Bài 1:

2^\(x\) = 4

2\(^x\) = 2²

 \(x=2\)

Vậy \(x=2\)

Bài 2:

2\(^x\) = 8

2\(^x\) = 2³

\(x=3\)

Vậy \(x=3\)

Vũ Hồng Linh bạn check lại bài đầu dùm =_=" 

\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)

\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)

\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)

\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)

\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)

\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)

=> Làm nốt 

Mấy bài kia cũng làm tương tự

(- \(\dfrac{1}{3}\))³.\(x\) = \(\dfrac{1}{81}\)

          \(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))³

          \(x\) =  - (\(\dfrac{1}{3}\))⁴ :(\(\dfrac{1}{3}\))³

           \(x=-\dfrac{1}{3}\)

Vậy \(x=-\dfrac{1}{3}\)

1: \(\dfrac{16^{11}\cdot5^{40}}{10^{41}}=\dfrac{2^{44}\cdot5^{40}}{2^{41}\cdot5^{41}}=\dfrac{2^3}{5^1}=\dfrac{8}{5}\)

2: \(\dfrac{3^7\cdot8^5}{6^6\cdot\left(-2\right)^{12}}=\dfrac{3^7\cdot2^{15}}{2^6\cdot3^6\cdot2^{12}}=\dfrac{3}{2^3}=\dfrac{3}{8}\)

a, 27³ : 3⁵  =  ( 3³)³ : 3⁵ = 3⁹ : 3⁵ = 3⁴

b, 7² . 343 . 49³⁰ = 7². 7³.(7²)³  = 7¹¹

c, 625 : 5³ = 5⁴ : 5³ =  5

d, 1 000 000 : 10³ = 10⁶ . 10³ = 10³

e, 11⁵ : 121= 11⁵ : 11² = 11³

f, 8⁷ : 64 :8 = 8⁷ : 8² : 8¹ = 8⁴

i, 1024 . 16 : 2⁶  = 2¹⁰ . 2³ : 2⁶ = 2⁷

B2:

 số chính phương là:

4 ; 121 ; 196 ; 225.

mình cũng hỏi về cái này gấp nha ai biết trả lời câu nào thì trả

7: \(\left(xy+4\right)^2-\left(2x+2y\right)^2\)

=(xy+4-2x-2y)(xy+4+2x+2y)

=[x(y-2)-2(y-2)][x(y+2)+2(y+2)]

=(x-2)(y-2)(x+2)(y+2)

8: \(81x^2-64y^2=\left(9x\right)^2-\left(8y\right)^2\)

=(9x-8y)(9x+8y)

9: \(\left(a^2+b^2+5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2+5\right)^2-\left(2ab+4\right)^2\)

\(=\left(a^2+b^2+5-2ab-4\right)\left(a^2+b^2+5+2ab+4\right)\)

\(=\left\lbrack\left(a-b\right)^2+1\right\rbrack\left\lbrack\left(a+b\right)^2+9\right\rbrack\)

10: \(\left(x-1\right)^2-\left(x+1\right)^2\)

=(x-1-x-1)(x-1+x+1)

=2x*(-2)=-4x

11: \(8x^3-\frac18=\left(2x\right)^3-\left(\frac12\right)^3=\left(2x-\frac12\right)\left(4x^2+2x\cdot\frac12+\frac14\right)\)

\(=\left(2x-\frac12\right)\left(4x^2+x+\frac14\right)\)

12: \(\frac{1}{25}x^2-64y^2=\left(\frac15x\right)^2-\left(8y\right)^2=\left(\frac15x-8y\right)\left(\frac15x+8y\right)\)

13: \(x^3+\frac{1}{27}=x^3+\left(\frac13\right)^3=\left(x+\frac13\right)\left(x^2-\frac13x+\frac19\right)\)

a, \(\frac{6^5\cdot27^2}{7^3\cdot9^5}=\frac{2^5\cdot3^5\cdot\left(3^3\right)^2}{7^3\cdot\left(3^2\right)^5}=\frac{2^5\cdot3^5\cdot3^6}{7^3\cdot3^{10}}=\frac{2^5\cdot3^{11}}{7^3\cdot3^{10}}=\frac{2^5\cdot3}{7^3}\)

b, \(\frac{12^7\cdot9^3}{8^5\cdot27^3}=\frac{3^7\cdot2^{12}\cdot3^6}{2^{15}\cdot3^9}=\frac{2^{12}\cdot3^{13}}{2^{15}\cdot3^9}=\frac{3^4}{2^3}\)

c, \(\frac{20^6\cdot8^2}{16^3\cdot25^3}=\frac{2^{12}\cdot5^6\cdot2^6}{2^{12}\cdot5^6}=2^6\)

a,5mũ 36=(5mũ3)mũ12=125 mũ12

11^24=(11^2)12=121^12

vì 121<125 nên 5^36>11^24

cảm ơn nha

\(4^8.2^{20}=2^{16}.2^{20}=2^{36}\)

\(9^{12}.27^5.81^4=3^{24}.3^{15}.3^{16}=3^{55}\)

mk chỉnh đề

\(64^3.4^5.16^2=4^9.4^5.4^4=4^{18}\)

\(25^{20}.125^4=5^{40}.5^{12}=5^{52}\)

\(x^7.x^4.x^3=x^{14}\)

cái này dễ thế mà cũng phải hỏi à bạn