Bài 3

a,26/100+0,009+41/100+0,24

0,26+0,09+0,41+0,24

(0,26+0,24)+(0,09+0,41)

0,5+0,5

=1

b,9+1/4+6+2/7+7+3/5+8+2/3+2/5+1/3+5/7+3/4

(9+6+7+8)+(2/7+5/7)+(1/4+3/4)+(3/5+2/5)+(2/3+1/3)

30+1+1+1+1

=34

Bài 4,5 khó quá mik ko bít lamf^^))

Bài 5: vì \(\dfrac{3}{11}\) = \(\dfrac{3\times5}{11\times5}\) = \(\dfrac{15}{55}\)

Vậy Khi giữ nguyên tử số thì số cần thêm vào mẫu số là: 

              55 - 39 = 16

Đáp số: 16 

Bài 4: a, \(\dfrac{2008}{2009}\) < 1; \(\dfrac{10}{9}\) > 1

           \(\dfrac{2008}{2009}\) < \(\dfrac{10}{9}\)

         b, \(\dfrac{1}{a+1}\) và \(\dfrac{1}{a-1}\)

Ta có: a + 1 > a - 1 ⇒ \(\dfrac{1}{a+1}\) < \(\dfrac{1}{a-1}\)

\(49,=\dfrac{38}{5}-\left(\dfrac{19}{7}+\dfrac{28}{5}\right)\)

\(=\dfrac{38}{5}-\dfrac{19}{7}-\dfrac{28}{5}\)

\(=\left(\dfrac{38}{5}-\dfrac{28}{5}\right)-\dfrac{19}{7}\)

\(=2-\dfrac{19}{7}=-\dfrac{5}{7}\)

\(50,=\dfrac{25}{81}.\dfrac{15}{22}=\dfrac{125}{594}\)

Câu 50 tus sửa đề 

\(-\dfrac{1}{9}.\dfrac{15}{22}:\dfrac{-25}{9}=-\dfrac{1}{9}.\dfrac{15}{22}.\dfrac{-9}{25}=\dfrac{15}{22}.\dfrac{1}{25}=\dfrac{3}{110}\)

Bài 1 :

Câu 49 . = 38/5 - ( 19/7 + 28/5 )

               = 38/5 - 19/7 - 28/5 

               = ( 38/5 - 28/5 ) - 19/7 

               = 2 - 19/7 = -5/7

Câu 50 . = ( -1/9 : -25/9 ) . 15/22

               = 1/25 . 15/22 = 3/110 

Đúng 100 % nha bạn 🤍

\(a,=\dfrac{13}{50}\cdot\dfrac{50}{13}\cdot\left(-\dfrac{31}{2}\right)\cdot\dfrac{169}{2}=-\dfrac{5239}{2}\\ b,=\dfrac{-\dfrac{49}{100}\cdot\left(-125\right)}{-\dfrac{343}{27}\cdot\dfrac{81}{16}\cdot\left(-1\right)}=\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{245}{4}\cdot\dfrac{16}{1029}=\dfrac{20}{21}\)

a) \(\dfrac{13}{50}.\left(-15.5\right):\dfrac{13}{50}.84\dfrac{1}{2}=\dfrac{13}{50}.-75:\dfrac{13}{50}.\dfrac{169}{2}=-\dfrac{75.169}{2}=-\dfrac{12675}{2}\)

b) \(\dfrac{\left(-0,7\right)^2.\left(-5\right)^3}{\left(-2\dfrac{1}{3}\right)^3.\left(1\dfrac{1}{2}\right)^4.\left(-1\right)^5}=\dfrac{0,49.\left(-125\right)}{-\dfrac{343}{27}.\dfrac{81}{16}.\left(-1\right)}=-\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{20}{21}\)

\(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)

= \(\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}\)

= \(\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}\)

= \(\dfrac{40}{14}=\dfrac{20}{7}\)

\(4\dfrac{1}{2}+\dfrac{1}{2}\div5\dfrac{1}{2}\)

=\(\dfrac{9}{2}+\dfrac{1}{2}\div\dfrac{11}{2}\)

=\(\dfrac{9}{2}+\dfrac{1}{2}\times\dfrac{2}{11}\)

=\(\dfrac{9}{2}+\dfrac{1}{11}\)

=\(\dfrac{101}{22}\)

\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{10}{3}\div\dfrac{17}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{10}{3}\times\dfrac{4}{17}\)

\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)

\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)

\(x=\dfrac{40}{51}\times\dfrac{3}{10}\)

\(x=\dfrac{120}{510}=\dfrac{12}{51}=\dfrac{4}{7}\)

\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)

\(\dfrac{17}{3}\div x=\dfrac{11}{3}-\dfrac{5}{2}\)

\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\times\dfrac{6}{7}\)

\(x=\dfrac{102}{21}=\dfrac{34}{7}\)

A= \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{9}{11}=\dfrac{1}{3}-\dfrac{7}{9}=\dfrac{3}{9}-\dfrac{7}{9}=-\dfrac{4}{9}\)

\(B=\left(\dfrac{1}{5}+\dfrac{2}{15}+\dfrac{2}{3}\right)+\left(-\dfrac{2}{7}+\dfrac{1}{42}-\dfrac{13}{28}-\dfrac{1}{4}\right)\)

\(=\dfrac{3+2+10}{15}+\dfrac{-2\cdot12+2-13\cdot3-21}{84}\)

=1-82/84

=2/84=1/42

\(C=\dfrac{1}{50}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\right)\)

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

\(=\dfrac{1}{50}-1+\dfrac{1}{50}=\dfrac{1}{25}-1=-\dfrac{24}{25}\)

\(D=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)

Bài 7: Tính

a) \(4\dfrac{2}{5}\times8\dfrac{3}{4}-2\dfrac{3}{4}\)

\(=\dfrac{22}{5}\times\dfrac{35}{4}-\dfrac{11}{4}\)

\(=\dfrac{77}{2}-\dfrac{11}{4}\)

\(=\dfrac{143}{4}\)

b) \(2\dfrac{2}{3}+1\dfrac{2}{5}-\dfrac{2}{15}\)

\(=\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{2}{15}\)

\(=\dfrac{47}{15}-\dfrac{2}{15}\)

\(=\dfrac{45}{15}=3\)

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

\(=\dfrac{10}{3}-\dfrac{8}{3}+\dfrac{11}{6}\)

\(=\dfrac{2}{3}+\dfrac{11}{6}\)

\(=\dfrac{15}{6}\)

a) \(...=\dfrac{19}{8}:\dfrac{15}{4}x\dfrac{8}{3}=\dfrac{19}{8}x\dfrac{4}{15}x\dfrac{8}{3}=\dfrac{76}{45}\)

b) \(...=\dfrac{3}{2}:\dfrac{7}{3}:\dfrac{17}{6}=\dfrac{3}{2}x\dfrac{3}{7}x\dfrac{6}{17}=\dfrac{27}{119}\)

c) \(...=\dfrac{14}{3}-\dfrac{7}{4}:\dfrac{12}{5}=\dfrac{14}{3}-\dfrac{7}{4}x\dfrac{5}{12}=\dfrac{14}{3}-\dfrac{35}{48}=\dfrac{14x16}{48}-\dfrac{35}{48}=\dfrac{224}{48}-\dfrac{35}{48}=\dfrac{189}{48}=\dfrac{63}{16}\)

\(a,2\dfrac{3}{8}:3\dfrac{3}{4}\times2\dfrac{2}{3}\\ =\dfrac{2\times8+3}{8}:\dfrac{3\times4+3}{4}\times\dfrac{2\times3+2}{3}\\ =\dfrac{19}{8}:\dfrac{15}{4}\times\dfrac{8}{3}\\ =\dfrac{19\times4\times8}{8\times15\times3}=\dfrac{76}{45}\)

\(b,1\dfrac{1}{2}:\dfrac{7}{3}:2\dfrac{5}{6}\\ =\dfrac{3}{2}:\dfrac{7}{3}:\dfrac{2\times6+5}{6}\\ =\dfrac{3}{2}\times\dfrac{3}{7}\times\dfrac{6}{17}\\ =\dfrac{54}{238}=\dfrac{27}{119}\)

\(c,4\dfrac{2}{3}-1\dfrac{3}{4}:2\dfrac{2}{5}\\ =\dfrac{4\times3+2}{3}-\dfrac{1\times4+3}{4}:\dfrac{2\times5+2}{5}\\ =\dfrac{14}{3}-\dfrac{7}{4}:\dfrac{12}{5}\\ =\dfrac{14}{3}-\dfrac{7}{4}.\dfrac{5}{12}\\ =\dfrac{14}{3}-\dfrac{35}{48}\\ =\dfrac{14\times16-35}{48}=\dfrac{189}{48}=\dfrac{63}{16}\)

Bài 1:

a.

$|x+\frac{7}{4}|=\frac{1}{2}$

\(\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.\)

b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$

$|2x+1|=\frac{11}{15}$

\(\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.\)

c.

$3x(x+\frac{2}{3})=0$

\(\Leftrightarrow \left[\begin{matrix} 3x=0\\ x+\frac{2}{3}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\frac{-3}{2}\end{matrix}\right.\)

d.

$x+\frac{1}{3}=\frac{2}{5}-(\frac{-1}{3})=\frac{2}{5}+\frac{1}{3}$

$\Leftrightarrow x=\frac{2}{5}$

Bài 2:

$\frac{1}{100}-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}$

$=\frac{99}{100}$

$\Rightarrow A=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}=\frac{-49}{50}$

Bài 1: 

a) Ta có: \(\left|x+\dfrac{7}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{7}{4}=\dfrac{1}{2}\\x+\dfrac{7}{4}=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)

b) Ta có: \(\left|2x+1\right|-\dfrac{2}{5}=\dfrac{1}{3}\)

\(\Leftrightarrow\left|2x+1\right|=\dfrac{11}{15}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{11}{15}\\2x+1=\dfrac{-11}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-4}{15}\\2x=\dfrac{-26}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{15}\\x=\dfrac{-13}{15}\end{matrix}\right.\)

c) Ta có: \(3x\left(x+\dfrac{2}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)

` a/`

` 2 - 1 5/6 + 2 2/3 = 2 - 11/6 - 8/3 = 1/6+ 8/3 = 1/6 + 16/6 = 17/6 `

`b/`

`5/9 xx ( 2 5/6 - 1 2/3 ) = 5/9 xx ( 17/6 - 5/3 ) = 5/9 xx 7/6 = 35/54 `

`c/`

` 1 1/3 : ( 2 + 1 1/6 : 2 5/6 ) `

`= 4/3 :  ( 2 + 7/6 : 17/6 ) `

`= 4/3 : ( 2 + 7/6 xx 6/17 )`

`= 4/3 : ( 2 + 7/17 ) `

`= 4/3 : ( 34/17 + 7/17 ) `

`= 4/3 : 41/17 `

`= 4/3 xx 17/41 `

`= 68/123`

` d/` 

` 2 3/5 : 3/4 xx 1 4/5 = 13/5 xx 4/3 xx 9/5 =52/15 xx 9/5 = 156/25`