chịu em mới học lớp 5

Bài làm

a)\(\text{ -19+84+19+16 }\) 

     \(=\left(-19+19\right)+\left(84+16\right)\)

\(=0+100=100\)

\(\left(437-25\right)-\left(175+437\right)\)

\(=437-437-\left(25+175\right)=-200\)

\(c\))\(2^3.\left(-7\right).125\)

\(=\left(8.125\right).\left(-7\right)\)

\(=1000.\left(-7\right)=-7000\)

a). -19+84+19+16= 19+ (-84) +19+16=  19+ ( -84 + 16) = 19 + (-68 )= - 49
c). 2³. (-7) . 125 = 8. (-7) . 125 = (125.8). (-7) = 1000. (-7 ) = -7000 
d). (-17) . 39 + 17.( -161) = 17. (-39 ) + 17. (-161) = 17. \([\left(-39\right)+\left(-161\right)]\) = 17. (-200)= -3400
nhớ k cho mik nha <3
 

`Answer:`

Bài 1:

\(S_1=2+2^3+2^5+...+2^{51}\)

\(\Rightarrow4S_1=2^3+2^5+2^7+...+2^{53}\)

\(\Rightarrow4S-S=\left(2^3+2^5+2^7+...+2^{53}\right)-\left(2+2^3+2^5+...+2^{51}\right)\)

\(\Rightarrow3S=2^{53}-2\)

\(\Rightarrow S=\frac{2^{53}-2}{3}\)

Bài 2:

\(S_2=\frac{1}{3}+\left(\frac{1}{3}\right)^3+\left(\frac{1}{3}\right)^5+...+\left(\frac{1}{3}\right)^{99}\)

\(\Rightarrow3^2.S_2=3^2.\left(\frac{1}{3}+\left(\frac{1}{3}\right)^3+\left(\frac{1}{3}\right)^5+...+\left(\frac{1}{3}\right)^{99}\right)\)

\(\Rightarrow9.S_2=3+\frac{1}{3}+\left(\frac{1}{3}\right)^3+\left(\frac{1}{3}\right)^5+...+\left(\frac{1}{3}\right)^{97}\)

\(\Rightarrow9.S_2-S_2=\left(3+\frac{1}{3}+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{3}\right)^{97}\right)-\left(\frac{1}{3}+\left(\frac{1}{3}\right)^3+\left(\frac{1}{3}\right)^5+...+\left(\frac{1}{3}\right)^{99}\right)\)

\(\Rightarrow8.S_2=3+\frac{1}{3}+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{3}\right)^{97}-\frac{1}{3}-\left(\frac{1}{3}\right)^3-\left(\frac{1}{3}\right)^5-...-\left(\frac{1}{3}\right)^{99}\)

\(\Rightarrow8.S^2=3-\left(\frac{1}{3}\right)^{99}\)

\(\Rightarrow8.S_2=\frac{3^{100}}{3^{99}}-\frac{1}{3^{99}}\)

\(\Rightarrow S_2=\frac{3^{100}-1}{3^{99}.8}\)

Đáp án:   

A. Phân số \(\frac{-3}{5}< \frac{-4}{5}\) là sai.

*Giải thích:

Vì \(-3>-4\) nên \(\frac{-3}{5}>\frac{-4}{5}\) , còn \(\frac{-3}{5}< \frac{-4}{5}\) là sai nha

\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{99}{100}\)

cảm ơn nha