Giả thiết tương đương:

\(C_{2n+1}^{n+1}+C_{2n+1}^{n+2}+...+C_{2n+1}^{2n}+C_{2n+1}^{2n+1}=2^{100}\) (thay \(1=C_{2n+1}^{2n+1}\))

Mặt khác:

\(C_{2n+1}^{2n+1}=C_{2n+1}^0\)

\(C_{2n+1}^{2n}=C_{2n+1}^1\)

....

\(C_{2n+1}^{n+1}=C_{2n+1}^n\)

Cộng vế:

\(\Rightarrow C_{2n+1}^{n+1}+C_{2n+1}^{n+2}+...+C_{2n+1}^{2n+1}=C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^n\)

\(\Rightarrow2\left(C_{2n+1}^{n+1}+...+C_{2n+1}^{2n+1}\right)=C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^{2n+1}\)

\(\Rightarrow2.2^{100}=2^{2n+1}\) (đẳng thức cơ bản: \(\sum\limits^n_{k=0}C_n^k=2^n\))

\(\Leftrightarrow2^{101}=2^{2n+1}\)

\(\Rightarrow2n+1=101\)

\(\Rightarrow n=50\)

SHTQ trong khai triển: \(C_{50}^k.\left(x^{-3}\right)^k.\left(x^2\right)^{50-k}=C_{50}^kx^{100-5k}\)

\(100-5k=20\Rightarrow k=16\)

Hệ số: \(C_{50}^{16}\)

a: \(4x^2\left(3x^{n+1}-2x^n\right)\)

\(=4x^2\cdot3x^{n+1}-4x^2\cdot2x^n\)

\(=12x^{n+3}-8x^{n+2}\)

b: \(2\left(x^{2n}+2x^ny^n+y^{2n}\right)-y^n\left(4x^n+2y^n\right)\)

\(=2x^{2n}+4x^ny^n+2y^{2n}-4x^ny^n-2y^{2n}\)

\(=2x^{2n}\)

c: \(=\left(x^{3n}-y^{3n}\right)\left(x^{3n}+y^{3n}\right)\)

\(=x^{6n}-y^{6n}\)

d: \(=4^n\cdot4-3\cdot4^n=4^n\)

a: 4x2(3xn+1−2xn)4x2(3xn+1−2xn)

=4x2⋅3xn+1−4x2⋅2xn=4x2⋅3xn+1−4x2⋅2xn

=12xn+3−8xn+2=12xn+3−8xn+2

b: 2(x2n+2xnyn+y2n)−yn(4xn+2yn)2(x2n+2xnyn+y2n)−yn(4xn+2yn)

=2x2n+4xnyn+2y2n−4xnyn−2y2n=2x2n+4xnyn+2y2n−4xnyn−2y2n

=2x2n=2x2n

c: =(x3n−y3n)(x3n+y3n)=(x3n−y3n)(x3n+y3n)

=x6n−y6n=x6n−y6n

d: 

Chj có thê ấn vào phân tìm kiếm đê có đáp án ah

#lâurôi

====ZU====

@EMĐÂY

Với \(n\in\mathbb{N^*}\), ta có: \(\left\{{}\begin{matrix}\left(x+1\right)^{2n}\ge0\forall x\\\left(y-1\right)^{2n}\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x+1\right)^{2n}+\left(y-1\right)^{2n}\ge0\forall x,y\)

Mà: \(\left(x+1\right)^{2n}+\left(y-1\right)^{2n}=0\)

nên: \(\left\{{}\begin{matrix}x+1=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

Vậy ...

e ko bt

ta có

  1+m =  \(\frac{2x^n}{x^n+\frac{1}{x^n}}\), 1-m = \(\frac{2}{x^n\left(x^n+\frac{1}{x^x}\right)}\)

=> \(\frac{1+m}{1-m}\)= x²ⁿ

do đó P = \(\frac{\frac{1+m}{1-m}-\frac{1-m}{1+m}}{\frac{1+m}{1-m}+\frac{1-m}{1+m}}\)= \(\frac{\left(1+m\right)^2-\left(1-m\right)^2}{\left(1-m\right)\left(1+m\right)}\). \(\frac{\left(1-m\right)\left(1+m\right)}{\left(1+m\right)^2+\left(1-m\right)^2}\)

= \(\frac{2m}{1+m^2}\)

Đặt x​ ²ⁿ = a ta có

\(\frac{x^n-x^{-n}}{x^n+x^{-n}}=\frac{x^{2n}-1}{x^{2n}+1}=\frac{a-1}{a+1}=m\)

\(\Leftrightarrow a-1=m\left(a+1\right)\)

\(\Leftrightarrow a\left(1-m\right)=1+m\)

\(\Leftrightarrow a=\frac{1+m}{1-m}\)

Ta lại có

\(\frac{x^{2n}-x^{-2n}}{x^{2n}+x^{-2n}}=\frac{x^{4n}-1}{1+x^{4n}}=\frac{a^2-1}{1+a^2}\)

Tới đây thì e chỉ cần thế vô rồi rút gọn là ra nhé

\(\Leftrightarrow!m!< 1\)

\(\frac{x^n-x^{-n}}{x^n+x^{-n}}=\frac{\left(x^{2n}-1\right)}{\left(x^{2n}+1\right)}=x^{2n}=\frac{m+1}{1-m}=>x^2=\sqrt[n]{\frac{m+1}{1-m}}\)

\(P=\frac{x^{4n}-1}{x^{4n}+1}=\frac{\left(\frac{m+1}{1-m}\right)^2-1}{\left(\frac{m+1}{1-m}\right)^2+1}=\frac{\left(m+1\right)^2-\left(1-m\right)^2}{\left(m+1\right)^2+\left(1-m\right)^2}=\frac{2m}{m^2+1}\\ \)