\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{1}{99}\)

\(B=\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+\frac{100}{4.5}+\frac{100}{5.6}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{100}{99.100}\)

\(B=100\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(B=100\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(B=100\left(1-\frac{1}{100}\right)\)

\(B=100.\frac{99}{100}=99\)

Đúng òi nên xin chúc mừng nha !!!

trời ??? này mà là toán lớp 5 what???

I donnot know

I DONTS NO

What !!!!!!! đây bài lớp 6 hả man
 

Bài 1 :

    \(a+b=3.\left(a-b\right)=\)\(2\frac{a}{b}\)

\(\Rightarrow a+b=3.\left(a-b\right)\)

\(\Rightarrow a+b=3a-3b\)

\(\Rightarrow3a-3b-a-b=0\)

\(\Rightarrow2a-4b=0\)

\(\Rightarrow2.\left(a-2b\right)=0\)

\(\Rightarrow\hept{\begin{cases}a-2b=0\\a=2b\end{cases}}\)

Ta có : \(a+b=\frac{2a}{b}\)

Thay \(a=2b\) vào 

\(\Rightarrow2b+b=\frac{2.23}{b}\)

\(\Rightarrow3b=\frac{4b}{b}\Rightarrow3b=4\)

\(\Rightarrow b=\frac{4}{3}\Rightarrow a=2.\frac{4}{3}=\frac{8}{3}\)

Vậy \(a=\frac{8}{3}\) và \(b=\frac{4}{3}\)

Chúc bạn học tốt ( -_- )

Bài 2 :

\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{5}+\frac{100}{6.7}+...+\)\(\frac{100}{98.99}+\frac{1}{99}\)

\(B=\frac{100}{2}+\frac{100}{6}+\frac{100}{12}+\frac{100}{20}+\frac{100}{30}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{100}{9900}\)

\(B=\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+\frac{100}{4.5}+\frac{100}{5.6}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{100}{99.100}\)

\(B=100.\frac{100}{2}+\frac{100}{2}-\frac{1}{3}+\frac{100}{3}-\frac{100}{4}+\frac{100}{4}-\frac{100}{5}+\frac{100}{5}-\frac{100}{6}+\frac{100}{6}\)\(-\frac{100}{7}+...+\frac{100}{98}+\frac{100}{99}+\frac{100}{99}-1\)

\(B=100-1\)

\(B=99\)

Chúc bạn học tốt ( -_- )

a) 100.75

b)\(\frac{-7}{6}\)

n=ghi lộn nhé !!

a)\(10.\sqrt{0,01.\sqrt{ }\frac{16}{9}}+3\sqrt{49-\frac{1}{6}}\sqrt{4}\)

a, \(10.\sqrt{100}-\sqrt{\frac{1}{16}}+\left(\frac{1}{3}\right)^0\)

= 10 . 10 - \(\frac{1}{4}\) + 1

= 100 - \(\frac{1}{4}+1\)

= 99,75 + 1 = 100,75

b, \(\left(\frac{1}{3}\right)^{50}.\left(-9\right)^{25}-\frac{2}{3}:4\)

= \(\left(\frac{1}{9}\right)^{25}.\left(-9\right)^{25}-\frac{1}{6}\)

= \(\left(\frac{1}{9}.-9\right)^{25}-\frac{1}{6}\)

\(\left(-1\right)^{25}-\frac{1}{6}\)

= \(-1-\frac{1}{6}=\frac{-7}{6}\)