\(5.3^2-32:4^2=43\)

trình bày tận gốc 

A.x=11

B.=47

bạn làm chi tiết đi

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

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

=\(375:25=15\)

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

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

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

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

\(=375:25\)

\(=15\)

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

a) \(2^x=32\)=>\(2^x=2^5\)=>\(x=5\)

b) \(64\times4^x=4^5\)

\(4^3.4^x=4^5\)

\(4^{3+x}=4^5\)

=>\(3+x=5\)

=> x=2

c)\(2^x-15=17\)

\(2^x=17+15\)

\(2^x=32\)

=>\(2^x=2^5\)

=>\(x=5\)

a, 2 mũ x = 32

    2 mũ x = 2 mũ 5

    x = 5

b, 64 x 4 mũ x = 4 mũ 5

    64 x 4 mũ x = 1024

     4 mũ x = 1024 : 64

      4 mũ x = 16

       4 mũ x = 4 mũ 2

       x = 2

   c, 2 mũ x - 15 = 17 

       2 mũ x = 17 + 15

       2 mũ x = 32

       2 mũ x = 2 mũ 5

       x = 5

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

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

1: \(A=2+2^2+2^3+\cdots+2^{100}\)

=>\(2A=2^2+2^3+2^4+\cdots+2^{101}\)

=>\(2A-A=2^2+2^3+2^4+\cdots+2^{101}-2-2^2-2^3-\cdots-2^{100}\)

=>\(A=2^{101}-2\)

2: \(B=1+5+5^2+5^3+\cdots+5^{150}\)

=>\(5B=5+5^2+5^3+\cdots+5^{151}\)

=>\(5B-B=5+5^2+5^3+\cdots+5^{151}-1-5-5^2-\cdots-5^{150}\)

=>\(4B=5^{151}-1\)

=>\(B=\frac{5^{151}-1}{4}\)

3: \(C=3+3^2+\cdots+3^{1000}\)

=>\(3C=3^2+3^3+\cdots+3^{1001}\)

=>\(3C-C=3^2+3^3+\cdots+3^{1001}-3-3^2-\cdots-3^{1000}\)

=>\(2C=3^{1001}-3\)

=>\(C=\frac{3^{1001}-3}{2}\)

Câu 1:

A = 2 + 2\(^2\) + 2\(^3\) + ... + 2\(^{100}\)

2A = 2\(^2\) + 2\(^3\) + ... + 2\(^{100}\) + 2\(^{101}\)

2A - A = (2\(^2\) + 2\(^3\) + ... + 2\(^{100}\)+ 2\(^{101}\)) -(2 + 2\(^2\) + 2\(^3\) + ... + 2\(^{100}\))

A = 2\(^2\) + 2\(^3\) + ... + 2\(^{100}\)+ 2\(^{101}\) - 2 - 2\(^2\) -2\(^3\) - ... - 2\(^{100}\)

A = (2\(^2\) - 2\(^2\)) + (2\(^3\) - 2\(^3\)) + ... + (2\(^{100}\) - 2\(^{100}\)) + (2\(^{101}\) - 2)

A = 0 + 0 + 0 + ... + 0 + 2\(^{101}\) - 2

A = 2\(^{101}\) - 2

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

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

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

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

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

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

\(=810+25=835\)

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

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

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

\(=129\)

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

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

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

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

\(=1\)