Cho F = − 9 25 . 53 3 − − 3 5 2 . 22 3 và P = 1 − 1 2 . 1 − 1 3 . 1 − 1 4 ... 1 − 1 999 . 1 − 1 1000 . Khi đó tổng F + P bằng
A. − 624 125
B. 3133 1000
C. − 8999 1000
D. 314 100
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a) \(3.5^2+15.2^2-26\div2\)
= 3.25 + 15.4 - 13
= 75 + 60 - 13
= 135 - 13
= 122
b) \(5^3.2-100\div4+2^3.5\)
= 125.2 - 25 + 8.5
= 250 - 25 + 40
= 225 + 40
= 265
c)\(6^2\div9+50.2-3^3.33\)
= 36 : 9 + 100 - 9.33
= 4 + 100 - 297
= 104 - 297
= -193
d)\(3^2.5+2^3.10-81\div3\)
= 9.5 + 8.10 - 27
= 45 + 80 - 27
= 125 - 27
= 98
e) \(5^{13}\div5^{10}-25.2^2\)
= 53 - 25.4
= 125 - 100
= 25
f) \(20\div2^2+5^9\div5^8\)
= 20 : 4 + 5
= 5 + 5
= 10
Tính các tổng sau:
1, S=1-2+3_4+..+25-26
S =-1+3-5+7-...-53+55 ( có 28 số hạng )
= (-1+3)+(-5+7)+...+(-53+55) ( có 28:2=14 nhóm )
= 2+2+...+2
= 2 . 14
= 28
\(Q=\left(\frac{-9}{25}\right).\frac{53}{3}-\left(\frac{-3}{5}\right)^2.\frac{22}{3}\)
\(Q=\left(\frac{-9}{25}\right).\frac{53}{3}-\left(\frac{9}{25}\right).\frac{22}{3}\)
\(Q=\left(\frac{9}{25}\right).\frac{-53}{3}-\left(\frac{9}{25}\right).\frac{22}{3}\)
\(Q=\frac{9}{25}.\left(\frac{-53}{3}-\frac{22}{3}\right)\)
\(Q=\frac{9}{25}.\frac{-75}{3}\)
\(Q=\frac{9}{25}.\left(-25\right)\)
\(Q=\frac{-225}{25}=-9\)
Q= \(\left(\frac{-9}{25}\right).\frac{53}{3}-\left(\frac{-3}{5}\right)^2.\frac{22}{3}\)
Q=\(\left(\frac{-9}{25}\right).\frac{53}{3}-\frac{9}{25}.\frac{22}{3}\)
Q=\(\frac{9}{25}.\left(\frac{-53}{3}\right)-\frac{9}{25}.\frac{22}{3}\)
Q=\(\frac{9}{25}.\left(\frac{-53}{3}-\frac{22}{3}\right)\)
Q=\(\frac{9}{25}.\left(-25\right)\)=\(-9\)
1+2+3+4+5+6+7+8+9+10=55
11+12+13+14+15+16+17+18+19+20=155
1+2+3+4+5+6+7+8+9+10+11+12+13+14 +15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30-50-53=362
Đáp án cần chọn là: C
F = − 9 25 . 53 3 − − 3 5 2 . 22 3 = − 9 25 . 53 3 − 9 25 . 22 3 = − 9 25 . 25 = − 9
P = 1 − 1 2 . 1 − 1 3 . 1 − 1 4 ... ... 1 − 1 999 . 1 − 1 1000 = 1 2 . 2 3 . 3 4 ..... 998 999 . 999 1000 = 1.2.3.....999 2.3.4...999.1000 = 1 1000
F + P = 1 1000 + ( − 9 ) = − 8999 1000