Bài 6: Lũy thừa của một số hữu tỉ (tiếp theo...)

\(a,\left(2x+3\right)^2=36\\ \left[{}\begin{matrix}2x+3=6\\2x+3=-6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{9}{2}\end{matrix}\right.\\ b,\left(3x+6\right)^2=\left(x-10\right)^2\\ 3x+6=x-10\\ 3x-x+6+10=0\\ 2x+16=0\\ 2x=-16\\ x=-8\\ c,\left[{}\begin{matrix}x+1=0\\2x-4=0\end{matrix}\right.=>x=-1;x=2\\ d,\left(2x+3\right)^3=64\\ 2x+3=4\\ 2x=1\\ x=\dfrac{1}{2}\\ b,x+6=2x-1\\ x-2x=-1-6\\ -x=-7\\ x=7\)

a: =>(2x+3+6)(2x+3-6)=0

=>(2x+9)(2x-3)=0

=>x=3/2 hoặc x=-9/2

b: =>(3x+6+x-10)(3x+6-x+10)=0

=>(4x-4)(2x+16)=0

=>x=-8 hoặc x=1

c: =>(x+1)=0 hoặc 2x-4=0

=>x=2 hoặc x=-1

d: =>(2x+3+8)(2x+3-8)=0

=>(2x+11)(2x-5)=0

=>x=5/2 hoặc x=-11/2

\(\dfrac{2^7}{8^2}=\dfrac{2^7}{\left(2^3\right)^2}=\dfrac{2^7}{2^6}=2\)

\(=\dfrac{2^{4+3}}{\left(2^3\right)^2}=\dfrac{2^7}{2^6}=2\)

`= (2^7)/(2^6)`

`= 2`

`a, A = (2^51 . 5^23 . 3^23)/(2^50 . 3^25 . 5^22)`

`= (2.5)/3^2`

`= 10/9`

`b, B = (3^12)/(2^2 . 3^12)`

`= 1/9`

`c, = (5^13 . 3^13 . 2^24)/(3^12 . 2^12 .2^13 . 5^13)`

`=3/2`

a: \(=\dfrac{2^{51}\cdot3^{23}\cdot5^{23}}{2^{50}\cdot3^{25}\cdot5^{22}}=6\cdot5=30\)

b: \(=\dfrac{3^6\cdot3^6}{3^2\cdot2^2\cdot3^{10}}=\dfrac{1}{4}\)

c: \(=\dfrac{3^{13}\cdot5^{13}\cdot2^{24}}{3^{12}\cdot2^{12}\cdot2^{13}\cdot5^{13}}=\dfrac{3}{2}\)

a: \(=\dfrac{5^8\cdot2^8}{5^{10}\cdot2^{10}}=\dfrac{1}{100}\)

b: \(=\dfrac{3^7\cdot3^6}{3^4\cdot3^6}=3^3=27\)

c: \(=\dfrac{5^{10}\cdot5^{20}\cdot3^{20}}{5^{30}\cdot3^{15}}=3^5=243\)

a: \(=\dfrac{2^3\cdot3^{13}}{2^3\cdot3^{12}}=3\)

b: \(=\dfrac{3^{15}\cdot5^{12}}{5^{13}\cdot3^{17}}=\dfrac{1}{9\cdot5}=\dfrac{1}{45}\)

c: \(=\dfrac{3^{15}\cdot5^{12}}{5^{11}\cdot3^{19}}=\dfrac{5}{81}\)

`a, = (2^20 . 5^14)/(5^14 . 2^18)`

`= 2^2 = 4`

`b, = (2^44 . 5^40)/(10^41)`

`= 16/5`

`d, = (5^14 . 2^15)/(2^15 . 5^12)`

`= 5^2 = 25`

a: \(=\dfrac{2^{24}\cdot3^{14}}{3^{12}\cdot2^{10}}=2^{14}\cdot3^2\)

b: \(=\dfrac{3^{16}\cdot2^{18}}{2^{16}\cdot3^{17}}=\dfrac{2}{3}\)

c: \(=\dfrac{2^{11}\cdot3^4}{3^5\cdot2^8}=\dfrac{2^3}{3}=\dfrac{8}{3}\)

\(\dfrac{8^3.6^5}{2^{12}.27^2}=\dfrac{\left(2^3\right)^3.2^5.3^5}{2^{12}.\left(3^3\right)^2}=\dfrac{2^9.2^5.3^5}{2^{12}.3^6}=\dfrac{2^2}{3}=\dfrac{4}{3}\)

\(\left(x+1\right)^3-3=24\)

\(\Leftrightarrow\left(x+1\right)^3=27=3^3\)

\(\Leftrightarrow x+1=3\)

\(\Leftrightarrow x=2\)

a: \(\Leftrightarrow\left(\dfrac{1}{2}\right)^x\left(1+\dfrac{1}{16}\right)=17\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^x=16\)

hay x=-4

b: =>3<x<5

=>x=4