\(C=\dfrac{5122512}{2^2}-512\left(\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}}\right)\)

Đặt BT trong ngoặc đơn là B

\(\Rightarrow2B=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\)

\(B=2B-B=\dfrac{1}{2^2}-\dfrac{1}{2^{10}}\)

\(\Rightarrow C=\dfrac{5120512+2000}{2^2}-512\left(\dfrac{1}{2^2}-\dfrac{1}{2^{10}}\right)=\)

\(=\dfrac{512.10001+2^2.500}{2^2}-512\left(\dfrac{1}{2^2}-\dfrac{1}{2^{10}}\right)=\)

\(=\dfrac{2^9.10001+2^2.500}{2^2}-2^9\left(\dfrac{1}{2^2}-\dfrac{1}{2^{10}}\right)=\)

\(=2^7.10001+500-2^7+\dfrac{1}{2}=\)

\(=2^7.10000+500+0,5=1280000+500+0,5=1280500,5\)

uses crt;

var x,n,i:integer;

lt:real;

begin

clrscr;

write(\'nhap co so x:\'); readln(x);

write(\'nhap so mu n:\'); readln(n);

lt:=1;

for i:=1 to n do

lt:=lt*x;

writeln(x,\'^\',n,\'=\',lt:4:0);

readln;

end.

1) 125.(-24) + 24.225.

\(=24\left(-125+225\right)=24.100=2400.\)

2) 512.(2 – 128) – 128. (-512).

\(=512\left(2-128+128\right)=512.2=1024.\)

(-50+19+143)-(-79+25+48)

= 112 - -6

= 118

(-25)(75-45)-75(45-25)

=(-25) 30 - 75 20

= -750 - 1500

= -2250

(-37-17)(-9)+35(-9-11)

= 54 (-9)+ 35-20

= 486 + -700

= -214

(187-23)-(20-180)

= 164 - -160

= 324

512(2-128)-128(-512)

=512 (-126)-128(-512)

= 64512 - (-65536)

= 1024

17(-25)+25 21

= -425 + 525

= 100.

a)

(x - 3) : 2 =  5 14 : 5 12

(x - 3) : 2 = 5 2

(x - 3) : 2 = 25

(x - 3) = 25.2

x = 50 + 3

x = 53