\(\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{25}{8}+...+\dfrac{897}{128}\)
\(=\dfrac{3+9+25+...+897}{2+4+8+...+128}\)
\(=\dfrac{\left(897-3\right)\div6+1}{\left(128-2\right)\div2+1}\)
\(=\dfrac{150}{64}\)
\(=\dfrac{75}{32}\)
\(\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{25}{8}+...+\dfrac{897}{128}\)
\(=\dfrac{3+9+25+...+897}{2+4+8+...+128}\)
\(=\dfrac{\left(897-3\right)\div6+1}{\left(128-2\right)\div2+1}\)
\(=\dfrac{150}{64}\)
\(=\dfrac{75}{32}\)
3/2+9/4+25/8+...+897/128
3/2+9/4+25/8+...+897/128
Bài 2: Tính hợp lí
1) 125.(-24) + 24.225
2) 512.(2 – 128) – 128. (-512)
3) (187 -23) – (20 – 180)
4) (-37 – 17). (-9) + 35. (-9 – 11)
5) (-50 +19 +143) – (-79 + 25 + 48)
6) (-8).25.(-2). 4. (-5).125
7) 3784 + 23 – 3785 – 15
8) 215 +(-38) – (-58) –15
9) 5.(-3)2 –14.(-8)+(-40)
10) 215 + (-38) – (- 58) + 90 – 852)
Bài 3*: Tìm các cặp số nguyên (x;y) thỏa mãn xy2 + 2x – y2 =
cho P=1+5+9+13+...+25;Q=1+2+4+8+...+128.tính P.Q
cho P = 1 + 5 + 9 + 13 + .... + 25 ; Q = 1 + 2 + 4 + 8 + 16 + ...... + 128
P=1+5+9+........+25
Q=1+2+4+8+16+..........+128
(-25).8.(-4).(-125).3
27.(-17)+17.(-73)
512(2-128)-128(-512)
Cho P = 1+5+9+13+...+25;Q=1+2+4+8+16+...+128.Gí trị của P*Q là ?
tính hợp lý
a)5.(-3)^2-14.(-8)+(-40)
b)31-[26-(209+35)]
c)237+26-(209+26)
d)21+22+23+24-11-12-13-14
e)(-12).46-12.54
g)(-8).25.(-2).4.(-5).125
h)(-50+19+143)-(-79+25+48)
i)(-37-17).(-9)+35.(-9-11)
k)512.(2-128)-128.(-512)
l)160:{|-17|+[3^2.5-(14+2^11:2^8)]
m)|-6|+(-17)+|-4|+(-3)
n)5^6:5^3-5^0.5^1.5^2
o)100:{31-[24:(18-7.2)]}
k)