a)\(\frac{1}{8}.\frac{3}{5} = \frac{{1.3}}{{8.5}} = \frac{3}{{40}}\)           

b)\(\frac{{ - 6}}{7}:\left( { - \frac{5}{3}} \right) = \frac{{ - 6}}{7}.\frac{{ - 3}}{5} = \frac{{18}}{{35}}\)           

c)\(0,6.\left( { - 0,15} \right) = \frac{6}{{10}}.\frac{{ - 15}}{{100}} = \frac{{ - 90}}{{1000}} = \frac{{ - 9}}{{100}}\).

1. a) Ta có BCNN(12, 15) = 60 nên ta lấy mẫu chung của hai phân số là 60. 

Thừa số phụ:

60:12 =5; 60:15=4

Ta được:

\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\)

\(\frac{7}{{15}} = \frac{{7.4}}{{15.4}} = \frac{{28}}{{60}}\)

 b) Ta có BCNN(7, 9, 12) = 252 nên ta lấy mẫu chung của ba phân số là 252. 

Thừa số phụ:

252:7 = 36; 252:9 = 28; 252:12 = 21

Ta được:

\(\frac{2}{7} = \frac{{2.36}}{{7.36}} = \frac{{72}}{{252}}\)

\(\frac{4}{9} = \frac{{4.28}}{{9.28}} = \frac{{112}}{{252}}\)

\(\frac{7}{{12}} = \frac{{7.21}}{{12.21}} = \frac{{147}}{{252}}\)

2. a) Ta có BCNN(8, 24) = 24 nên:

\(\frac{3}{8} + \frac{5}{{24}} = \frac{{3.3}}{{8.3}} + \frac{5}{{24}} = \frac{9}{{24}} + \frac{5}{{24}} = \frac{{14}}{{24}} = \frac{7}{{12}}\)

 b) Ta có BCNN(12, 16) = 48 nên:

\(\frac{7}{{16}} - \frac{5}{{12}} = \frac{{7.3}}{{16.3}} - \frac{{5.4}}{{12.4}} = \frac{{21}}{{48}} - \frac{{20}}{{48}} = \frac{1}{{48}}\).

a)-2

b)-7

c)12

e)4

c)-15

g)8

a) -2
b) -7
c) 12
e) 4
c) -15
g) 8

a) 6 – 8 = -2

d) 0 – 7; = -7

b) 3 – (-9); = 12

e) 4 – 0; = 4

c) (-5) – 10; = -15

g) (-2) – (-10). = 8

a)

b) (x – 1)(x + 1)(x² + 1)

= [x .(x + 1) – 1 .(x + 1)] . (x² + 1)

= {x.x + x.1 + (-1).x + (-1).1}. (x² + 1)

= (x² + x – x – 1) . (x² + 1)

= (x² – 1) . (x² + 1)

= x² . (x² +1) – 1.(x² + 1)

= x² . x² + x² . 1 – (1.x² + 1.1)

= x⁴ + x² – (x² + 1)

= x⁴ + x² – x² – 1

= x⁴ – 1

a,5/9+(-3/5)  +( -5/9)

=5/9+(-5/9)+(-3/5)

=0+(-3/5)

=-3/5

b, b,-3/7. 2/11- (-3/7).9/11

=-3/7.(2/11-9/11)

=-3/7.(-7/11)

=3/11

c,,2/3 + 5/6 : 5 - 1/18 . (-3)

=2/3+(1/6-(-1/6))

=2/3+1/3

=3/3

=1

tick nếu đúng nha

Dễ mà, dùng máy tính là xong

a,5/9+(-3/5)  +( -5/9)

=5/9+(-5/9)+(-3/5)

=0+(-3/5)

=-3/5

b, b,-3/7. 2/11- (-3/7).9/11

=-3/7.(2/11-9/11)

=-3/7.(-7/11)

=3/11

c,,2/3 + 5/6 : 5 - 1/18 . (-3)

=2/3+(1/6-(-1/6))

=2/3+1/3

=3/3

=1

a)\(\dfrac{19}{48}-\dfrac{3}{40}=\dfrac{95}{240}-\dfrac{18}{240}=\dfrac{77}{240}\)

b)\(\dfrac{1}{6}+\dfrac{7}{27}+\dfrac{5}{18}=\dfrac{9}{54}+\dfrac{14}{54}+\dfrac{15}{54}=\dfrac{38}{54}=\dfrac{19}{27}\)

a) 19/48 - 3/40= 77/240

b) 1/6 + 7/27 + 5/18= 23/54 + 5/18= 19/27

a)\(\dfrac{19}{48}\) - \(\dfrac{3}{40}\)=\(\dfrac{77}{240}\)

b) \(\dfrac{1}{6}\)+\(\dfrac{7}{27}\)+\(\dfrac{5}{18}\)=\(\dfrac{23}{54}\)+\(\dfrac{5}{18}\)=\(\dfrac{19}{27}\)

a) \(6 - 8 = 6 + \left( { - 8} \right) =  - \left( {8 - 6} \right) =  - 2\)

b) \(3 - \left( { - 9} \right) = 3 + 9 = 12\)

c) \(\left( { - 5} \right) - 10 = \left( { - 5} \right) + \left( { - 10} \right)\)\( =  - \left( {5 + 10} \right) =  - 15\)

d) \(0 - 7 = 0 + \left( { - 7} \right) =  - 7\)

e) \(4 - 0 = 4 + 0 = 4\) (vì số đối của 0 là 0)

g) \(\left( { - 2} \right) - \left( { - 10} \right) = \left( { - 2} \right) + 10\)\( = 10 - 2 = 8\).

a)\(\frac{{ - 2}}{5} + \frac{3}{7} = \frac{{ - 14}}{{35}} + \frac{{15}}{{35}} = \frac{1}{{35}}\)           

b)\(0,123 - 0,234 =  - \left( {0,234 - 0,123} \right) =  - 0,111.\)

a/ \(\left(2x+3\right)\left(x-5\right)-\left(x-1\right)^2+x\left(7-x\right)\)

\(=2x^2-2x-15-x^2+2x-1+7x-x^2\)

\(=7x-16\)

b, = x² - 16 - ( x³ - 3³ ) : ( x - 3 )

= x² - 16 - \([\) ( x - 3 ) ( x² + 3x + 9 ) \(]\) : ( x - 3 )

= x² - 16 - ( x² + 3x + 9 )

= x² - 16 - x² - 3x - 9

= -25 - 3x

\(a,=2x^2-10x+3x-15-x^2+2x-1+7x-x^2=2x-16\\ b,=x^2-16-\left(x-3\right)\left(x^2+3x+9\right):\left(x-3\right)\\ =x^2-16-x^2-3x-9=-3x-25\)

1.

\(\sqrt{50}-3\sqrt{8}+\sqrt{32}=5\sqrt{2}-6\sqrt{2}+4\sqrt{2}=3\sqrt{2}\)

2. 

a, ĐK: \(x\in R\)

\(pt\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\)

\(\Leftrightarrow\left|x-2\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

b, ĐK: \(x\ge3\)

\(pt\Leftrightarrow\sqrt{x-3}\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(l\right)\end{matrix}\right.\)