Hỏi đáp Phân số

ờ ok toán lớp 5 đi

SORY TÔI MỚI LỚP BA

2/3-(1\6+3/8)

Tôi không biết 

\(a)\)

\(M=\frac{3}{8}+\frac{3}{15}+\frac{3}{7}\)

\(=3\left(\frac{1}{8}+\frac{1}{15}+\frac{1}{7}\right)\)

\(=3\frac{105+56+120}{8.15.7}\)

\(=3.\frac{281}{3.5.8.7}\)

\(=\frac{281}{280}>1\)

Vậy ...

\(b)\)

\(N=\frac{19}{60}+\frac{29}{100}+\frac{39}{150}+\frac{49}{300}\)

\(=\frac{95}{300}+\frac{87}{300}+\frac{78}{300}-\frac{49}{300}\)

\(=\frac{95+87+78+49}{300}\)

\(=\frac{320}{300}>1\)

Vậy ...

\(c)\)

\(B=\frac{41}{90}+\frac{31}{72}+\frac{21}{40}-\frac{11}{45}-\frac{1}{36}\)

\(=\frac{293}{180}>\frac{180}{180}\)

Vậy ...

\(a,\frac{3}{4}:\frac{4}{7}=\frac{21}{16}\)

\(b,\frac{6}{11}\cdot\frac{2}{3}=\frac{12}{33}\)

\(c,\frac{2}{3}\cdot\frac{4}{7}=\frac{8}{21}\)

\(d,\frac{2}{3}\cdot\frac{4}{7}:\frac{1}{5}=\frac{8}{21}:\frac{1}{5}=\frac{40}{21}\)

\(\frac{3}{4}:\frac{4}{7}=\frac{3}{4}\times\frac{7}{4}=\frac{21}{28}\)

\(\frac{6}{11}\times\frac{2}{3}=\frac{4}{11}\)

\(\frac{2}{3}\times\frac{4}{7}:\frac{1}{5}=\frac{2}{3}\times\frac{4}{7}\times5=\frac{2\times4\times5}{3\times7}=\frac{40}{21}\)

A.\(\frac{3}{4}\)+\(\frac{4}{7}\)

=\(\frac{21}{28}\)+\(\frac{16}{28}\)

=\(\frac{37}{28}\)

B.\(\frac{6}{11}\)x\(\frac{2}{3}\)

=\(\frac{4}{11}\)

C.\(\frac{2}{3}\)x\(\frac{4}{7}\)+\(\frac{1}{5}\)

=\(\frac{8}{21}\)+\(\frac{1}{5}\)

=\(\frac{40}{105}\)+\(\frac{21}{105}\)

=\(\frac{61}{105}\)

Học tốt

Nhìn mà chẳng mún làm .

1/101+1/102+..+1/200=(1+1/2+1/3+...+1/100)+1/101+1/102+1/103+...+1/200-(1+1/2+1/3+...+1/100)

=(1/2+1/4+1/6+...+1/200)+(1+1/3+1/5+...+1/199)-2(1/2+1/4+1/6+...+1/200)

=(1+1/3+1/5+...+1/199)-(1/2+1/4+1/6+...+1/200)

=1-1/2+1/3-1/4+1/5-1/6+...+1/19 

=>ĐPCM

A~~~ trái đất này tròn wá @~@

có j thắc mắc ib mk nhé

Gọi d là ƯCLN của 10n + 1 và 15n + 2 ( d \(\in\)N* ) 

\(\Rightarrow\hept{\begin{cases}10n+1⋮d\\15n+2⋮d\end{cases}\Rightarrow\hept{\begin{cases}3\left(10n+1\right)⋮d\\2\left(15n+2\right)⋮d\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}30n+3⋮d\\30n+4⋮d\end{cases}\Rightarrow\left(30n+4\right)-\left(30n+3\right)⋮d}\)

\(\Rightarrow1⋮d\Rightarrow d=1\)

Vậy \(\frac{10n+1}{15n+2}\)là p/s tối giải.

ko biết tự làm đi

ghi lại đề cái

sao phải ghi lại vậy

\(\text{Theo đề ta có :}\)

    \(\frac{\left(-1\right)^6\cdot3^5\cdot4^3}{9^2\cdot2^5}\)

= \(\frac{1\cdot3^5\cdot\left(2^2\right)^3}{\left(3^2\right)^2\cdot2^5}\) =  \(\frac{3^5\cdot2^6}{3^4\cdot2^5}=\frac{3^4\cdot3\cdot2^5\cdot2}{3^4\cdot2^5}=3\cdot2=6\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=\frac{1\times16}{2\times16}+\frac{1\times8}{4\times8}+\frac{1\times4}{8\times4}+\frac{1\times2}{16\times2}+\frac{1}{32}\)

                                                         \(=\frac{16}{32}+\frac{8}{32}+\frac{4}{32}+\frac{2}{32}+\frac{1}{32}\)

                                                         \(=\frac{16+8+4+2+1}{32}\)\(=\frac{31}{32}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\) 

\(A\cdot2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)

\(A\cdot2-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)

\(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}\)

\(A=1-\frac{1}{32}\)

\(A=\frac{31}{32}\)

mình ko giải theo cách của thương . xin lỗi bạn

mình đang cần vội

a, Cho mk sửa đề

\(\frac{145145}{607607}=\frac{145.1001}{107.1001}=\frac{145}{607}\)

b, \(\frac{19491949}{20062006}=\frac{1949.10001}{2006.10001}=\frac{1949}{2006}\)

\(B=\frac{3737.43-4343.37}{2+4+6+...+100}\)

\(=\frac{37.101.43-43.101.37}{2+4+6+...+100}\)

\(=\frac{0}{2+4+6+...+100}\)

\(=0\)

\(B\)\(=\)\(\frac{3737.43-4343.37}{2+4+6+...+100}\)\(=\)\(\frac{37.101.43-4343.37}{2+4+6+...+100}\)

       \(=\)\(\frac{37.4343-4343.37}{2+4+6+...+100}\)\(=\)\(\frac{0}{2+4+6+...+100}\)\(=\)\(0\)

Xét tử ta có: 

\(101+100+99+98+...........+3+2+1\)

\(=1+2+3+..........+99+100+101\)

\(=\frac{101.102}{2}=5151\)

Xét mẫu ta có:

\(101-100+99-98+.......+3-2+1\)

\(=\left(101-100\right)+\left(99-98\right)+.......+\left(3-2\right)+1\)

\(=1+1+.......+1+1=51\)

\(\Rightarrow A=\frac{5151}{51}=101\)

a, \(\left|x-\frac{2}{3}\right|=\frac{1}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{2}{3}=\frac{1}{2}\\x-\frac{2}{3}=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}+\frac{2}{3}\\x=\frac{2}{3}-\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{6}\\x=\frac{1}{6}\end{cases}}}\)

b, \(\left|x+\frac{7}{20}\right|=\frac{3}{15}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{7}{20}=\frac{1}{5}\\x+\frac{7}{20}=-\frac{1}{5}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}-\frac{7}{20}\\x=-\frac{1}{5}-\frac{7}{20}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-3}{20}\\x=\frac{-11}{20}\end{cases}}}\)

c, \(\left|3x+2\right|=\left|7x-4\right|\)

\(\Leftrightarrow\orbr{\begin{cases}3x+2=7-4x\\3x+2=4x-7\end{cases}\Leftrightarrow\orbr{\begin{cases}7x=5\\x=9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{7}\\x=9\end{cases}}}\)

d, \(\left|5-2x\right|=\left|2x-5\right|\)

\(\Leftrightarrow\orbr{\begin{cases}5-2x=2x-5\\5-2x=5-2x\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\0x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x\in Q\end{cases}}}\)

=> Có vô số x thỏa mãn \(\left|5-2x\right|=\left|2x-5\right|\)

e, \(\left|-5-6x\right|=\left|-x-5\right|\)

\(\Leftrightarrow\orbr{\begin{cases}-5-6x=-x-5\\-5-6x=x+5\end{cases}\Leftrightarrow\orbr{\begin{cases}-5x=0\\-7x=10\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=-\frac{10}{7}\end{cases}}}\)

f, \(\left|-x+5\right|=\left|12-3x\right|\) đúng ko ???

\(\Leftrightarrow\orbr{\begin{cases}-x-5=12-3x\\-x+5=3x-12\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\-4x=17\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{17}{4}\end{cases}}}\)

2 học sinh nữ chiếm số phần là 

\(\frac{1}{3}-\frac{1}{4}=\frac{1}{12}\) 

Số học sinh lớp 5A là 

\(2:\frac{1}{12}=24\) ( học sinh ) 

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