a; \(\dfrac{3}{11}\) + \(\dfrac{5}{-9}\) + \(\dfrac{4}{11}\) - \(\dfrac{4}{9}\) + \(\dfrac{3}{17}\) + \(\dfrac{15}{11}\)

= (\(\dfrac{3}{11}\) + \(\dfrac{4}{11}\) + \(\dfrac{15}{11}\)) - (\(\dfrac{5}{9}\) + \(\dfrac{4}{9}\)) + \(\dfrac{3}{17}\)

= 2 - 1 + \(\dfrac{3}{17}\)

= 1 + \(\dfrac{3}{17}\)

= \(\dfrac{20}{17}\) 

c; N = \(\dfrac{\dfrac{5}{7}-\dfrac{5}{9}-\dfrac{5}{11}}{\dfrac{15}{7}+\dfrac{15}{9}+\dfrac{15}{11}}\)

   Phải là - \(\dfrac{5}{7}\) chỗ tử số mới đúng em nhé!

d; \(\dfrac{4+\dfrac{4}{73}\times\dfrac{1}{23}}{5+\dfrac{5}{73}\times\dfrac{1}{23}}\)

= \(\dfrac{4\times\left(1+\dfrac{1}{73\times23}\right)}{5\times\left(1+\dfrac{1}{73\times23}\right)}\)

=  \(\dfrac{4}{5}\)

hong ai giúp mik dọ? ;-;

Bài 46:

11: Ta có: \(-4\left|x-2\right|=-8\)

\(\Leftrightarrow\left|x-2\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)

Vậy: x∈{0;4}

12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)

\(\Leftrightarrow5\left|x+2\right|=20\)

\(\Leftrightarrow\left|x+2\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

Vậy: x∈{-6;2}

13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)

\(\Leftrightarrow6\left|x-2\right|=-6\)(1)

Ta có: \(\left|x-2\right|\ge0\forall x\)

\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)

Ta có: -6<0(3)

Từ (1), (2) và (3) suy ra x∈∅

Vậy: x∈∅

14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)

\(\Leftrightarrow-7\left|x+4\right|=-7\)

\(\Leftrightarrow\left|x+4\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

Vậy: x∈{-5;-3}

15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)

\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)

\(\Leftrightarrow4\left|x+1\right|=24\)

\(\Leftrightarrow\left|x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)

Vậy: x∈{-7;5}

16: Ta có: \(3\left|x+5\right|=-9\)(4)

Ta có: |x+5|≥0∀x

⇒3|x+5|≥0∀x(5)

Ta có: -9<0(6)

Từ (4), (5) và (6) suy ra x∈∅

Vậy: x∈∅

17: Ta có: \(-8\left|x-3\right|=24-16:2\)

\(\Leftrightarrow-8\left|x-3\right|=16\)

\(\Leftrightarrow\left|x-3\right|=-2\)

mà |x-3|≥0>-2∀x

nên x∈∅

Vậy: x∈∅

18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)

\(\Leftrightarrow-3\left|x+6\right|=3\)

\(\Leftrightarrow\left|x+6\right|=-1\)

mà |x+6|≥0>-1∀x

nên x∈∅

Vậy: x∈∅

19: Ta có: \(5-\left|x+7\right|=4\)

\(\Leftrightarrow\left|x+7\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=-1\\x+7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-6\end{matrix}\right.\)

Vậy: x∈{-8;-6}

20: Ta có: \(12-\left|x+8\right|=10\)

\(\Leftrightarrow\left|x+8\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)

Vậy: x∈{-10;-6}

Mk chịu!

-_- mik ko bít

troi cac anh chi ko biet sao : 1 - (49/9 + x - 133/18 ) :63 /4 = 0

( 49/9 -133/18 ) :63/4 =1 ( boi vi 1 -1 = 0 ma cho nen ta coi phan nay = 1 )

49/9+ x - 133/18 = 63/4

49/9 + x = 63/4 + 133/18

49/9 + x = 833/36

x = 833/36 - 49/9

x =637/36

giúp mik zứi😭😭😭

các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha

a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)

              -\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)

              -6 < \(x\) < -4

             vì \(x\) \(\in\) Z nên \(x\) = -5

- \(\dfrac{19}{6}+\dfrac{-15}{2}+\dfrac{11}{3}\) < \(x\) < - \(\dfrac{5}{4}\) + \(\dfrac{19}{12}\) - \(\dfrac{10}{3}\)

- \(\dfrac{32}{3}+\dfrac{11}{3}< x< \) \(\dfrac{1}{3}\) - \(\dfrac{10}{3}\)

-7 < \(x\) < - 3

Vì \(x\in\) Z nên \(x\in\) {-6;-5;-4}

Giải chi tiết cho mình nha

\(\dfrac{2}{5}+\dfrac{11}{15}=\dfrac{6}{15}+\dfrac{11}{15}=\dfrac{17}{15}\)

\(\dfrac{7}{8}-\dfrac{7}{9}=\dfrac{63}{72}-\dfrac{56}{72}=\dfrac{7}{72}\)

\(\dfrac{11}{13}\times\dfrac{26}{31}=\dfrac{22}{31}\)

\(\dfrac{16}{24}:4=\dfrac{16}{24}\times\dfrac{1}{4}=\dfrac{4}{24}=\dfrac{1}{6}\)

\(30:\dfrac{6}{5}=30\times\dfrac{5}{6}=25\)

\(\dfrac{9}{16}:4=\dfrac{9}{16}\times\dfrac{1}{4}=\dfrac{9}{64}\)

\(\dfrac{1}{2}:\dfrac{1}{3}\times\dfrac{8}{15}=\dfrac{1}{2}\times3\times\dfrac{8}{15}=\dfrac{4}{5}\)

2/5 + 11/15 = 17/15
7/8 - 7/9 = 7/72
11/13 x 26/31 = 22/31
16/24 : 4 = 1/6
30 : 6/5 = 25
9/16 : 4 = 9/64
1/2 : 1/3 x 8/15 = 3/2 x 8/15 = 4/5

(8 x 7 - 16 - 5 x 8) x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

= [8 x 7 - 8 x 2 - 5 x 8] x [(1 + 9) + (2 + 8 ) + (3 + 7) + (4 + 6) + 5]

= [8 x (7 - 2 - 5)] x [10 + 10 + 10 + 10 + 10 + 5]

= (8 x 0) x 45

= 0

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

= (1 + 9) + (3 + 7) + (5 + 15) + (11 + 19) + (13 + 17)

= 10 + 10 + 20 + 30 + 30

= 100

a] 4/12 ; 5/12 ; 11/2 ; 1/4

b] 1 ; 9/12; 12/5 ; 11/3

c] 8/21; 6/11;8/7

d] 2/3 ;2/7 15/2

a]1/3  5/12     11/2       1/4

b]1      3/4         12/5         33/9

c]8/21      6/11        8/7

d]2/3        2/7          15/2

a) 1/3 ; 5/12 ; 11/2 ; 1/14

b) 1 ; 9/12 ; 12/5 ; 11/3

c) 8/21 ; 6/11 ; 8/7

d) 2/3 ; 2/7 ; 15/2