a: =>3^x=3^4*3=3^5

=>x=5

b: =>\(2^{x+1}=2^5\)

=>x+1=5

=>x=4

c: \(\Leftrightarrow3^{x+2-3}=3\)

=>x-1=1

=>x=2

d: \(\Leftrightarrow x^2=\dfrac{32}{2}=16\)

=>x=4 hoặc x=-4

e: (2x-1)^4=81

=>2x-1=3 hoặc 2x-1=-3

=>2x=4 hoặc 2x=-2

=>x=-1 hoặc x=2

f: (2x-6)^4=0

=>2x-6=0

=>x-3=0

=>x=3

a) \(3^x=81\cdot3\)

\(\Rightarrow3^x=3^4\cdot3\)

\(\Rightarrow3^x=3^5\)

\(\Rightarrow x=5\)

b) \(2^{x+1}=32\)

\(\Rightarrow2^{x+1}=2^5\)

\(\Rightarrow x+1=5\)

\(\Rightarrow x=4\)

c) \(3^{x+2}:27=3\)

\(\Rightarrow3^{x+2}:3^3=3\)

\(\Rightarrow3^{x+2-3}=3\)

\(\Rightarrow3^{x-1}=3\)

\(\Rightarrow x-1=1\)

\(\Rightarrow x=2\)

d) \(2x^2=32\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x^2=4^2\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

e) \(\left(2x-1\right)^4=81\)

\(\Rightarrow\left(2x-1\right)^4=3^4\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

f)  \(\left(2x-6\right)^4=0\)

\(\Rightarrow2x-6=0\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=6:2\)

\(\Rightarrow x=3\)

\(a,3^x=81\cdot3\\ \Leftrightarrow3^x=3^4\cdot3\\ \Leftrightarrow3^x=3^5\\ \Leftrightarrow x=5\\ d,2^{x+1}=32\\ \Leftrightarrow x+1=5\\ \Leftrightarrow x=4\\ c,3^{x+2}:27=3\\ \Leftrightarrow3^{x+2}:3^3=3\\ \Leftrightarrow3^{x-1}=3\\ \Leftrightarrow x-1=1\\ \Leftrightarrow x=2\\ d,2x^2=32\\ \Leftrightarrow x^2=16\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ e,\left(2x-1\right)^4=81\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ f,\left(2x-6\right)^4=0\\ \Leftrightarrow2x-6=0\\ \Leftrightarrow x=3\)

\(a)4\cdot8\cdot16^2=2^2\cdot2^3\cdot\left(2^4\right)^2=2^5\cdot2^8=2^{13}\\ b)4^3\cdot64=\left(2^2\right)^3\cdot2^6=2^6\cdot2^6=2^{12}\\ c)3^4\cdot9\cdot27^2=3^4\cdot3^2\cdot\left(3^3\right)^2=3^4\cdot3^2\cdot3^6=3^{12}\\ d)16^2\cdot9^4=\left(2^4\right)^2\cdot\left(3^2\right)^4=2^8\cdot3^8=\left(2\cdot3\right) ^8=6^8\\ e)25\cdot5\cdot125=5^2\cdot5\cdot5^3=5^6\\ g)81\cdot3^3=3^4\cdot3^3=3^7\\ h)125^2\cdot25^4=\left(5^2\right)^2\cdot\left(5^2\right)^4=5^4\cdot5^8=5^{12}\\ i)4^2\cdot25=4^2\cdot5^2=\left(4\cdot5\right)^2=20^2\)

a)4⋅8⋅162=22⋅23⋅(24)2=25⋅28=213b)43⋅64=(22)3⋅26=26⋅26=212c)34⋅9⋅272=34⋅32⋅(33)2=34⋅32⋅36=312d)162⋅94=(24)2⋅(32)4=28⋅38=(2⋅3)8=68e)25⋅5⋅125=52⋅5⋅53=56g)81⋅33=34⋅33=37h)1252⋅254=(52)2⋅(52)4=54⋅58=512i)42⋅25=42⋅52=(4⋅5)2=202

a) 2/9 : 2/3 x 1/2

= 2/9 x 3/2 x 1/2

= 6/18 x 1/2

= 6/36

b) 2 + 1/4 x 4/3

= 2 + 4/7

= 2/1₍₇₎ + 4/7

= 14/7 + 4/7

= 18/7

c) 3 x 1/2 x 1/4

= 3/1 x 1/2 x 1/4

= 3/2 x 1/4

= 3/8

#Hok_tốt

Thông báo,Te2qAMq.png (1280×800)

Các bạn vô link trên sẽ xem đc độ dell có văn hóa của con aTRẦN LÊ KIM MAI,link nick https://olm.vn/thanhvien/kimmai123az

a: \(=\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{21+10}{35}=\dfrac{31}{35}\)

b: \(=\dfrac{6}{5}+\dfrac{3}{2}\cdot\dfrac{1}{3}=\dfrac{6}{5}+\dfrac{1}{2}=\dfrac{12+5}{10}=\dfrac{17}{10}\)

c: \(=\dfrac{5\cdot3}{9}=\dfrac{5}{3}\)

d: =5-2/3*6/5

=5-12/15

=5-4/5

=21/5

a) \(5x^3y\left(x-x^2y\right)=5x^4y-5x^5y^2\)

b) \(\left(27x^3-1\right):\left(9x^2+3x+1\right)\)

\(=\left(3x-1\right)\left(9x^2+3x+1\right):\left(9x^2+3x+1\right)\)

\(=3x-1\)

Đề câu b phải như này nha !

Lời giải của các bạn đều thỏa mãn yêu cầu đề bài là phân tích đa thức thành nhân tử

\(\dfrac{1}{2}+\dfrac{4}{3}\times\dfrac{1}{6}=\dfrac{1}{2}+\dfrac{4}{18}=\dfrac{9}{18}+\dfrac{4}{18}=\dfrac{13}{18}\)

1/2 + 2/9 = 13/18

13/18

13/42

\(\dfrac{7}{3}\times\dfrac{1}{2}+\dfrac{1}{7}=\dfrac{7}{6}+\dfrac{1}{7}=\dfrac{49}{42}+\dfrac{6}{42}=\dfrac{55}{42}\)

13/42

\(\frac{1}{2}\)- \(\frac{1}{3}\)x \(\frac{1}{4}\)

= \(\frac{1}{2}\)- \(\frac{1}{12}\)

= \(\frac{6}{12}\)- \(\frac{1}{12}\)

= \(\frac{5}{12}\)

tk nhé

5

12

 kb và k nha

1/2-1/3x1/4=1/24

k cho mình nhé