1)-73

2)-5

3)-64

4)-56

5)9

6)-20

7)6

8)-2

Tick nha

mk ra kết quả thôi nha

a,6

b,-200

c,1900

d,-900000

k mình nha 

mình cần gấp

 16 . ( 38 - 2 ) - 38 . ( 16 - 1 )

=16 . 38 - 2 .16 - 38 .16 - 1.38

=16.38.(2.16-1.38)

=608.(32-38)

=608.(-6)

=-3648

\(a,\left(-5\right)+11+\left(-15\right)+21+\left(-25\right)+31+...+\left(-95\right)+101\\ =\left[\left(-5\right)+11\right]+\left[\left(-15\right)+21\right]+\left[\left(-25\right)+31\right]+...+\left[\left(-95\right)+101\right]\\ =6+6+6+...+6\left(10\text{ số }6\right)\\ =6\cdot10\\ =60\)

\(b,3+\left(-12\right)+13+\left(-22\right)+23+\left(-32\right)+...+93+\left(-102\right)\\ =\left[3+\left(-12\right)\right]+\left[13+\left(-22\right)\right]+\left[23+\left(-32\right)\right]+...+\left[93+\left(-102\right)\right]\\ =\left(-9\right)+\left(-9\right)+\left(-9\right)+...+\left(-9\right)\left(10\text{ số }-9\right)\\ =\left(-9\right)\cdot10\\ =-90\)

\(a,\left(-5\right)+11+\left(-15\right)+21+\left(-25\right)+31+...+\left(-95\right)+101\)

\(=\left[\left(-5\right)+11\right]+\left[\left(-15\right)+21\right]+\left[\left(-25\right)+31\right]+...+\left[\left(-95\right)+101\right]\)

\(=6+6+6+...+6\) (10 số 6)

\(=6.10=60\)

\(\)

Ta có: \(\frac{4}{3\cdot5}+\frac{8}{5\cdot9}+\frac{12}{9\cdot15}+\cdots+\frac{32}{n\left(n+16\right)}=\frac{16}{25}\)

=>\(\frac{2}{3\cdot5}+\frac{4}{5\cdot9}+\frac{6}{9\cdot15}+\cdots+\frac{16}{n\left(n+16\right)}=\frac{8}{25}\)

=>\(\frac13-\frac15+\frac15-\frac19+\cdots+\frac{1}{n}-\frac{1}{n+16}=\frac{8}{25}\)

=>\(\frac13-\frac{1}{n+16}=\frac{8}{25}\)

=>\(\frac{1}{n+16}=\frac13-\frac{8}{25}=\frac{25-24}{75}=\frac{1}{75}\)

=>n+16=75

=>n=75-16=59

5: \(16=2^4;32=2^5;112=2^4\cdot7\)

=>ƯCLN(16;32;112)=\(2^4=16\) ; BCNN(16;32;112)=\(2^5\cdot7=32\cdot7=224\)

6: \(25=5^2;55=5\cdot11;75=5^2\cdot3\)

=>ƯCLN(25;55;75)=5; BCNN(25;55;75)=\(5^2\cdot3\cdot11=33\cdot25=825\)

7: \(150=2\cdot3\cdot5^2;84=2^2\cdot3\cdot7;30=2\cdot3\cdot5\)

=>ƯCLN(150;84;30)=\(2\cdot3=6\) ; BCNN(150;84;30)=\(2^2\cdot3\cdot5^2\cdot7=10^2\cdot3\cdot7=2100\)

8: \(12=2^2\cdot3;15=3\cdot5;10=2\cdot5\)

=>ƯCLN(12;15;10)=1; BCNN(12;15;10)=\(2^2\cdot3\cdot5=4\cdot3\cdot5=60\)