ho tro dum mk trong 1tuan toi nha hom nay 16\3\2019

\(10^2\cdot10^3\cdot10^4\cdot10^5\cdot10^6\)

\(=10^{2+3+4+5+6}\)\

\(=10^{20}\)

Trả lời:

10² . 10³ . 10^{4 .} 10⁵ . 10⁶

= 10^2+3+4+5+6

= 10²⁰

Hok tốt!

b: \(\left(2x+1\right)^2=25\)

=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c: \(\left(1-3x\right)^3=64\)

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

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1

d: \(\left(4-x\right)^3=-27\)

=>\(\left(4-x\right)^3=\left(-3\right)^3\)

=>4-x=-3

=>x=4+3=7

e: \(x^2-5x=0\)

=>\(x\left(x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

\(40:\left(-5\right)-6.\left(-6\right)+\left(-2\right)^3\)

\(=\left(-8\right)-\left(-36\right)+\left(-8\right)\)

\(=\left(-8\right)+36+\left(-8\right)\)

\(=28+\left(-8\right)\)

\(=20\)

Chúc bạn học tốt !!!! 

\(40:\left(-5\right)-6.\left(-6\right)+\left(-2\right)^3\)

\(=-8+36+\left(-8\right)\)

\(=28+\left(-8\right)=20\)

học tốt nhé!!

\(10^2.10^3.10^5=10^{2+3+5}=10^{10}\)

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

\(a^3.a^2.a^5=a^{3+2+5}=a^{10}\)

mk nhanh nhất nha!

10^2 . 10^3 . 10^5 = 10^2+3+5 = 10^10

x . x^5 = x^5+1 = x^6

a^3 . a^2 . a^5 = a^3+2+5 = a^10

đầu hàng tớ đi

10².10³.10⁵=10²⁺³⁺⁵=10¹⁰

x.x⁵=x¹.x⁵=x¹⁺⁵=x⁶

a³.a².a⁵=a³⁺²⁺⁵=a¹⁰

a) \(5-\left(-3\right)^2-14.\left(-8\right)+\left(-31\right)\)

\(=5-9-\left(-112\right)+\left(-31\right)\)

\(=-4+112-31\)

\(=108-31\)

\(=77\)

b) \(6.\left(-2\right)^3+5.\left(-4\right)-\left(-12\right)\)

\(=6.\left(-8\right)+\left(-20\right)+12\)

\(=-48+\left(-20\right)+12\)

\(=-56\)

Ta có :

\(\left(2x+1\right)^3=27\)

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

\(\Rightarrow2x+1=3\)

\(\Rightarrow2x=3-1=2\)

\(\Rightarrow x=2:2=1\)

Vậy \(x=1\)

Ta có : \(\left(2x+1\right)^3=27\)

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

2x + 1 = 3

2x = 2

x = 1

Vậy x = 1

\(\Rightarrow2\cdot x+1=3\\ \Rightarrow2\cdot x=2\\ \Rightarrow x=1\)

b: \(x\in\left\{4;-4\right\}\)

\(a,x^2=16\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ b,\left(2x-1\right)^3=-27\\ \Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\\ \Rightarrow2x-1=-3\\ \Rightarrow2x=-2\\ \Rightarrow x=-1\)