`@` `\text {Ans}`

`\downarrow`

`j)`

`-3/4 + 2/7 + (-1)/4 + 3/5 + 5/7`

`= (-3/4 - 1/4) + (2/7 + 5/7) + 3/5`

`= -1 + 1 + 3/5`

`= 3/5`

`k)`

`-2/17 + 15/23 + (-15)/17 + 4/19 + 8/23`

`= (-2/17 - 15/17) + (15/23 + 8/23) + 4/19`

`= -1 + 1 + 4/19`

`= 4/19`

$#KDN040510$

j: =-3/4-1/4+2/7+5/7+3/5

=-1+1+3/5

=3/5

k: =-2/17-15/17+15/23+8/23+4/19

=-1+1+4/19

=4/19

\(j)\dfrac{-3}{4}+\dfrac{2}{7}+\dfrac{-1}{4}+\dfrac{3}{5}+\dfrac{5}{7}\\ =\left(\dfrac{-3}{4}+\dfrac{-1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{3}{5}\\ =\dfrac{-4}{4}+\dfrac{7}{7}+\dfrac{3}{5}\\ =-1+1+\dfrac{3}{5}\\ =0+\dfrac{3}{5}\\ =\dfrac{3}{5}\\ k)\dfrac{-2}{17}+\dfrac{15}{23}+\dfrac{-15}{17}+\dfrac{4}{19}+\dfrac{8}{23}\\ =\left(\dfrac{-2}{17}+\dfrac{-15}{17}\right)+\left(\dfrac{15}{23}+\dfrac{8}{23}\right)+\dfrac{4}{19}\\ =\dfrac{-17}{17}+\dfrac{23}{23}+\dfrac{4}{19}\\ =-1+1+\dfrac{4}{9}\\ =0+\dfrac{4}{19}\\ =\dfrac{4}{19}\)

\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)

\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)

\(=\dfrac{-6}{5}.4:\dfrac{4}{5}\)

\(=\dfrac{-6.4.5}{5.4}=-6\)

= -4 đúng không

\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)

\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)

\(=\dfrac{-6}{5}.\dfrac{4}{3}:\dfrac{4}{5}\)

\(=\dfrac{-6.4.5}{5.3.4}=\dfrac{-6}{3}=-2\)

Vậy...

dễ mà

\(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)thay x = 1;x=3 vào ra kq thui

giúp mk vs mn ơi , mai cô giáo ktra mk r

a: \(=\dfrac{8}{9}\cdot\dfrac{9}{4}\cdot\dfrac{12}{19}\cdot\dfrac{19}{24}=\dfrac{1}{2}\cdot2=1\)

b: \(=\dfrac{5}{16}\cdot\dfrac{17}{15}\cdot\dfrac{8}{17}=\dfrac{5}{16}\cdot\dfrac{8}{15}=\dfrac{40}{240}=\dfrac{1}{6}\)

c: \(=\dfrac{4}{13}\left(\dfrac{2}{7}+\dfrac{5}{7}\right)-\dfrac{3}{26}=\dfrac{4}{13}-\dfrac{3}{26}=\dfrac{5}{26}\)

c: \(=\dfrac{3}{4}\left(\dfrac{6}{11}+\dfrac{5}{11}\right)-\dfrac{1}{5}=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{11}{20}\)

Câu 1:

\(=\dfrac{15}{34}+\dfrac{19}{34}-1-\dfrac{15}{17}+\dfrac{1}{3}+\dfrac{3}{5}\)

\(=-\dfrac{15}{17}+\dfrac{14}{15}=\dfrac{13}{255}\)

Câu 2: 

\(=\dfrac{5^4\cdot5^4\cdot2^8}{4^4\cdot6^4\cdot3^2\cdot5}=\dfrac{5^7}{6^4\cdot3^2}\)

Ta có:B=1\(\dfrac{6}{41}\)( \(\dfrac{12+\dfrac{12}{19}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2006}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2006}}\) )

B=\(\dfrac{47}{41}\) [\(\dfrac{12\left(1+\dfrac{1}{19}-\dfrac{1}{37}-\dfrac{1}{53}\right)}{3\left(1+\dfrac{1}{3}-\dfrac{1}{37}-\dfrac{1}{53}\right)}:\dfrac{4\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}\) B = \(\dfrac{47}{41}\) [ \(\dfrac{12}{3}:\dfrac{4}{5}\)]

B = \(\dfrac{47}{41}\)[ 4 . \(\dfrac{5}{4}\)]

B = \(\dfrac{47}{41}.5\)

B = \(\dfrac{235}{41}\)

Chúc bn hc tốt!!!

mk có thắc mắc là bạn để 3 ra ngoài sao 1/3 vẫn giữ nguyên vậy phải bằng 1/9 mới  đúng chứ'

\(C=\dfrac{4}{9}\times\dfrac{13}{17}+\dfrac{4}{17}\times\dfrac{4}{9}+\dfrac{2}{9}\\ =\dfrac{4}{9}\times\left(\dfrac{13}{17}+\dfrac{4}{17}\right)+\dfrac{2}{9}\\ =\dfrac{4}{9}\times1+\dfrac{2}{9}\\ =\dfrac{4}{9}+\dfrac{2}{9}\\ =\dfrac{6}{9}=\dfrac{2}{3}\)

\(D=\dfrac{8}{19}\times\dfrac{5}{11}+\dfrac{7}{11}\times\dfrac{8}{19}+\dfrac{12}{11}\times\dfrac{11}{19}\\ =\dfrac{8}{19}\times\left(\dfrac{5}{11}+\dfrac{7}{11}\right)+\dfrac{12}{11}\times\dfrac{11}{19}\\ =\dfrac{8}{19}\times\dfrac{12}{11}+\dfrac{12}{11}\times\dfrac{11}{19}\\ =\dfrac{12}{11}\times\left(\dfrac{8}{19}+\dfrac{11}{19}\right)\\ =\dfrac{12}{11}\times19\\ =\dfrac{12}{11}\)

\(C=\dfrac{4}{9}\cdot\dfrac{13}{17}+\dfrac{4}{17}\cdot\dfrac{4}{9}+\dfrac{2}{9}\)

\(C=\dfrac{4}{9}\cdot\left(\dfrac{13}{17}+\dfrac{4}{17}\right)+\dfrac{2}{9}\)

\(C=\dfrac{4}{9}\cdot\dfrac{13+4}{17}+\dfrac{2}{9}\)

\(C=\dfrac{4}{9}\cdot\dfrac{17}{17}+\dfrac{9}{2}\)

\(C=\dfrac{4}{9}\cdot1+\dfrac{2}{9}\)

\(C=\dfrac{4}{9}+\dfrac{2}{9}\)

\(C=\dfrac{4+2}{9}\)

\(C=\dfrac{6}{9}\)

\(C=\dfrac{2}{3}\)

\(D=\dfrac{8}{19}\cdot\dfrac{5}{11}+\dfrac{7}{11}\cdot\dfrac{8}{19}+\dfrac{12}{11}\cdot\dfrac{11}{19}\)

\(D=\dfrac{8}{19}\cdot\left(\dfrac{5}{11}+\dfrac{7}{11}\right)+\dfrac{12}{11}\cdot\dfrac{11}{19}\)

\(D=\dfrac{8}{19}\cdot\dfrac{12}{11}+\dfrac{12}{11}\cdot\dfrac{11}{19}\)

\(D=\dfrac{12}{11}\cdot\left(\dfrac{8}{19}+\dfrac{11}{19}\right)\)

\(D=\dfrac{12}{11}\cdot\dfrac{19}{19}\)

\(D=\dfrac{12}{11}\cdot1\)

\(D=\dfrac{12}{11}\)

ai giúp mik ik T_T

a, \(\dfrac{7}{8}\) \(\times\) \(\dfrac{3}{13}\) + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{4}{13}\)

= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{21}{8}\) + \(\dfrac{16}{9}\))

= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{189}{72}\) + \(\dfrac{128}{72}\))

= \(\dfrac{1}{13}\) \(\times\)  \(\dfrac{317}{73}\)

= \(\dfrac{317}{949}\)

b, \(\dfrac{6}{5}\) + \(\dfrac{7}{3}\) + \(\dfrac{8}{9}\)

=   \(\dfrac{54}{45}\) + \(\dfrac{105}{45}\) + \(\dfrac{40}{45}\)

= \(\dfrac{199}{45}\)

c, 23 : \(\dfrac{5}{14}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)

=   \(\dfrac{322}{5}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)

= \(\dfrac{20286}{315}\) + \(\dfrac{270}{315}\) + \(\dfrac{140}{315}\)

= \(\dfrac{20696}{315}\)

d, 4\(\dfrac{1}{4}\) + 7\(\dfrac{3}{7}\) - 2\(\dfrac{4}{17}\)

= 4 + \(\dfrac{1}{4}\) + 7 + \(\dfrac{3}{7}\) - 2 - \(\dfrac{4}{17}\)

= (4+7-2) + (\(\dfrac{1}{4}\) + \(\dfrac{3}{7}\) - \(\dfrac{4}{17}\))

= 9 + \(\dfrac{119}{476}\) + \(\dfrac{204}{476}\) - \(\dfrac{112}{476}\)

= 9\(\dfrac{211}{476}\) = \(\dfrac{4495}{476}\)

e, 8 - (9\(\dfrac{2}{11}\) + \(\dfrac{8}{33}\))

= 8 - 9 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)

=  -1 - \(\dfrac{2}{11}\)  - \(\dfrac{8}{33}\)

= \(\dfrac{-33}{33}\) - \(\dfrac{-6}{33}\) -  \(\dfrac{8}{33}\)

= - \(\dfrac{47}{33}\)

Muốn làm những bài toán có chứa hỗn số thì em cần nhớ:

Bước 1: Chuyển các hỗn số có trong phép tính thành phân số.

Bước 2: Thực hiện phép tính theo thứ tự thực hiện phép tính

a, \(\dfrac{8\dfrac{1}{2}}{15:\dfrac{5}{17}}\)

= \(\dfrac{\dfrac{17}{2}}{15\times\dfrac{17}{5}}\)

= \(\dfrac{\dfrac{17}{2}}{3\times17}\)

= \(\dfrac{17}{2}\) x \(\dfrac{1}{3\times17}\)

= \(\dfrac{1}{6}\)

b,\(\dfrac{4\dfrac{4}{5}:\dfrac{4}{17}}{3\dfrac{2}{5}}\)

= \(\dfrac{\dfrac{24}{5}:\dfrac{4}{17}}{\dfrac{17}{5}}\)

= \(\dfrac{\dfrac{24}{5}\times\dfrac{17}{4}}{\dfrac{17}{5}}\)

= \(\dfrac{6\times\dfrac{17}{5}}{\dfrac{17}{5}}\)

= 6

    \(\dfrac{\dfrac{28}{29}:\dfrac{4}{29}}{\dfrac{7}{9}:\dfrac{1}{9}}\)

=    \(\dfrac{\dfrac{28}{29}\times\dfrac{29}{4}}{\dfrac{7}{9}\times\dfrac{9}{1}}\)

= \(\dfrac{7}{7}\)

= 1