1. Phn bit ma trn v mng?

2. To mt vect vi cc phn t l tt c cc s chn t 31 n 753. Trnh by cch to vect x vi cc phn t sau ...:1, 1/8, 1/27, 1/64, 1/125.4. Cho t t 1 n 100 (100 phn t), vit biu thc tnh:ln(2 + t + sin(t)t2)

et(1 + tcos(t2))

5. Vit c php v trn cng mt th cc hm: x, x2, sin(x) trong khong 0:100. Lm sao thy c ng sin(x).6. Hy cho bit kt qu (gii thch) ca tng dng lnh sau:A = [1:3;4:6]

B = [A A;A A]

C = B(1:2,3:4)D=[A ; A(end,:)] (A(:))7. A = [ 2 4 1 1 ; 6 7 2 2 ; 3 5 9 3], vit cu lnh :To B l ma trn ly ct 2, 3 ca A

To C l ma trn ly ct 1, 4 ca A

C^D

Tnh sai s chun ( lch chun trung bnh) mi ct ca A

8. Cho x = 1:10 v y = [3 1 5 6 8 2 9 4 7 0], gii thch cc cu lnh sau:x(y > 5)

x( (x < 2) | (y >= 8) )

gii thch s khc bit gia: y(rem(x,2)) vs. y(logical(rem(x,2)))

9. Cho x = [3 15 9 12 -1 0 -12 9 6 1], vit cc cu lnh

a. Chuyn cc phn t dng thnh zero

b. Nhn cc phn t no l s l vi 5

10. Hy cho bit kt qu v gii thch khi chy on chng trnh sau:a = [1 2 3 4; 4 5 6 7; 7 8 9 10];

m=size(a,2);

for i = 1:m

disp(a(:,i));

end11. Vit on chng trnh tnh tng ca n s t nhin, vi n c nhp t bn phm12. Vit mt hm tinhtong.m c:Nhn vo gi tr n

Tr v gi tr tng cc tch 2 s lin tip t 1 n n

1*2 + 2*3 + 3*4 + .. + (n-1)*n13. Khi s dng cc vng lp lm sao bit vng lp c b lp v hn hay khng v nu v hn th dng li bng cch no?

14. Nu cch gii h phng trnh tuyn tnh [A]X=[B]15. Hy chy on code sau v gii thch ngha ca n, nu b lnh continue v break th s nh th no, ti sao?:

for m=3:1:7

for n=2:1:m-1

if mod(m, n) ~=0

continue;

end

fprintf('%2d khng l mt s nguyn t !\n',m)

break;

end

if n==m-1

fprintf('%2d l mt s nguyn t !\n',m)

end

end