^2\right)^2\cdot\left(\frac{1}{x}\right)^3+C_5^4\cdot x^2\cdot\left(\frac{1}{x}\right)^4+\left(\frac{1}{x}\right)^5\)
\(=x^{10}+5x^7+10x^4+10x+\frac{5}{x^2}+\frac{1}{x^5}\)
d: \(\left(x^3-\frac{2}{x}\right)^6\)
\(=\left(x^3\right)^6-C_6^1\cdot\left(x^3\right)^5\cdot\left(\frac{2}{x}\right)^1+C_6^2\cdot\left(x^3\right)^4\cdot\left(\frac{2}{x}\right)^2-C_6^3\left(x^3\right)^3\cdot\left(\frac{2}{x}\right)^3+C_6^4\cdot\left(x^3\right)^2\cdot\left(\frac{2}{x}\right)^4-C_6^5\cdot\left(x^3\right)\cdot\left(\frac{2}{x}\right)^5+\left(\frac{2}{x}\right)^6\)
\(=x^{18}-12\cdot x^{14}+60x^{10}-160x^6+960x^2-\frac{192}{x^2}+\frac{64}{x^6}\)