a: 

=\(18x^{2n-3}+3x^n-18^{2n-3}+2x^n\)

\(=5x^n\)

b: \(=5^n\cdot5-4\cdot5^n=5^n\)

c: \(=6^6-4^3\cdot3^6+4^3\)

\(=2^6\cdot3^6-2^6\cdot3^6+64=64\)

\(B=\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

\(B=\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(B=\left|\sqrt{5}+2\right|+\left|\sqrt{5}-2\right|\)

\(B=\sqrt{5}+2+\sqrt{5}-2\)

\(B=2\sqrt{5}\)

\(A=\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}\)

\(A=\left(\dfrac{\sqrt{12}-\sqrt{6}}{2\sqrt{2}-2}-\dfrac{6\sqrt{6}}{3}\right).\dfrac{1}{\sqrt{6}}\)

\(A=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)

\(A=\left(\sqrt{6}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)

\(A=-\sqrt{6}.\dfrac{1}{\sqrt{6}}\)

\(A=-1\)

a: \(A=3-\left|3-x\right|\)

=3-|x-3|

TH1: x>=3

=>x-3>=0

A=3-|x-3|

=3-(x-3)

=3-x+3

=6-x

TH2: x<3

=>x-3<0

A=3-|x-3|

=3-(3-x)

=3-3+x

=x

b: \(B=\left|x-6\right|+\left|6-x\right|-2\)

=2|x-6|-2

TH1: x>=6

=>x-6>=0

B=2|x-6|-2

=2(x-6)-2

=2x-12-2

=2x-14

TH2: x<6

=>x-6<0

B=2|x-6|-2

=2(6-x)-2

=12-2x-2

=10-2x

c: \(C=\left|-x-1\right|+\left|-x-5\right|-x\)

\(=\left|x+5\right|+\left|x+1\right|-x\)

TH1: x<-5

=>x+5<0; x+1<0

\(C=\left|x+5\right|+\left|x+1\right|-x\)

=-x-5-x-1-x

=-3x-6

TH2: -5<=x<-1

=>x+5>=0; x+1<0

\(C=\left|x+5\right|+\left|x+1\right|-x\)

=x+5-x-1-x

=-x+4

TH3: x>=-1

=>x+5>0; x+1>=0

\(C=\left|x+5\right|+\left|x+1\right|-x\)

=x+5+x+1-x

=x+6

a: \(A=3-\left|3-x\right|\)

=3-|x-3|

TH1: x>=3

=>x-3>=0

A=3-|x-3|

=3-(x-3)

=3-x+3

=6-x

TH2: x<3

=>x-3<0

A=3-|x-3|

=3-(3-x)

=3-3+x

=x

b: \(B=\left|x-6\right|+\left|6-x\right|-2\)

=2|x-6|-2

TH1: x>=6

=>x-6>=0

B=2|x-6|-2

=2(x-6)-2

=2x-12-2

=2x-14

TH2: x<6

=>x-6<0

B=2|x-6|-2

=2(6-x)-2

=12-2x-2

=10-2x

c: \(C=\left|-x-1\right|+\left|-x-5\right|-x\)

\(=\left|x+5\right|+\left|x+1\right|-x\)

TH1: x<-5

=>x+5<0; x+1<0

\(C=\left|x+5\right|+\left|x+1\right|-x\)

=-x-5-x-1-x

=-3x-6

TH2: -5<=x<-1

=>x+5>=0; x+1<0

\(C=\left|x+5\right|+\left|x+1\right|-x\)

=x+5-x-1-x

=-x+4

TH3: x>=-1

=>x+5>0; x+1>=0

\(C=\left|x+5\right|+\left|x+1\right|-x\)

=x+5+x+1-x

=x+6

b)11.6^x-1+2.6^x+1=11.6¹¹+2.6¹³

11.6^x-1+2.6^x+1 = 11. 6^12-1+ 2. 6¹²⁺¹

=> x= 12

c) 2^4-x / 16⁵ = 32⁶

2^4-x / 2²⁰= 2³⁰

2^4-x = 2⁵⁰

=> 4-x = 50

x= -46

c)      \(\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)

       \(2^{4-x}:2^{20}=\left(2^5\right)^6\)

       \(2^{4-x}=2^{30}.2^{20}\)

       \(2^{4-x}=2^{50}\)

=>    \(4-x=50\)

=>       \(x=4-50=-46\)

vậy x = -46

A)3.5ⁿ.5²+4.5ⁿ:5³=19.9765625

5ⁿ(3.5²+4:5³)=185546875

5ⁿ.\(\frac{12}{5}\)=185546875

b) mình chắc chắn x = 12 không cần tk đâu

d: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\left(3-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\dfrac{5}{6}\cdot\dfrac{6}{5}-17=1-17=-16\)

h: \(\dfrac{\left(-1\right)^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left|-\dfrac{5}{6}\right|\)

\(=-\dfrac{1}{15}+\dfrac{-8}{27}:\dfrac{8}{3}-\dfrac{5}{6}\)

\(=-\dfrac{1}{15}-\dfrac{1}{9}-\dfrac{5}{6}\)

\(=\dfrac{-6-10-75}{90}=\dfrac{-91}{90}\)

k: \(\dfrac{2\cdot6^9-2^5\cdot18^4}{2^2\cdot6^8}\)
\(=\dfrac{2^{10}\cdot3^9-2^5\cdot2^4\cdot3^8}{2^2\cdot2^8\cdot3^8}\)

\(=\dfrac{2^{10}\cdot3^9-2^9\cdot3^8}{2^{10}\cdot3^8}=\dfrac{2^9\cdot3^8\left(2\cdot3-1\right)}{2^{10}\cdot3^8}\)

\(=\dfrac{5}{2}\)

n: \(3-\left(-\dfrac{7}{8}\right)^0+\left(\dfrac{1}{2}\right)^3\cdot16\)

\(=3-1+\dfrac{1}{8}\cdot16\)

=2+2

=4