Question

[Lê Yến Nhi]

Bài 1: Đổi các biểu thức toán sau sang ngôn ngữ Pascal:

a) \(\dfrac{1}{b+2}\left(a^2+c\right)=5\) k)\(x\ge\dfrac{m+5}{2a}\)

b)\(k^2+\left(k+1\right)^2\ne\left(k+2\right)^2\) l) \(3,14R^2>a^2\)

c)\(8x-7>1\)

d)\(b^2-4ac\ge0\)

e)\(\dfrac{1}{n}.\dfrac{1}{n+1}.\dfrac{1}{n+2}< 0,01\)

f)\(\left(a-3\right)\left(a+5\right)=0\)

g)\(\dfrac{\left(a+c\right)h}{2}\ne1\)

h)\(2x+3\le25y\)

Answer 1

[Tú Nguyễn]

a) (1/(b+2))*(a*a+c)=5

b) (k*k)+((k+1)*(k+1)) < > (k+2)*(k+2)

c) 8*x-7>1

d) (b*b)-(4*a*c) >=0

e) (1/n)*(1/(n+i))*(1/(n+2)) <0,01

f) (a-3)*(a+5)=0

g) ((a+c)*h)/2 < > 1

h) 2*x+3 <= 25*y;

k) x >= (m+5)/(2*a);

l)3,14*(R*R) > a*a

Answer 2

[Lieu Loo]

a) ((1/(b+2))*(a*a+c)=5*k*x)>=(m+5)/2*a

b)(k*k)+((k+1)*(k+1)) <>((k+2)*(k+2)) .(dấu | ) mình không hiểu lắm )

c)8*x-7>1

d) (b*b)-(4*a*c)>=0

e) (1/n)*(1/(n+1))*(1/(n+2)<0,01

f) (a-3)*(a+5)=0

g)((a+c)*h)/2<>1

h)2*x+3<=25*y