\(\dfrac{1}{2}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=\dfrac{1}{2^x}\)

=>\(\dfrac{2}{2}\cdot\dfrac{3}{6}\cdot\dfrac{4}{8}\cdot...\cdot\dfrac{30}{60}\cdot\dfrac{31}{62}\cdot\dfrac{1}{64}=\dfrac{1}{2^x}\)

=>\(\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot...\cdot\dfrac{1}{2}\cdot\dfrac{1}{64}=\dfrac{1}{2^x}\)

=>\(\dfrac{1}{2^{29}}\cdot\dfrac{1}{2^6}=\dfrac{1}{2^x}\)

=>x=29+6=35

?

a: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)

\(=1-\dfrac{1}{6}=\dfrac{5}{6}\)

b: \(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{10100}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{100\cdot101}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{100}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}=\dfrac{100}{101}\)

c: \(A=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}\)

d: \(A=\dfrac{3}{10}+\dfrac{3}{40}+...+\dfrac{3}{340}\)

\(=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{17\cdot20}\)

\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\)

\(=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\)

Bài 10:

Số học sinh giỏi ngoại ngữ chiếm:

\(\dfrac{1}{3}:\dfrac{4}{5}=\dfrac{1}{3}\cdot\dfrac{5}{4}=\dfrac{5}{12}\)(tổng số học sinh)

Số học sinh giỏi Văn là:

\(1-\dfrac{1}{3}-\dfrac{5}{12}=\dfrac{12-4-5}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)(tổng số học sinh)

Tổng số học sinh là: \(6:\dfrac{1}{4}=24\left(bạn\right)\)

Số học sinh giỏi toán là \(24\cdot\dfrac{1}{3}=8\left(bạn\right)\)

Số học sinh giỏi ngoại ngữ là 24-8-6=10(bạn)

Bài 11:

a: Để A là phân số thì \(x+2\ne0\)

=>\(x\ne-2\)

b: Để A là số nguyên thì \(2x-1⋮x+2\)

=>\(2x+4-5⋮x+2\)

=>\(-5⋮x+2\)

=>\(x+2\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{-1;-3;3;-7\right\}\)

a: \(\dfrac{2}{5}\cdot4-\dfrac{3}{5}\cdot7\cdot\dfrac{3}{2}\)

\(=\dfrac{8}{5}-\dfrac{63}{10}\)

\(=\dfrac{16}{10}-\dfrac{63}{10}=-\dfrac{47}{10}\)

b: \(-\dfrac{21}{10}+\dfrac{21}{10}\cdot\dfrac{3}{4}-\dfrac{3}{4}\)

\(=\dfrac{-21}{10}\left(1-\dfrac{3}{4}\right)-\dfrac{3}{4}\)

\(=-\dfrac{21}{10}\cdot\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-21}{40}-\dfrac{3}{4}=\dfrac{-51}{40}\)

c: \(\dfrac{-2}{5}\cdot\left(-6\right)+\dfrac{3}{4}\cdot\dfrac{4}{-10}\)

\(=\dfrac{12}{5}-\dfrac{3}{10}\)

\(=\dfrac{24}{10}-\dfrac{3}{10}=\dfrac{21}{10}\)

a)2/5.4-3/5.7.3/2                           b)-21/10 + 21/10 . 3/4-34

=8/5-63/10                                       =-21/10 +(21/10 . 3/4) -3/4

=-47/10                                             =-21/10+63/40-3/4

                                                        =-56/40

\(A=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dots+\dfrac{2}{99\cdot101}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dots+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}=\dfrac{100}{101}\)

thx bn ngheng :)

58

đây mà là toán lớp 6 á

thôi kệ đáp án là:58

\(\dfrac{3}{13}\cdot\dfrac{6}{11}+\dfrac{3}{13}-\dfrac{9}{11}\cdot\dfrac{9}{11}-\dfrac{3}{13}\cdot\dfrac{4}{11}\)
\(=\dfrac{3}{13}\left(\dfrac{6}{11}+1-\dfrac{4}{11}\right)-\dfrac{81}{121}\)
\(=\dfrac{3}{13}\cdot\dfrac{13}{11}-\dfrac{81}{121}\)
\(=\dfrac{3}{11}-\dfrac{81}{121}=-\dfrac{48}{121}\)

cgvgvgvg

a/Lãi suất tiết kiệm:
\(6030000:6000000-100\%=0,005=0,5\%\)
b/Số tiền gốc và lãi rút ra được tất cả trong 2 tháng:
\(6030000+6030000\cdot0,5\%=6060150\left(đ\right)\)

a,Lãi Suất là:

[6030000-6000000]:6000000 nhân 100=0,5 phần trăm

b,Số tiền lãi sau $2$ tháng là :

$30000.0,5=60000$ ( đồng )

Trong $2$ tháng thì rút ra tiền gốc và lãi được tất cả số tiền là :

$6000000+60000=6060000$ ( đồng )

\(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{10100}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{100\cdot101}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{100}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}=\dfrac{100}{101}\)

a: \(\dfrac{-14}{21}< 0\)

\(\dfrac{-60}{-72}=\dfrac{60}{72}>0\)

Do đó: \(\dfrac{-14}{21}< \dfrac{-60}{-72}\)

a) Ta có \(\dfrac{-60}{-72}\)=\(\dfrac{60}{72}\)

Vì \(\dfrac{-14}{21}\) < 0 Mà  \(\dfrac{-60}{-72}\)=\(\dfrac{60}{72}\) > 0

=> \(\dfrac{60}{72}\) > \(\dfrac{-14}{21}\) => \(\dfrac{-60}{-72}\) >\(\dfrac{-14}{21}\)

Vậy \(\dfrac{-60}{-72}\) > \(\dfrac{-14}{21}\)

b)

  +) Ta có BCNN(24,18) = 72

  +) 72 : 24 = 3

       72 : 18 = 4

  +)\(\dfrac{5}{24}\) = \(\dfrac{5.3}{24.3}\) = \(\dfrac{15}{72}\)

      \(\dfrac{7}{18}\) = \(\dfrac{7.4}{18.4}\) = \(\dfrac{28}{72}\)

  +) Vì 15 < 28 nên \(\dfrac{15}{72}\) <  \(\dfrac{28}{72}\) => \(\dfrac{5}{24}\) < \(\dfrac{7}{18}\)

Vậy \(\dfrac{5}{24}\) < \(\dfrac{7}{18}\)