Tính hợp lý 3 2 . 1 243 . 81 2 . 1 3 2
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Đặt A= 1/3+1/9+1/27+1/81+1/243
A= 1/3+1/3^2+1/3^3+1/3^4+1/3^5
3A=1+1/3+1/3^2+1/3^3+1/3^4
3A-A=1+1/3+1/3^2+1/3^3+1/3^4-1/3-1/3^2-1/3^3-1/3^4-1/3^5
2A=1-1/3^5
2A=242/243
A=121/243
\(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ =\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}\\ =>3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}\\ =>3A-A=2A=1-\dfrac{1}{3^5}\\ =>A=\dfrac{1-\dfrac{1}{3^5}}{2}=\dfrac{3^5-1}{2.3^5}\)
\(1-\frac{2}{3}-\frac{2}{9}-\frac{2}{27}-\frac{2}{81}-\frac{2}{243}\)
\(=\frac{243}{243}-\frac{162}{243}-\frac{54}{243}-\frac{6}{243}-\frac{2}{243}=\frac{1}{243}\)
\(x\) \(\times\) \(\dfrac{1}{4}\) = 6 : 1 : 2
\(x\) \(\times\) \(\dfrac{1}{4}\) = 6:2
\(x\) \(\times\) \(\dfrac{1}{4}\) = 3
\(x\) = 3 : \(\dfrac{1}{4}\)
\(x\) = 12
Q = (1 + 2 - 3 - 4) + (5 + 6 - 7- 8) + ... + (77 + 78 - 79 - 80) - 81
Q = (-4) . 20 - 81
Q = -161
Nhân 20 vì có 80 số hạng, mỗi cặp 4 số hạng nên có 20 cặp.
`@` `\text {Ans}`
`\downarrow`
\(3^2\cdot\dfrac{1}{243}\cdot81^2\cdot\dfrac{1}{3^3}\)
`=`\(\dfrac{3^2}{243}\cdot\dfrac{81^2}{3^3}\)
`=`\(\dfrac{3^2}{3^5}\cdot\dfrac{3^8}{3^3}=\dfrac{1}{3^3}\cdot3^5=\dfrac{3^5}{3^3}=3^2=9\)
\(3^2\cdot\dfrac{1}{243}\cdot81^2\cdot\dfrac{1}{3^2}\)
\(=3^2\cdot\dfrac{1}{3^5}\cdot\left(3^4\right)^2\cdot\dfrac{1}{3^2}\)
\(=3^2\cdot\dfrac{1}{3^5}\cdot3^8\cdot\dfrac{1}{3^2}\)
\(=\dfrac{3^2}{3^5}\cdot\dfrac{3^8}{3^2}\)
\(=\dfrac{3^2}{3^5}\cdot3^6\)
\(=\dfrac{3^2\cdot3^6}{3^5}\)
\(=3^2\cdot3\)
\(=3^3\)
\(=27\)
\(\frac{2^{12}.243-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3-1\right)}\)
\(=\frac{1}{3}\)
HT
3 2 . 1 243 . 81 2 . 1 3 2 = 3 2 . 1 3 5 . 3 4 2 . 1 3 2 = 3 10 3 7 = 3 3 = 27