\(\frac{(3^4-3^3)^3}{27^3}=\frac{3{^3}^3(3-1)}{27^3}=\frac{27^3.2}{27^3}=2 chia hết cho 2 \)

Ta có:3ⁿ⁺³+3ⁿ⁺¹+2ⁿ⁺³+2ⁿ⁺²=3ⁿ⁺¹(3²+1)+2ⁿ⁺²(2+1)=10.3ⁿ⁺¹+2ⁿ⁺²3=3.2.(5.3ⁿ⁺¹+2ⁿ⁺¹)chia hết cho 6

Vậy...

Ta có: \(\frac{a+b}{c}=\frac{b+c}a{}=\frac{a+c}{b}=\frac{a+b+b+c+c+a}{a+b+c}=2 \)

=> a+b=2c

b+c=2a

a+c=2b

=> P=\((1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a})=\frac{(a+b)(b+c)(c+a) }{bca} =\frac{2a2b2c}{abc} =8\)

Pt<=>(x-2)(m-1)x+2(x-1)=0

<=>(xm-2m-x+2)x+2x-1=0

<=>\(x^2m-2mx-2x^2+2x+2x-1=0\)

<=>\(x^2(m-2)-2x(m-4)-1=0\)

Để Pt vô nghiệm=> \((m-4)^2+m-2<0\)

<=> \(m^2-4m+4+m-2<0\)

<=> \(m^2-3m+2<0\)

<=>(m-2)(m-1)<0

<=>1<m<2

Vậy ....

\(\sqrt{x+2\sqrt{x-1} } -\sqrt{x-2\sqrt{x-1} } =2\)(ĐKXĐ x>1)

<=>\(\sqrt{x-1+2\sqrt{x-1} +1} +\sqrt{x-1-2\sqrt{x-1}+1 } =2\)

<=>\(\sqrt{x-1}+1-\sqrt{x-1}+1=2 \)

<=>2=2

Vậy Pt có nghiệm với mọi x >1

Tích chéo 2 hpt ta có:

\(6x^4-6y^4=15 x^3y-15y^3x\)

<=>\(6x^4-6y^4-15x^3y+15y^3x=0\)

<=> \(6(x^2-y^2)(x^2+y^2)-15xy (x^2-y^2)=0\)

<=>\((x^2-y^2)(6x^2+6y^2+15)=0\)

=> x²=y²

^{=> x=y hoặc x=-y}

^{(*)x=y=>vô nghiệm}

^{(*)x=-y=> vô no}

^{Vậy hpt vô nghiệm}

a,Ta có \(16<2^n\le2^3.32\)

<=>\(2^4<2^n\le2^3,2^5\)

<=> \(2^4<2^n\le2^8\)

<=> \(4 < n \le 8\)

=> \(n \in{5,6,7,8}\)

b, \(25<5^n<625\)

<=>\(5^2 < 5^n<5^4\)

<=> 2<n<4

=> n=3

\(\frac{2^{2015}+3^2-1}{2^{2012}+1}=\frac{2^{2015}+8}{2^{2012}+1 }=\frac{2^3(2^{2012}+1) }{2^{2012}+1} =2^3=8\)

\(\frac{2^{2017}+2^2}{2^{2015}+1}=\frac{2^2(2^{2015}+1) }{2^{2015}+1} =2^2=4\)

8>4

=>....

Áp dụng dãy tỉ số bằng nhau ta có:

\(\frac{a}{2b+c}=\frac{b}{2c+a}=\frac{c}{2a+b} =\frac{a+b+c}{3(a+b+c)}=\frac{1}{3} \)

=>a=3(2b+c)

=>b=3(2c+a)

=>c=3(2a+b)

=> A=\(\frac{2b+c}{a}+\frac{2c+a}{b}+\frac{2a+b}{c}=\frac{2b+c}{3(2b+c)} +\frac{2c+a}{3(2c+a)}+\frac{2a+b}{3(2a+b)} \)=\(\frac{1}{3}+\frac{1}3{}+\frac{1}3{} \)=1

Ta có: \(43^4=***1\)

=>(43⁴)¹⁰= *****1

=>(43⁴)¹⁰.4³=******7

17⁴=***1

=>(17⁴)⁴=*****1

=>(17⁴)⁴.17=*****7

=>A=********7-*****7=********0

=>A chia hết cho 10