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)

15.B

16.C

17.A

18.D

19.A

còn câu 20,21 mình sợ mình làm sai nên k ghi đáp án sorry bạn nha:(

Điều kiện: \(x\ge5\)

\(\dfrac{4\left(x-1\right)!}{4!.\left(x-5\right)!}-\dfrac{4\left(x-1\right)!}{3!\left(x-4\right)!}< \dfrac{5\left(x-2\right)!}{\left(x-4\right)!}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-4\right)}{6}-\dfrac{2\left(x-1\right)}{3}< 5\)

\(\Leftrightarrow x^2-9x-22< 0\)

\(\Rightarrow-2< x< 11\)

\(\Rightarrow x=\left\{5;6;7;8;9;10\right\}\)

25ⁿ = 25²

n = 2

Chọn A

A

m là 2

n cũng là 2

vì 3^2=9

2^2=4

9-4=5

m la 2 

n cung la 2

vi 3*2=9

2*2 = 4

9-4 = 5

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