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DQ
3
HP
5 tháng 3 2016
B=23+43+63+....+2003
B=(1.2)3+(2.2)3+(2.3)3+....+(2.100)3
B=13.23+23.23+23.33+....+23.1003
B=23.(13+23+33+....+1003)
\(\Rightarrow\frac{B}{A}=\frac{2^3.\left(1^3+2^3+3^3+....+100^3\right)}{1^3+2^3+3^3+...+100^3}=2^3=8\)
Miu Ti làm vớ vẩn
5 tháng 3 2016
A=.......ghi lại cái đề
B=..............ghi lại cái đề=2.A
=> B/A=2
Theo mình là vậy nhưng ko bít đùng hay ko!
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
25 tháng 11 2019
\(\frac{A}{3}=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{100}{3^{101}}.\)
\(\frac{2A}{3}=A-\frac{A}{3}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{5^5}+...+\frac{1}{3^{100}}-\frac{100}{3^{101}}\)
Đặt \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(\frac{B}{3}=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{101}}\)
\(\frac{2B}{3}=B-\frac{B}{3}=\frac{1}{3}-\frac{1}{3^{101}}\Rightarrow B=\frac{3^{100}-1}{2}\)
\(\frac{2A}{3}=B-\frac{100}{3^{101}}=\frac{3^{100}-1}{2}-\frac{100}{3^{101}}=\frac{3^{201}-3^{101}-200}{2.3^{101}}\)
\(\Rightarrow A=\frac{3^{201}-3^{101}-200}{4.3^{100}}\)
27 tháng 9 2016
- \(\frac{4^6.3^4.9^5}{6^{12}}=\frac{\left(2^2\right)^6.3^4.\left(3^2\right)^5}{\left(2.3\right)^{12}}=\frac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}=\frac{2^{12}.3^{14}}{2^{12}.3^{12}}=3^2=9\)
- \(\frac{3^{10}.11+9^5.5}{3^9.2^4}=\frac{3^{10}.11+\left(3^2\right)^5.5}{3^9.16}=\frac{3^{10}.11+3^{10}.5}{3^9.16}=\frac{3^{10}.\left(11+5\right)}{3^9.16}=\frac{3^{10}.16}{3^9.16}=3\)
- 2100 - 299 - 298 - ... - 22 - 2
= 2100 - (299 + 298 + ... + 22 + 2)
Đặt A = 299 + 298 + ... + 22 + 2
2A = 2100 + 299 + ... + 23 + 22
2A - A = (2100 + 299 + ... + 23 + 22) - (299 + 298 + ... + 22 + 2)
A = 2100 - 2
Ta có:
2100 - 299 - 298 - ... - 22 - 2
= 2100 - (2100 - 2)
= 2100 - 2100 + 2
= 0 + 2
= 2
- 38 : 36 + (22)4 : 29
= 32 + 28 : 29
\(=9+\frac{1}{2}\)
\(=\frac{18}{2}+\frac{1}{2}=\frac{19}{2}\)
16 tháng 10 2022
a: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)
=>4^x=4^8
=>x=8
b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
=>2^x=2^11
=>x=11
c: =>1/6*6^x+6^x*36=6^15(1+6^3)
=>6^x=6*6^15
=>x=16
d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>x=9
11 tháng 9 2019
\(a,\frac{(-10)^5}{3\cdot(-6)^4}=\frac{(-2\cdot5)^5}{3\cdot(-2\cdot3)^4}=\frac{(-2)^5\cdot5^5}{3\cdot(-2)^4\cdot3^4}=\frac{(-2)^5\cdot5^5}{(-2)^4\cdot3^5}=-2\cdot\frac{5^5}{3^5}=\frac{-6250}{243}\)
\(b,\frac{2^{15}\cdot9^4}{6^6\cdot8^3}=\frac{\left[2^3\right]^5\cdot\left[3^2\right]^4}{\left[3\cdot2\right]^6\cdot\left[2^3\right]^3}=\frac{2^{15}\cdot3^8}{3^6\cdot2^6\cdot2^9}=\frac{2^{15}\cdot3^8}{3^6\cdot2^{15}}=\frac{3^8}{3^6}=3^2=9\)
\(c,\left[1+\frac{2}{3}-\frac{1}{4}\right]\cdot\left[\frac{4}{5}-\frac{3}{4}\right]^2\)
\(=\left[\frac{12}{12}+\frac{8}{12}-\frac{3}{12}\right]\cdot\left[\frac{16}{20}-\frac{15}{20}\right]^2\)
\(=\frac{17}{12}\cdot\left[\frac{1}{20}\right]^2=\frac{17}{12}\cdot\frac{1^2}{20^2}=\frac{17}{12}\cdot\frac{1}{400}=\frac{17}{4800}\)
\(d,2^3+3\cdot\left[\frac{1}{2}\right]^0+\left[(-2)^2:\frac{1}{2}\right]\)
\(=8+3\cdot\frac{1^0}{2^0}+\left[4:\frac{1}{2}\right]\)
\(=8+3\cdot1+8=8+3+8=19\)
AH
Akai Haruma
Giáo viên
19 tháng 3 2019
1.
\(-3x^5y^4+3x^2y^3-7x^2y^3+5x^5y^4\)
\(=(-3x^5y^4+5x^5y^4)+(3x^2y^3-7x^2y^3)\)
\(=2x^5y^4-4x^2y^3\)
2.
\(\frac{1}{2}x^4y-\frac{3}{2}x^3y^4+\frac{5}{3}x^4y-x^3y^4\)
\(=(\frac{1}{2}x^4y+\frac{5}{3}x^4y)-(\frac{3}{2}x^3y^4+x^3y^4)\)
\(=\frac{13}{6}x^4y-\frac{5}{2}x^3y^4\)
3.
\(5x-7xy^2+3x-\frac{1}{2}xy^2\)
\(=(5x+3x)-(7xy^2+\frac{1}{2}xy^2)\)
\(=8x-\frac{15}{2}xy^2\)
AH
Akai Haruma
Giáo viên
19 tháng 3 2019
4.
\(\frac{-1}{5}x^4y^3+\frac{3}{4}x^2y-\frac{1}{2}x^2y+x^4y^3\)
\(=(\frac{-1}{5}x^4y^3+x^4y^3)+(\frac{3}{4}x^2y-\frac{1}{2}x^2y)\)
\(=\frac{4}{5}x^4y^3+\frac{1}{4}x^2y\)
5.
\(\frac{7}{4}x^5y^7-\frac{3}{2}x^2y^6+\frac{1}{5}x^5y^7+\frac{2}{3}x^2y^6\)
\(=(\frac{7}{4}x^5y^7+\frac{1}{5}x^5y^7)+(-\frac{3}{2}x^2y^6+\frac{2}{3}x^2y^6)\)
\(=\frac{39}{20}x^5y^7-\frac{5}{6}x^2y^6\)
6.
\(\frac{1}{3}x^2y^5(-\frac{3}{5}x^3y)+x^5y^6=(\frac{1}{3}.\frac{-3}{5})(x^2.x^3)(y^5.y)+x^5y^6\)
\(=\frac{-1}{5}x^5y^6+x^5y^6=\frac{4}{5}x^5y^6\)
\(A=1+3^2+3^4+...+3^{102}\)
\(9A=3^2+3^4+...+3^{102}+3^{104}\)
\(\Rightarrow9A-A=3^{104}-1\)
\(\Rightarrow8A=3^{104}-1\)
\(\Rightarrow A=\dfrac{3^{104}-1}{8}\)