Tìm m biết : 23 * xm+2 = 512
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\(C=\dfrac{5122512}{2^2}-512\left(\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}}\right)\)
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;
\n\nvar x,n,i:integer;
\n\nlt:real;
\n\nbegin
\n\nclrscr;
\n\nwrite(\'nhap co so x:\'); readln(x);
\n\nwrite(\'nhap so mu n:\'); readln(n);
\n\nlt:=1;
\n\nfor i:=1 to n do
\n\nlt:=lt*x;
\n\nwriteln(x,\'^\',n,\'=\',lt:4:0);
\n\nreadln;
\n\nend.
\n1) 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
23 * xm+2 = 512
8 * xm+2 = 512
xm+2 = 512 : 8
xm+2 = 64
xm+2= 82
=> m+2 = 2
m = 2 -2
m = 0
vậy m =0