A.x=11

B.=47

\(a,\Rightarrow7\left(x-5\right)=100-58=42\\ \Rightarrow x-5=6\\ \Rightarrow x=11\\ b,=5\cdot9-2^5:2^4=45-2=43\)

\(375:\left\{32-\left[4+\left(5.3^2-42\right)\right]\right\}\)

\(=375:\left\{32-\left[4+\left(45-42\right)\right]\right\}=375:\left\{32-\left[4+3\right]\right\}=375:\left\{32-7\right\}\)

=\(375:25=15\)

\(375:\left\{32-\left[4+\left(5.3^2-42\right)\right]\right\}\)

\(=375:\left\{32-\left[4+\left(45-42\right)\right]\right\}\)

\(=375:\left\{32-\left[4+3\right]\right\}\)

\(=375:\left\{32-7\right\}\)

\(=375:25\)

\(=15\)

Đề em ghi bị thiếu dấu ngoặc nhé, sửa lại đề và giải như vầy:

375 : {32 - [4 + (5.3² - 42)]} - 14

= 375 : {32 - [4 + (5.9 - 42)]} - 14

= 375 : {32 - [4 + (45 - 42)]} - 14

= 375 : [32 - (4 + 3)] - 14

= 375 : (32 - 7) - 14

= 375 : 25 - 14

= 15 - 14

= 1

:)?

giúp e với em cần gấp ạ

\(A=3^2\cdot4^3-3^2+333\)

\(=3^2\cdot63+333\)

\(=3^2\cdot\left(63+37\right)\)

\(=30^2\)

\(B=5\cdot3^2+4\cdot3^2=3^2\cdot9=9^2\)

Là sao

{x^2−[6^2−(8^2−9⋅7)^3−7⋅5]^3−5⋅3}^3=1

⇒x^2−[36−(64−63)^3−35]^3−15=1

⇒x^2−[36−35−1^3]^3=16

⇒x^2−0^3=16

⇒x^2=16

⇒x=±4

Hok tốt

d: Ta có: \(D=5^3+6^3+59\)

\(=125+216+59\)

\(=400=20^2\)

Ông ơi

thằng  trung anh ơiiiiiiiiiiiii

5.3^x=405

3^x=405:5

3^x=81

3^x=3⁴

^{Vậy x=4}

2^x:8=4

2^x=4.8

2^x=32

2^x=2⁵

^{Vậy x=5}

x²⁸=x⁵

x^28-x^5=0

x^5.x^23-x^5.1=0

x^5.(x^23-1)=0

suy ra x^5=0 hoặc x^23-1=0 suy ra x^5=0^5 hoặc x^23=0+1=1 suy ra x=0 hoặc x^23=1^23 suy ra x=0 hoăc x=1

9

(x-2)^4=256

(x-2)^4=4^4

x-2=4

x=4+2=6

(x+1)^3=125

(x+1)^3=5^3

x+1=5

x=5-1=4

a,\(2^4\cdot3^5:6^4\)

\(=\frac{2^4\cdot3^6}{\left(2\cdot3\right)^4}\)

\(=\frac{2^4\cdot3^6}{2^4\cdot3^4}\)

\(=3^2\)

Bài 2

\(a,5^3\cdot8=5^3\cdot2^3=10^3=1000\)

\(b,2^5-2019^0=32-1=31\)

\(c,3^3+2^5-1^{10}=27+32-1=58\).

\(d,9^2\cdot33-81\cdot23+5^2=81\cdot33-81\cdot23+25\)

\(=81\cdot\left(33-23\right)+25\)

\(=810+25=835\)

\(g,\left[2^2+6^2\right]:5+11^2\)

\(=\left[4+36\right]:5+121\)

\(=40:5+121=8+121\)

\(=129\)

\(d,\frac{14\cdot3^{10}-5\cdot3^{10}}{3^{12}}\)

\(=\frac{3^{10}\cdot\left(14-5\right)}{3^{12}}\)

\(=\frac{3^{10}\cdot9}{3^{12}}\)

\(=\frac{3^{10}\cdot3^2}{3^{12}}=\frac{3^{12}}{3^{12}}\)

\(=1\)

còn bài 1 nưa mà bạn

a: \(7\cdot3^x=5\cdot3^7+2\cdot3^7\)

\(\Leftrightarrow7\cdot3^x=7\cdot3^7\)

=>3^x=3⁷

hay x=7

b: \(4^{x+3}-3\cdot4^{x+1}=13\cdot4^{11}\)

\(\Leftrightarrow4^{x+1}\left(4^2-3\right)=13\cdot4^{11}\)

=>x+1=11

hay x=10

d: \(\left(x-1\right)^{13}=\left(x-1\right)^{12}\)

\(\Leftrightarrow\left(x-1\right)^{12}\left(x-2\right)=0\)

hay \(x\in\left\{1;2\right\}\)