sao?

\(\dfrac{2^8\cdot9^3}{6^4\cdot4^3}=\dfrac{2^8\cdot3^6}{2^4\cdot3^4\cdot2^6}=\dfrac{2^8\cdot3^6}{2^{10}\cdot3^4}=\dfrac{3^2}{2^2}=\dfrac{9}{4}\)

\(12^5\div\left(2^6\cdot3^8\right)=2^{10}\cdot3^5\div\left(2^6\cdot3^8\right)=\dfrac{2^{10}\cdot3^5}{2^6\cdot3^8}=\dfrac{2^4}{3^3}=\dfrac{16}{27}\)

\(\dfrac{3^{12}\cdot2^{14}\cdot5^5}{10^5\cdot6^8\cdot12^4}=\)\(\dfrac{3^{12}\cdot2^{14}\cdot5^5}{5^5\cdot2^5\cdot2^8\cdot3^8\cdot2^8\cdot3^4}=\)\(\dfrac{3^{12}\cdot2^{14}\cdot5^5}{5^5\cdot2^{21}\cdot3^{12}}=\dfrac{1}{2^7}=\dfrac{1}{128}\)

1) \(\dfrac{2^8\cdot9^3}{6^4\cdot4^3}=\dfrac{2^8\cdot3^6}{2^4\cdot2^6\cdot3^4}=\dfrac{3^2}{2^2}=\dfrac{9}{4}\)

2) \(12^5:\left(2^6\cdot3^8\right)=\dfrac{2^{10}\cdot3^5}{2^6\cdot3^8}=\dfrac{2^4}{3^3}=\dfrac{16}{27}\)

\(\dfrac{3^{12}\cdot2^{14}\cdot5^5}{10^5\cdot6^8\cdot12^4}=\dfrac{3^{12}\cdot2^{14}\cdot5^5}{5^5\cdot2^{21}\cdot3^{12}}=\dfrac{1}{128}\)

mình chỉ có sách lớp 6 thui

bn gửi đè đi mk làm cho nha

Bn có thể gửi đề qua tin nhắn cho mk ko? Mk thử làm xem sao.

hãy trả lời rồi kết bạn với mk nhé

xⁿ-1=(x-1)(x^n-1+x^n-2+x^n-3+,,,,+1+1)

/x-5/=4 . TH1: x-5 = 4 => x=9

TH2: x-5 = - 4 => x= 1

Vậy....

\(\left|x-5\right|=4\)

\(\Rightarrow\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=9\\x=1\end{cases}}\)

\(\left|x-5\right|=4\)

\(\Rightarrow\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=9\\x=1\end{cases}}\)

Vậy x = 9 hoặc 1 

   `x^2-5x-2x^3+x^4+1 + (-5x^3) - 3+8x^4+x^2`

`= ( x^4 + 8x^4 ) - ( 2x^3 + 5x^3 ) + ( x^2 + x^2 ) - 5x + ( 1 - 3 )`

`= 9x^4 - 7x^3 + 2x^2 - 5x - 2`

= x²-5x-2x³+x⁴+1+(-5x³)-3+8x⁴+x²

=(x²+x²)+(-2x³-5x³)+(x⁴+8x⁴)-5x+(1-3)

=2x²+(-7x³)+9x⁴-5x+(-2)