a,\(25-\left(-75\right)+32-\left(32+75\right)\))

=\(25+75+32-32-75\)

=\(\left(75-75\right)+\left(32-32\right)+25\)

=\(0+0+25=25\)

\(b,\left|-127\right|-18.\left(5-6\right)\)

\(=127-18.\left(-1\right)\)

\(=127-\left(-18\right)\)

\(=145\)

\(c,17-25+55-17\)

\(=\left(17-17\right)+\left(55-25\right)\)

\(=0+30\)

\(=30\)

a)

25 - (-75) + 32 - ( 32 + 75 )

= 25 + 75 + 32 - 107

=100+32 - 107

=132 - 107

=25

b)

| - 127| - 18 . ( 5 - 6)

=127 - 18 . ( -1)

=127 - (-18)

=145

a)  (-25).(75-45)-75.(45-25)=(-25).30-75.20

\(a,\left(-25\right).\left(75-45\right)-75.\left(45-25\right)\)

\(=\left(-25\right).20-75.20\)

\(=20.\left(-25-75\right)\)

\(=20.\left(-100\right)\)

\(=-2000\)

\(b,\frac{18.12-48.15}{-3.270-3.30}\)

\(=\frac{18.12-12.4.15}{-3.270-3.30}\)

\(=\frac{12.\left(18-60\right)}{-3.\left(270+30\right)}\)

\(=\frac{12.\left(-42\right)}{-3.300}\)

\(=\frac{14}{25}\)

\(c,\frac{2^5.7+2^5}{2^5.5^2-2^5.3}\)

\(=\frac{2^5.\left(7+1\right)}{2^5\left(25-3\right)}\)

\(=\frac{2^5.8}{2^5.22}\)

\(=\frac{8}{22}\)

\(=\frac{4}{11}\)

a) \(\frac{53}{101}.\frac{-13}{97}+\frac{53}{101}.\frac{-84}{97}\)

\(=\frac{53}{101}\left(\frac{-13}{97}+\frac{-84}{97}\right)\)

\(=\frac{53}{101}.\frac{-97}{97}\)

\(=\frac{53}{101}.\left(-1\right)\)

\(=\frac{-53}{101}\)

b) \(\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)

\(=\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)

\(=\left(\frac{1}{57}-\frac{1}{5757}\right).0\)

\(=0\)

c) \(\frac{3^2}{25}.\frac{75}{-21}.\frac{50}{35}\)

\(=\frac{3^2.75.50}{25.\left(-21\right).35}\)

\(=\frac{3.3.25.3.5.5.2}{25.3.\left(-7\right).5.7}\)

\(=\frac{3.3.5.2}{\left(-7\right).7}\)

\(=\frac{90}{-49}\)

d) \(\frac{25.48-25.18}{20.5^3}\)

\(=\frac{25\left(48-18\right)}{10.2.125}\)

\(=\frac{25.10.3}{10.2.25.5}\)

\(=\frac{3}{10}\)

 \(\frac{53}{101}.\frac{-13}{97}+\frac{53}{101}.\frac{-84}{97}\)

\(=\frac{53}{101}\left(\frac{-13}{97}+\frac{-84}{97}\right)\)

\(=\frac{53}{101}.\left(-1\right)=\frac{-53}{101}\)

a/ A = 100² - 99² + 98² -...+2² - 1²

= (100² - 99²) + (98² - 97²) +...+ (2² - 1²)

= 199 + 195 + 191 + ... + 1

= (\(\frac{199-1}{4}+1\))(\(\frac{199+1}{2}\)) = 5050

b/ Y chang câu a luôn nha

c/ \(C=\frac{780^2-220^2}{125^2+150.125+75^2}=\frac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}\)

\(=\frac{560.1000}{200^2}=14\)

d/ 2(a + 4)(a + 5) - (a + 5)² - (a² - 9a - 17) 

= 17a + 32 = 17(a + 1) + 15 = 1700 + 15 = 1715

\(B=\left[1+\left(-3\right)\right]+\left[5+\left(-7\right)\right]+........+\left[97+\left(-99\right)\right]+101\)

\(=\left(-2\right)+\left(-2\right)+.......+\left(-2\right)+101\)( có 25 số -2)

\(=\left(-2\right).25+101\)

\(=\left(-50\right)+101\)

\(=51\)

\(B=1+\left(-3\right)+5+\left(-7\right)+...+97+\left(-99\right)+101\)

\(\Rightarrow B=\left[1+\left(-3\right)\right]+\left[5+\left(-7\right)\right]+...+\left[97+\left(-99\right)\right]+101\)

\(\Rightarrow B=\left(-2\right)+\left(-2\right)+...+\left(-2\right)+101\)

                     có 25 số -2

\(\Rightarrow B=\left(-2\right).25+101\)

\(\Rightarrow B=-50+101\)

\(\Rightarrow B=51\)

Trả lời:

\(B=1+\left(-3\right)+5+\left(-7\right)+...+97+\left(-99\right)+101\)(Có 51 số)

\(B=\left[1+\left(-3\right)\right]+\left[5+\left(-7\right)\right]+...+\left[97+\left(-99\right)\right]+101\)

\(B=\left(-2\right)+\left(-2\right)+...+\left(-2\right)+101\)( Có 25 số (-2) )

\(B=\left(-2\right)\times25+101\)

\(B=\left(-50\right)+101\)

\(B=51\)

Vậy\(B=51\)

Hok tốt!

Good girl

Ta có: 8. 25 = 200

a) (-8). 25 = -200

b) 8. (-25) = -200

c) (-8). (-25) = 200

Ta có: 8 . 25 = 200

=> a) (- 8) . 25 = - 200.

     b) 8 . (- 25) = - 200.

     c) (- 8) . (- 25) = 200.

Ta có: 8. 25 = 200

a) (-8). 25 = -200

b) 8. (-25) = -200

c) (-8). (-25) = 200