a) \(\dfrac{7}{9}\times\dfrac{13}{25}\times\dfrac{9}{7}=\left(\dfrac{7}{9}\times\dfrac{9}{7}\right)\times\dfrac{13}{25}=1\times\dfrac{13}{25}=\dfrac{13}{25}\)

b) \(\dfrac{5}{9}\times\dfrac{21}{25}+\dfrac{21}{25}\times\dfrac{4}{9}\)

\(=\dfrac{21}{25}\times\left(\dfrac{5}{9}+\dfrac{4}{9}\right)\)

\(=\dfrac{21}{25}\times1=\dfrac{21}{25}\)

c) \(\dfrac{19}{5}\times\dfrac{11}{16}-\dfrac{9}{5}\times\dfrac{11}{16}\)

\(=\dfrac{11}{16}\times\left(\dfrac{19}{5}-\dfrac{9}{5}\right)\)

\(=\dfrac{11}{16}\times2=\dfrac{22}{16}=\dfrac{11}{8}\)

tách ra nha

Bạn ơi bạn làm đc bao nhiêu thì làm phụ mình nhé 

bài1

a, 

5 - 7 + 3 +(-8)

= -2 + 3 + ( - 8)

=     1    + (-8)= -7

a; - (-5) - (+7) + (+3) + (-8)

  = 5 - 7 + 3  - 8

= (5 + 3 - 8) - 7

= (8 - 8) - 7

= 0 - 7

= - 7

3) \(-12+2x-9+x=0\\ -21+3x=0\\ 3x=21\\ x=7\)

4)
\(11+\left(15-x\right)=1\)
\(15-x=1-11\)
\(15-x=-10\)
\(x=15-\left(-10\right)\)
\(x=25\)

5) 
\(4-\left(27-3\right)=x-\left(13-4\right)\)
\(4-24=x-9\)
\(x-9=-20\)
\(x=-20+9\)
\(x=-11\)
 

\(3.-12+\left(2x-9\right)+x=0.\)

\(\Leftrightarrow-12+2x-9+x=0.\Leftrightarrow3x=21.\Leftrightarrow x=7.\)

Vậy \(x=7.\)

\(4.11+\left(15-x\right)=1.\Leftrightarrow11+15-x=1.\Leftrightarrow26-x=1.\Leftrightarrow x=25.\)

Vậy \(x=25.\)

\(5.4-\left(27-3\right)=x-\left(13-4\right).\Leftrightarrow4-24=x-9.\Leftrightarrow-20=x-9.\Leftrightarrow x=-11.\)

Vậy \(x=-11.\)

\(6.8-\left(x-10\right)=23-\left(-4+12\right).\Leftrightarrow8-x+10=23-8.\Leftrightarrow18-x=15.\Leftrightarrow x=3.\)

Vậy \(x=3.\)

\(7.105-5\left(10-5x\right)=-20.\Leftrightarrow105-50+25x=-20.\Leftrightarrow25x=-75.\Leftrightarrow x=-3.\)

Vậy \(x=-3.\)

\(8.\left(x-1\right)\left(8-2x\right)\left(3x+123\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\8-2x=0.\\3x+123=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=4.\\x=-41.\end{matrix}\right.\)

Vậy \(x\in\left\{1;4;-41\right\}.\)

\(9.\left(x^2-25\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0.\\x+5=0.\\x+10=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\\x=-10.\end{matrix}\right.\)

Vậy \(x\in\left\{5;-5;-10\right\}.\)

\(10.x\left(x^2+5\right)=0.\Leftrightarrow x=0.\)

a)\(=\dfrac{3}{3}+\dfrac{4}{3}=\dfrac{7}{3}\)

b)\(=\dfrac{5}{9}\times\dfrac{3}{2}=\dfrac{15}{18}=\dfrac{5}{6}\)

d)\(=\left(\dfrac{12}{8}-\dfrac{3}{8}\right)\times2=\dfrac{9}{8}\times2=\dfrac{18}{8}=\dfrac{9}{4}\)

c)\(=\dfrac{4}{3}-\dfrac{5}{6}=\dfrac{8}{6}-\dfrac{5}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)

a) 1 + 4/3 = 7/3

b) 5/9 : 2/3 = 5/6

c ) 4/3 -1/3 x 5/2

= 1 x 5/2

= 5/2

d) ( 3/2 - 3/8) : 1/2

= 9/8 : 1/2

= 9/4

e) 15/16 : 3/8 x 3/4 

= 5/2 x 3/4

= 15/8

f) 7/19 x 1/3 x 7/19 x 2/3

= 7/19 x (1/3 x 2/3)

= 7/19 x 2/9

= 14/171

g) 3/5 x 8/27 x 25/3

= 3/5 x 25/3 x 8/27 

= 5 x 8/27

= 40/27

h) 1/5 + 4/11 + 4/5 + 7/11

= (1/5 + 4/5) + (4/11 + 7/11)

= 1 + 1

= 2

e)\(\dfrac{15}{16}\times\dfrac{8}{3}\times\dfrac{3}{4}=\dfrac{360}{192}=\dfrac{15}{8}\)

f) \(=\dfrac{7}{19}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)=\dfrac{7}{19}\times\dfrac{3}{3}=\dfrac{7}{19}\)

g)\(=\dfrac{600}{120}=\dfrac{600:120}{120:120}=\dfrac{5}{1}=5\)

h)\(=\left(\dfrac{1}{5}+\dfrac{4}{5}\right)+\left(\dfrac{4}{11}+\dfrac{7}{11}\right)=\dfrac{5}{5}+\dfrac{11}{11}=1+1=2\)

tách ra

\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)

\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)

1)10-x=5-7

x=10-(-2)

x=12

2)x+5=0+32

x=32-5

x=27

3) -12+2x-9+x=0

-12+(2x+x-9)=0

3x-9=0+12

3x=12+9

x=21:3

x=7

4) 15-x=1-11

x=15-(-10)

x=25

5) 4-(27+3)=x-9

4-30=x-9

-26+9=x

x=-17