Bài 1:

 a,\(\frac{3.21}{14.15}\)=\(\frac{1.3}{2.5}\)=\(\frac{3}{10}\)

 b,\(\frac{49+7.49}{49}\)=\(\frac{49\left(7+1\right)}{49}\)=\(\frac{1.8}{1}\)=8

\(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot.....\cdot\frac{899}{30^2}\)

\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot.....\cdot\frac{29\cdot31}{30\cdot30}\)

\(=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{3}{4}\cdot\frac{5}{4}\cdot....\cdot\frac{29}{30}\cdot\frac{31}{30}\)

\(=\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{29}{30}\right)\cdot\left(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot....\cdot\frac{31}{30}\right)\)

\(=\frac{1}{30}\cdot\frac{31}{2}\)

\(=\frac{31}{60}\)

b, \(A=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)

Ta có:

\(\frac{3}{15}< \frac{3}{10}=\frac{3}{10}\)

\(\frac{3}{15}< \frac{3}{11}< \frac{3}{10}\)

\(\frac{3}{15}< \frac{3}{12}< \frac{3}{10}\)

\(\frac{3}{15}< \frac{3}{13}< \frac{3}{10}\)

\(\frac{3}{15}< \frac{3}{14}< \frac{3}{10}\)

\(\Rightarrow\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}< \frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}\)

\(\Rightarrow\frac{3\cdot5}{15}< A< \frac{3\cdot5}{10}\)

\(\Rightarrow1< A< \frac{15}{10}=\frac{3}{2}\)

Mà \(\frac{3}{2}< 2\)

\(\Rightarrow1< A< 2\)

c ,Ta có

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{25}\right)+\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)

thanks!!!

\(c.x\left(\frac{2}{5}-\frac{3}{5}\right)=\frac{2}{35}\)

\(x=\frac{2}{35}:\frac{-1}{5}=-\frac{2}{7}\)

\(d.\left(2x+1\right)^2=49=7^2=\left(-7\right)^2\)

\(TH1:2x+1=7\Rightarrow x=3\)

\(TH2=2x+1=-7\Rightarrow x=-4\)

\(a.x=\frac{-3}{5}-\frac{4}{9}=\frac{-47}{45}\)

\(b.\frac{3}{5}:x=\frac{17}{10}-\frac{2}{5}\)

\(x=\frac{3}{5}:\frac{13}{10}=\frac{6}{13}\)

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Yêu cầu của bài là gì vậy. Tính A? hay Chứng minh A < 2 hoặc chứng minh A không phải là số nguyên

Chứng minh A < 2

\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)

\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)

\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=2-\frac{1}{50}< 2\)

Vậy A < 2

toán này ko phải toán lớp 6 rùi

dễ quá 

a,8/3x + 26/3= 10/3

8/3x = 10/3- 26/3 = -16/3

=>x = -16/3 : 8/3 = -2

b, (2/3-1/2)x = 5/12

1/6x = 5/12

=>x = 5/2

xong rùi đó 

nhớ tk nha

câu 1:

a 3/11 - 7/11 + 2/11 = 3 - 7 + 2 / 11

                             = -5 + 2/11

                             = -3/11

b 7/15 - 2/3 = 7/15 - 10/15

                 = -1/5

c 3/5.9/11 + 3/5.2/11 - 2 1/2

= 3/5.9/11 + 3/5.2/11 - 5/2

= 3/5.( 9/11 + 2/11 - 5/2 )

= 3/5.( 1 - 5/2 )

= 3/5 . -3/2

= -9/10

câu 2

a. x + 75% = 5/12

    x + 3/4   = 5/12

    x            = 5/12 - 3/4

    x            = -1/3

vậy x = -1/3

b.| x - 7/15 | = 3/10

     x - 7/15  = 3/10 => x = 3/10 + 7/15 => x = 23/30

     x - 7/15  = -3/10 => x = -3/10 + 7/15 => x = 1/6

vậy x = 23/30 hoặc 1/6

tk mk nhé mk làm đc nửa bài rùi

Câu  1 :

a) \(\frac{3}{11}-\frac{7}{11}+\frac{2}{11}=\frac{-4}{11}+\frac{2}{11}=\frac{-2}{11}\)

b) \(\frac{7}{15}+\frac{2}{3}=\frac{7}{15}+\frac{10}{15}=\frac{17}{15}\)

c) \(\frac{3}{5}.\frac{9}{11}+\frac{3}{5}.\frac{2}{11}-2\frac{1}{2}\)

\(=\left(\frac{3}{5}.\frac{9}{11}+\frac{3}{5}.\frac{2}{11}\right)-\frac{5}{2}\)

\(=\frac{3}{5}.\left(\frac{9}{11}+\frac{2}{11}\right)-\frac{5}{2}\)

\(=\frac{3}{5}.1-\frac{5}{2}\)

\(=\frac{3}{5}-\frac{5}{2}=\frac{6}{10}-\frac{25}{10}=\frac{-19}{10}=-1\frac{9}{10}\)

Câu 2 :

a) \(x+75\%=\frac{5}{12}\)

  \(x+\frac{3}{4}=\frac{5}{12}\)

 \(x=\frac{5}{12}-\frac{3}{4}\)

\(x=\frac{5}{12}-\frac{9}{12}=\frac{-4}{12}=\frac{-1}{3}\)

Vậy : \(x=\frac{-1}{3}\)

Câu 3 :            

                                        Giải :

Số HS trung bình là :

     \(48.\frac{5}{12}=20\left(HS\right)\)

Số HS còn lại là :

      \(48-20=28\)\(\left(HS\right)\)

Số HS khá là :

      \(28.\frac{4}{7}=16\left(HS\right)\)

Số HS giỏi là :

      \(28-16=12\left(HS\right)\) 

          Đ/S : 12 Học sinh

Bài 1 : Tính :

a, \(-\frac{8}{3}.\frac{6}{13}.\frac{7}{13}.\frac{-3}{8}+1\frac{3}{8}\)

\(=\left(-\frac{8}{3}.-\frac{3}{8}\right).\left(\frac{6}{13}.\frac{7}{13}\right)+1\frac{3}{8}\)

\(=1.\frac{42}{169}+1\frac{3}{8}\)

\(=\frac{2195}{1352}\)

b) \(75\%-\left(\frac{5}{2}+\frac{5}{3}\right)+\left(-\frac{1}{2}\right)^2\)

\(=\frac{3}{4}-\frac{25}{6}+\frac{1}{4}\)

\(=\frac{3}{4}+\frac{1}{4}-\frac{25}{6}\)

\(=1-\frac{25}{6}\)

\(=-\frac{19}{6}\)

~Hok tốt~

Bài 2 : 

a)\(\frac{3}{5}.x-\frac{2}{3}=\frac{1}{2}\)                 b) \(\left(\frac{1}{2}-x\right).\frac{2}{3}=\frac{1}{8}\)                  

\(\frac{3}{5}.x=\frac{1}{2}+\frac{2}{3}\)                        \(\frac{1}{2}-x=\frac{1}{8}:\frac{2}{3}\)

\(\frac{3}{5}.x=\frac{7}{6}\)                                    \(\frac{1}{2}-x=\frac{3}{16}\)

\(x=\frac{7}{6}:\frac{3}{5}\)                                     \(x=\frac{1}{2}-\frac{3}{16}\)

\(x=\frac{35}{18}\)                                         \(x=\frac{5}{16}\)

c) \(\left|2x-\frac{3}{7}\right|-\frac{1}{2}=\frac{3}{4}\)

\(\left|2x-\frac{3}{7}\right|=\frac{3}{4}+\frac{1}{2}\)

\(\left|2x-\frac{3}{7}\right|=\frac{5}{4}\)

\(\Rightarrow2x-\frac{3}{7}\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)

\(TH1:2x-\frac{3}{7}=\frac{5}{4}\)                  \(TH2:2x-\frac{3}{7}=-\frac{5}{4}\)

\(\Rightarrow2x=\frac{5}{4}+\frac{3}{7}\)                           \(\Rightarrow2x=-\frac{5}{4}+\frac{3}{7}\)

     \(2x=\frac{47}{28}\)                                       \(2x=-\frac{23}{28}\)

      \(x=\frac{47}{28}:2\)                                        \(2x=-\frac{23}{28}:2\)

    \(x=\frac{47}{56}\)                                         \(2x=-\frac{23}{56}\)

\(\Rightarrow x\in\left\{\frac{47}{56};-\frac{23}{56}\right\}\)

\(\frac{x}{5}=\frac23\)

\(x\) = \(\frac23\times5\)

\(x=\frac{10}{3}\)

Vậy \(x=\frac{10}{3}\)

\(\frac{x}{3}-\frac12=\frac15\)

\(\frac{x}{3}\) = \(\frac15\) + \(\frac12\)

\(\frac{x}{3}\) = \(\frac{2}{10}+\frac{5}{10}\)

\(\frac{x}{3}=\frac{7}{10}\)

\(x=\frac{7}{10}\times3\)

\(x=\frac{21}{10}\)

Vậy \(x=\frac{21}{10}\)

\(\frac{x}{5}+\frac12=\frac{6}{10}\)

\(\frac{x}{5}=\frac{6}{10}-\frac12\)

\(\frac{x}{5}=\frac{6}{10}-\frac{5}{10}\)

\(\frac{x}{5}=\frac{1}{10}\)

\(x=\frac{1}{10}\times5\)

\(x=\frac12\)

Vậy \(x=\frac12\)

\(\frac{x+3}{15}\) = \(\frac13\)

\(x+3=\frac13\times15\)

\(x+3=5\)

\(x=5-3\)

\(x=2\)

Vậy \(x=2\)