Bài luyện tập

a, 2.2.4.4 = $2.2.2^2.2^2$ = $2^{1+1+2+2}$ = $2^6$

b, 10.1000.100000 = $10.{10}^3.{10}^5$ = ${10}^{1+3+5}={10}^9$

c, $4^5.2^7$ = $\left(2^2\right).2^7=2^{2.5}.2^7=2^{10}.2^7$ = $2^{10+7}=2^7$

d, $3^3.18-3^3.15$ = $3^3(18-15)=3^3.3=3^{3+1}=3^4$

e, $6^3.36$ = $6^3.6^2=6^{3+2}=6^5$

f, ${49}^3.7^4$ = ${\left(7^2\right)}^3.7^4=7^{2.3}.7^4=7^6.7^4$ = $7^{6+4}=7^{10}$

g, $a^9.a^7=a^{9+7}=a^{16}$

h, ${\left(a.b\right)}^6.b^6$ = $\left(a^6.b^6\right).b^6=a^6.\left(b^6.b^6\right)$ = $a^6.b^{6+6}=a^6.b^{12}$

a, $8^2.{32}^4$ = ${\left(2^3\right)}^2.{\left(2^5\right)}^4=2^{3.2}.2^{5.4}$ = $2^6.2^{20}=2^{6+20}=2^{26}$

b, ${27}^4.9^3.243$ = ${\left(3^3\right)}^4.{\left(3^2\right)}^3.3^5$ = $3^{3.4}.3^{2.3}.3^5=3^{12}.3^6.3^5$ = $3^{12+6+5}=3^{23}$

c, ${13}^2-{12}^2$ = 169 $-$144 = 25 = $5^2$

d, $6^2+8^2$ = 36 + 64 = 100 = ${10}^2$

e, $\left(5^3+5^4+{125}^2\right).5^3$ = $\left(5^3+5^4+5^{3.2}\right).5^3$ = $\left(5^3+5^4+5^6\right).5^3$ = $5^3.\left(1+5+5^3\right).5^3=5^6.131$

f, $1^3+2^3+3^3+4^3+5^3$ = 1 + 8 + 27 + 64 + 125 = 225 = ${15}^2$

a, Ta có : $5^3$ = 5.5.5 = 125 và $3^5$ = 3.3.3.3.3 = 243

Vì 125 < 243 nên $5^3$<$3^5$

b, Ta có : $2^8$ = 2.2.2.2.2.2.2.2 = 256 và $3^5$ = 3.3.3.3.3 = 243

Vì 256>243 nên $2^8$>$3^5$

c, Ta có : $4^3$ = 4.4.4 = 64 và $8^2$ = 8.8 = 64

Vì 64 = 64 nên $4^3$ = $8^2$

d, Ta có :  ${25}^{45}={\left(5^2\right)}^{45}=5^{2.45}=5^{90}$ và ${125}^{30}={\left(5^3\right)}^{30}=5^{3.30}=5^{90}$

Vì $5^{90}=5^{90}$ nên ${25}^{45}$ = ${125}^{30}$

a, $3^n.3=243$ => $3^{n+1}=243$ => $3^{n+1}=3^5$

=> n + 1 = 5 => n = 4

Vậy n = 4

b, $4^3.2^{n+1}=1$

=> ${\left(2^2\right)}^3.2^{n+1}=1$

=> $2^{2.3}.2^{n+1}=1$ => $2^6.2^{n+1}=1$

=> $2^{6+n+1}=1$ => $2^{n+7}=2^0$

=> n + 7 = 0

Không tìm được số tự nhiên n thỏa mãn đầu bài

c, $2^n-15=17$

=> $2^n=32$ => $2^n=2^5$

=> n = 5

Vậy n = 5

d, $8\le 2^{n+1}\le 64$

=> $2^3\le 2^{n+1}\le 2^6$

=> 3$\le$n + 1 và n+1$\le$6

=> 2$\le$n và n$\le$5

=> 2$\le$n$\le$5

Vậy 2$\le$n$\le$5

e, $9<3^n<243$

=> $3^2<3^n<3^5$

=> 2<n<5

Vậy 2<n<5

a, $2^x.4=128$

=> $2^x=128:4$

=> $2^x=32$

=> $2^x=2^5$

=> x = 5

Vậy x = 5

b, $x^{15}=x$

=> x = 1 hoặc x = 0

Vì $1^{15}=1; 0^{15}=0$

c, ${\left(x-5\right)}^4={\left(x-5\right)}^6$

=> x – 5 = 1 hoặc x – 5 = 0

=> x = 6 hoặc x = 5

Vậy x = 6 hoặc x = 5

d, $x^{2018}=x^2$

=> x = 1 hoặc x = 0

Vì $1^{2018}=1^2=1; 0^{2018}=0^2=0$

a, ${\left(x-1\right)}^2=36$

ó ${\left(x-1\right)}^2=6^2$

ó x – 1 = 6

ó x = 7

Vậy x = 7

b, ${\left(7x-11\right)}^3=2^5.5^2+200$

=> ${\left(7x-11\right)}^3=800+200$

=> ${\left(7x-11\right)}^3=1000$

=> ${\left(7x-11\right)}^3={10}^3$

=> 7x – 11 = 10

=> 7x = 21

=> x = 3

Vậy x = 3

c, $x^{11}=x$

=> $x^{11}-x=0$

=> $x(x^{10}-1)=0$

=>

d, $x^2=2^3+3^2+4^3$

=> $x^2=81$

=> x = 9

e, $\frac{2x^3}{3^2}=48$

=> $\frac{2x^3}{9}=48$

=> $2x^3=48.9$

=> $x^3=216$

=> $x^3=6^3$

=> x = 6

f, ${\left(2x+1\right)}^3=125$

=> ${\left(2x+1\right)}^3=5^3$

=> 2x + 1 = 5

=> 2x = 4

=> x = 2