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\(A=1^2+2^2+3^2+....+10^2\\ A=1^{ }+\left(1+1\right)\cdot2+3\cdot\left(2+1\right)+.....+10\cdot\left(9+1\right)\\ A=1+2\cdot1+2+3\cdot2+3+....+10\cdot9+10\\ A=\left(1+2+3...+10\right)+\left(1\cdot2+3\cdot2+.....+10\cdot9\right)\)
Gọi 1+2+3+...+10 là P
Số số hạng là: (10 - 1) : 1 +1 = 10 (số)
P = (10+1) . 10 : 2 = 55
P = 55
Gọi \(1\cdot2+2\cdot3+....+9\cdot10\) là C
\(C=1\cdot2+2\cdot3+....+9\cdot10\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot3+....+9\cdot10\cdot3\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+....+9\cdot10\cdot\left(11-8\right)\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+.....+9\cdot10\cdot11-8\cdot9\cdot10\\ 3\cdot C=9\cdot10\cdot11\\ 3\cdot C=990\\ C=330\)
\(=>A=P+C\\ =>A=55+330\\ A=385\)
b)
\(B=5^2+10^2+15^2+...+50^2\\ B=5^2+\left(2\cdot5\right)^2+\left(3\cdot5\right)^2+....+\left(5\cdot10\right)^2\\ B=5^2+2^2\cdot5^2+3^2\cdot5^2+...+5^2\cdot10^2\\ B=5^2\cdot\left(1+2^2+3^2+....+10^2\right)\\ B=25\cdot\left(1+2^2+3^2+....+10^2\right)\)
\(\left(1+2^2+3^2+....+10^2\right)=A\)
\(=>B=25\cdot A\\ B=25\cdot385\\ B=9625\)
a)
\( 49.(51 - 4) - 51.(49 + 4)\)
\(=49.51-49.4-51.49-51.4\)
\(=49.(51-4-51)-51.4\)
\(=49.(-4)-51.4\)
\(=4.(-49-51)\)
\(=4.(-100)\)
\(=-400\)
b)
\( 71.64 + 32.(-7) - 13.32\)
\(=71.32.2+32.(-7)-13.32\)
\(=142.32+32.(-7)-13.32\)
\(=32.(142-7-13)\)
\(=32.122\)
\(=3904\)
c)
\( 11 + (-13) + 15 + (-17) + ... + 59 + (-61)\)
\(=(-2)+(-2)+...+(-2)\)
\(=(-2).13\)
\(=-26\)
a. 49.51- 49.4 -51.49 -51.4
= -49.4 - 51.4
=4.(-49-51)
=4.-100
=-400
b.71.64 + 32.(-7) -13.32
= 71.64 + 32.(-7-13)
=71.64 + 32.-20
=71.32.2 +32.-20
=32.(71x2-20)
=32. 122
= 3904
c. (11-13)+(15-17)+..+(59-61)
=-2 x 13
=-26
\(=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}\)
\(=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(=35\frac{0}{23}-5\frac{7}{32}\)
\(=\frac{805}{23}-\frac{167}{32}\)
\(\frac{25760}{736}-\frac{3841}{736}\)
\(=\frac{21919}{736}=\frac{953}{32}\)
\(P=\dfrac{32}{64}\cdot\dfrac{-57}{19}+\dfrac{35}{21}\cdot\dfrac{22}{44}=\dfrac{1}{2}\left(-3+\dfrac{5}{3}\right)=\dfrac{1}{2}\cdot\dfrac{-4}{3}=\dfrac{-2}{3}\)
\(Q=\dfrac{75}{125}\cdot\dfrac{82}{164}+\dfrac{49}{98}\cdot\dfrac{-35}{105}=\dfrac{1}{2}\left(\dfrac{3}{5}-\dfrac{1}{3}\right)=\dfrac{1}{2}\cdot\dfrac{4}{15}=\dfrac{2}{15}\)
bài 1 : Tính
2048 : 23 = 2048 : 8 = 256
bài 1.1 : Tính
625 : 52 . 40 = 625 : 25 . 40 = 25 . 40 = 1000
\(=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}\)
\(=49\frac{8}{23}-14\frac{8}{23}-5\frac{7}{12}\)
\(=35-5\frac{7}{32}\)
\(=35-\frac{167}{32}\)
\(=\frac{1120}{32}-\frac{167}{32}\)
\(=\frac{935}{32}\)
\(49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}\)
\(=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(=35-5\frac{7}{32}\)
\(=35-\frac{167}{32}\)
\(=\frac{953}{32}\)
=32x-49+32x102-102x49+102x32
=32(-49+102)+102(32-49)
=32x53+102x-17=1696+-1734=-38
CHÚC BẠN HỌC GIỎI!!!!!!!!