a, $4^8.4^{10}=4^{18}$

b, $2^{20}.2^7=2^{27}$

c, $5^{12}.5^5.5^4=5^{21}$

d, $4^3.4^5.4^5=4^{13}$

e, $8^6.8^5.8^5=8^{16}$

f, $x^7.x^4.x^3=x^{14}$

a, $4^5:4=4^4$

b, $2^{10}:2^3=2^7$

c, $x^9:x^3=x^6\left(x\ne 0\right)$

d, $5^{103}:5^3=5^{100}$

a, $2^x.2^2=32$

$2^{x+2}=2^5$

x + 2 = 5

x = 3

Vậy x = 3

b, $27.3^x=243$

$3^3.3^x=3^5$

$3^{3+x}=3^5$

x + 3 = 5

x = 2

Vậy x = 2

c, $2^x.2^4=1024$

$2^{x+4}=2^{10}$

x + 4 = 10

x = 6

Vậy x = 6

d, $49.7^x=2401$

$7^2.7^x=7^4$

$7^{2+x}=7^4$

2 + x = 4

x = 2

Vậy x = 2

a, $3^{x+1}:3^4=81$

$3^{x-3}=3^4$

x – 3 = 4

x = 7

Vậy x = 7

b, $3^{x+3}.3^{x+1}=729$

$3^{2x+4}=3^6$

2x + 4 = 6

x = 1

Vậy x = 1

c, $2^{x+3}.2^x=128$

$2^{2x+3}=2^7$

2x + 3 = 7

x = 2

Vậy x = 2

d, $23+3x=5^6:5^3$

$23+3x=5^3$

23 + 3x = 125

3x = 102

x = 34

Vậy x = 34

e, $2^x+2^{x+4}=272$

$2^x+2^x.2^4=272$

$2^x(1+2^4)=272$

$2^x.17=272$

$2^x=16$

$2^x=2^4$

x = 4

Vậy x = 4

a, $\left(2^{17}+{17}^2\right)\left(9^{15}-3^{15}\right)\left(2^4-4^2\right)$

= $\left(2^{17}+{17}^2\right)\left(9^{15}-3^{15}\right)\left(16-16\right)$

= $\left(2^{17}+{17}^2\right)\left(9^{15}-3^{15}\right).0$

= 0

b, $\left(1^2+2^3+3^4+4^5\right)$.$(1^3+2^3+3^3+4^3)(3^8-{81}^2)$

= $\left(1^2+2^3+3^4+4^5\right)$.$(1^3+2^3+3^3+4^3)(3^{4.2}-{81}^2)$

= $\left(1^2+2^3+3^4+4^5\right)$.$(1^3+2^3+3^3+4^3)({81}^2-{81}^2)$

= $\left(1^2+2^3+3^4+4^5\right)$.$(1^3+2^3+3^3+4^3).0$

= 0

c, $\left(7^{24}+7^{23}\right):7^{22}$

= $7^{24}:7^{22}+7^{23}:7^{22}$

= $7^{24-22}+7^{23-22}$

= $7^2+7^1$

= 49 + 7 = 56

$A=1+2+2^2+...+2^{2007}$

$2A=2+2^2+...+2^{2007}+2^{2008}$

A = 2A$-$A = $(2+2^2+...+2^{2007}+2^{2008})$$-$$(1+2+2^2+...+2^{2007})$ = $2^{2008}-1$

Vậy $A=2^{2008}-1$

a, $A=1+5^3+5^5+5^7+...+5^{99}$

$B=5^4+5^6+5^8+...+5^{100}$ = $5.(5^3+5^5+5^7+...+5^{99})$ = 5(A – 1)

A + B – 1 = $5^3+5^4+...+5^{100}$

5(A + B – 1) = $5^4+5^5+...+5^{100}+5^{101}$

4(A + B – 1) = 5(A + B – 1) – (A + B – 1) = $5^{101}-5^3$

=> A + B – 1 = $\frac{5^{101}-5^3}{4}$

=> A + 5(A – 1) –1 = $\frac{5^{101}-5^3}{4}$ => 6A – 6 = $\frac{5^{101}-5^3}{4}$

=> A – 1 = $\frac{5^{101}-5^3}{24}$

=> A = $\frac{5^{101}-5^3+24}{24}$

b, $A=1-2+2^2-...-2^{2007}$

$A=1+2^2+...+2^{2006}-\left(2+2^3+...+2^{2007}\right)$

$A=(1+2^2+...+2^{2006})-2.\left(1+2^2+...+2^{2006}\right)$

$A=-\left(1+2^2+...+2^{2006}\right)$

Đặt $B=-\left(2+2^3+...+2^{2007}\right)$ = $-2.\left(1+2^2+...+2^{2006}\right)$ = 2A

A + B = $-\left(1+2+2^2+...+2^{2006}+2^{2007}\right)$

2(A+B) = $-\left(2+2^2+...+2^{2006}+2^{2007}+2^{2008}\right)$

A+B = 2(A+B)–(A+B) = $-\left(2^{2008}-1\right)$

=> A+2A = $-\left(2^{2008}-1\right)$

=> 3A = $-\left(2^{2008}-1\right)$

=> A = $\frac{-(2^{2008}-1)}{3}$

c, $A=7+7^3+7^5+7^7+...+7^{1999}$

Đặt B = $7^2+7^4+7^6+...+7^{1999}+7^{2000}$ = $7(7+7^3+7^5+7^7+...+7^{1999})$ = 7A

A+B = $7+7^2+7^3+...+7^{1999}+7^{2000}$

7(A+B) = $7^2+7^3+...+7^{1999}+7^{2000}+7^{2001}$

7(A+B) – (A+B) = $(7^2+7^3+...+7^{1999}+7^{2000}+7^{2001})$ – $(7+7^2+7^3+...+7^{1999}+7^{2000})$

6(A+B) = $7^{2001}-7$

A+B = $\frac{7^{2001}-7}{6}$

=> A + 7A = $\frac{7^{2001}-7}{6}$ => 8A = $\frac{7^{2001}-7}{6}$ => A = $\frac{7^{2001}-7}{48}$