P = (10.10².10³.10⁴...10⁹) : (10⁵.10¹⁰.10²⁵)

= 10¹⁺²⁺³⁺⁴⁺⁵⁺⁶⁺⁷⁺⁸⁺⁹ 10⁵⁺¹⁰⁺²⁵

= 10⁴⁵ : 10⁴⁰

= 10⁴⁵⁻⁴⁰

= 10⁵

= 100000

\(P=\left(10.10^2.10^3.10^4.....10^9\right):\left(10^5.10^{10}.10^{25}\right)\)

\(P=10^{45}:10^{40}\)

\(P=10^5\)

a: y=0; x=3

😟😟

bÌ NÀO BN 

Bài 1: 

a: =204-7=197

b: =900-1+1=900

c: \(=155-12\cdot9\cdot2^8=-27493\)

Bài 2: 

a:=>x=405:27=15

phần nào vậy bạn

A                   =      \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\)+ \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)+......+\(\dfrac{1}{1024}\)

A \(\times\)2           = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+...+\(\dfrac{1}{512}\)

A\(\times\)2 - A     =   1 - \(\dfrac{1}{1024}\)

A               = \(\dfrac{1023}{1024}\)

B = \(\dfrac{1}{1\times2}\)+\(\dfrac{1}{2\times3}\)+\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{4\times5}\)+...+\(\dfrac{1}{98\times99}\)+\(\dfrac{1}{99\times100}\)

B = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+...+\(\dfrac{1}{98}\)-\(\dfrac{1}{99}\)+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)

B = 1 - \(\dfrac{1}{100}\)

B = \(\dfrac{99}{100}\)

C= \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\)+ \(\dfrac{1}{20}\)+ \(\dfrac{1}{30}\)+\(\dfrac{1}{42}\)+...+\(\dfrac{1}{90}\)+\(\dfrac{1}{110}\)

C = \(\dfrac{1}{2\times3}\) +\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+...+\(\dfrac{1}{9\times10}\)+\(\dfrac{1}{10\times11}\)

C = \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)-\(\dfrac{1}{11}\)

C = \(\dfrac{1}{2}\) - \(\dfrac{1}{11}\)

C = \(\dfrac{9}{22}\)

Bài 3:

a. \(R=R1+R2=15+30=45\Omega\)

b. \(\left\{{}\begin{matrix}I=U:R=9:45=0,2A\\I=I1=I2=0,2A\left(R1ntR2\right)\end{matrix}\right.\)

c. \(\left\{{}\begin{matrix}U1=R1.I1=15.0,2=3V\\U2=R2.I2=30.0,2=6V\end{matrix}\right.\)

Bài 4:

\(I1=U1:R1=6:3=2A\)

\(\Rightarrow I=I1=I2=2A\left(R1ntR2\right)\)

\(U=R.I=\left(3+15\right).2=36V\)

\(U2=R2.I2=15.2=30V\)