\(\frac{x^8}{27}\)= 243 àk ?

x⁸ = 243 x 27

x⁸ = 6561

x⁸ = 3⁸

\(\Rightarrow\)x = 3

`a)x^3=343=7^3`

`=>x=7`

Vậy `x=7`

`b)(x-2,5)^4=(x-2,5)^2`

`=>(x-2,5)^2[(x-2,5)^2-1]=0`

`+)(x-2,5)^2=0<=>x=2,5`

`+)(x-2,5)^2=1`

`TH1:x-2,5=1<=>x=3,5`

`th2:x-2,5=-1<=>x=1,5`

Vậy `x=0` hoặc `x=1,5` hoặc `x=3,5

`c)x^8/243=27`

`=>x^8=27.243`

`=>x^8=3^3*3^5=3^8`

`=>x=+-3`

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

⇔\(x^2-2.x.1,2+1,2^2=4\)

⇔\(x^2-2,4x+1,44=4\)

⇔\(x^2-2,4x=4-1,44\)

⇔\(x\left(x-2,4\right)=2,56\)

⇔\(x=2,56\) hoặc \(x-2,4=2,56\)

⇔\(x=2,56\) hoặc \(x=4,96\)

a) \(\left(x-1,2\right)^2=4=2^2\)

\(\Leftrightarrow x-1,2=4\)

\(\Leftrightarrow x=5,2\)

b) \(\left(x+1\right)^3=-125=\left(-5\right)^3\)

\(\Leftrightarrow x+1=-5\)

\(\Leftrightarrow x=-6\)

c) \(\left(x+1,5\right)^8+\left(2,7-y\right)^{10}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\2,7-y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\y=2,7\end{matrix}\right.\)

\(243^5\)và \(3\cdot27^8\)

ta có

\(243^5=\left(3^5\right)^5=3^{5\cdot5}=3^{25}\)

\(3\cdot27^8=3\cdot3^{24}=3^{25}\)

\(\Rightarrow3^{25}=3^{25}\)

hay 

\(243^5=3\cdot27^8\)

a, A= 5x4¹⁵ x 9⁹ - 4x3²⁰x 8⁹ = 5x2³⁰x3¹⁸-2²x 2²⁷x3²⁰

=2²⁹x3¹⁸(2x5-3²)=2²⁹x3¹⁸

B=5x2⁹x6¹⁹-7x2²⁹x27⁶=5x2⁹x2¹⁹x3¹⁹7x2²⁹x3¹⁸

=2³⁸x3¹⁸(5x3-7x2)=2²⁸-3¹⁸

^->A:B= 2²⁹x3¹⁸:2²⁸:3¹⁸=2

b sai nha

`@` `\text {Ans}`

`\downarrow`

`a,`

`4*8`

`= 2^2*2^3`

`=`\(2^{2+3}=2^5\)

`b,`

`4^5*4^4*4`

`=`\(4^{5+4+1}=4^{10}\)

`c,`

\(x^5\cdot x=x^{5+1}=x^6\)

`d,`

\(9\cdot27=3^2\cdot3^3=3^{2+3}=3^5\)

`e,`

\(243\div81=3^5\div3^4=3\)

`g,`

\(625\div25=5^4\div5^2=5^2\)

`@` `\text {Kaizuu lv uuu}`

a) \(4\cdot8=2^2\cdot2^3\)

\(=2^{2+3}=2^5=32\)

b) \(4^5\cdot4^4\cdot4\)

\(=4^{5+4+1}=4^{10}\)

c) \(x^5\cdot x\)

\(=x^{5+1}=x^6\)

d) \(9\cdot27=3^2\cdot3^3\)

\(=3^{2+3}=3^5\)

e) \(243:81=3^5:3^4\)

\(=3^{5-4}=3^1=3\)

g) \(625:25=5^4:5^2\)

\(=5^{4-2}=5^2=25\)