ĐKXĐ: n-1>=4

=>n>=5

Ta có: \(4\cdot C_{n-1}^4-4\cdot C_{n-1}^3<5\cdot A_{n-2}^2\)

=>\(4\cdot\frac{\left(n-1\right)!}{\left(n-1-4\right)!\cdot4!}-4\cdot\frac{\left(n-1\right)!}{\left(n-1-3\right)!\cdot3!}<5\cdot\frac{\left(n-2\right)!}{\left(n-2-2\right)!}\)

=>\(\frac{4\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)}{24}-\frac{4\left(n-1\right)\left(n-2\right)\left(n-3\right)}{6}\) <5*\(\frac{\left(n-2\right)\left(n-3\right)}{1}\)

=>\(\frac{\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)-4\cdot\left(n-1\right)\left(n-2\right)\left(n-3\right)}{6}<\frac{30\left(n-2\right)\left(n-3\right)}{6}\)

=>(n-1)(n-2)(n-3)(n-4-4)<30(n-2)(n-3)

=>(n-1)(n-2)(n-3)(n-8)-30(n-2)(n-3)<0

=>(n-3)(n-2)[(n-1)(n-8)-30]<0

=>(n-1)(n-8)-30<0

=>\(n^2-9n+8-30<0\)

=>\(n^2-9n-22<0\)

=>(n-11)(n+2)<0

=>n-11<0

=>n<11

mà n>=5

nên 5<=n<11

mà n là số nguyên dương

nên n∈{5;6;7;8;9;10}

=>Có 6 giá trị nguyên dương n thỏa mãn

ĐKXĐ: x+6>=3

=>x>=-3

Ta có: \(C_{x+8}^{x+3}=5\cdot A_{x+6}^3\)

=>\(\frac{\left(x+8\right)!}{\left(x+3\right)!\cdot\left(x+8-x-3\right)!}=5\cdot\frac{\left(x+6\right)!}{\left(x+6-3\right)!}\)

=>\(\frac{\left(x+8\right)\left(x+7\right)\left(x+6\right)\left(x+5\right)\left(x+4\right)}{5!}=\frac{5\cdot\left(x+6\right)\left(x+5\right)\left(x+4\right)}{1}\)

=>(x+8)(x+7)(x+6)(x+5)(x+4)=600(x+6)(x+5)(x+4)

=>(x+8)(x+7)=600

=>\(x^2+15x+56-600=0\)

=>\(x^2+15x-544=0\)

=>(x+32)(x-17)=0

=>x=-32(loại) hoặc x=17(nhận)

ĐKXĐ: x+6>=3

=>x>=-3

Ta có: \(C_{x+8}^{x+3}=5\cdot A_{x+6}^3\)

=>\(\frac{\left(x+8\right)!}{\left(x+3\right)!\cdot\left(x+8-x-3\right)!}=5\cdot\frac{\left(x+6\right)!}{\left(x+6-3\right)!}\)

=>\(\frac{\left(x+8\right)\left(x+7\right)\left(x+6\right)\left(x+5\right)\left(x+4\right)}{5!}=\frac{5\cdot\left(x+6\right)\left(x+5\right)\left(x+4\right)}{1}\)

=>(x+8)(x+7)(x+6)(x+5)(x+4)=600(x+6)(x+5)(x+4)

=>(x+8)(x+7)=600

=>\(x^2+15x+56-600=0\)

=>\(x^2+15x-544=0\)

=>(x+32)(x-17)=0

=>x=-32(loại) hoặc x=17(nhận)

a: \(\Leftrightarrow\left(2x+1\right)^3=8\cdot25-75=125\)

=>2x+1=5

hay x=2

c: x=2; y=0

ĐK: \(x\ge4\)

\(\dfrac{\left(x-2\right)!}{\left(x-4\right)!}+\dfrac{x!}{\left(x-2\right)!.2!}=101\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)+\dfrac{x\left(x-1\right)}{2}=101\)

\(\Leftrightarrow3x^2-11x-190=0\)

\(\Rightarrow x=10\)

a)x=1

b)x=2

c)x=3

a, x= 1

b, x=2

c,x= 3

Ta có:

\(2A_n^2=C_{n-1}^2+C_{n-1}^3\) \(\left(n\ge4\right)\)

\(\Rightarrow2\cdot\dfrac{n!}{\left(n-2\right)!}=\dfrac{\left(n-1\right)!}{2!\left(n-1-2\right)!}+\dfrac{\left(n-1\right)!}{3!\left(n-1-3\right)!}\)

\(\Rightarrow2\cdot n\left(n-1\right)=\dfrac{\left(n-1\right)\left(n-2\right)}{4}+\dfrac{\left(n-1\right)\left(n-2\right)\left(n-3\right)}{6}\)

\(\Rightarrow2n=\dfrac{n-2}{4}+\dfrac{\left(n-2\right)\left(n-3\right)}{6}\)

\(\Rightarrow n=14\) hoặc \(n=0\left(loại\right)\)

Với n=14 ta có khai triển:

\(\left(x^2-\dfrac{1}{x^2}\right)^{14}=\sum\limits^{14}_{k=0}\cdot C_{14}^k\cdot\left(x^2\right)^{14-k}\cdot\left(\dfrac{1}{x^2}\right)^k\)

                      \(=C_{14}^k\cdot x^{28-4k}\)

Số hạng không chứa x: \(\Rightarrow28-4k=0\Rightarrow k=7\)

Vậy số hạng không chứa x trong khai triển là:

\(C_{14}^7\cdot x^{28-4\cdot7}=C_{14}^7=3432\)