Rút gọn biểu thức : A=(2^3-1)/(2^3+1).(3^3-1)/(3^3+1)...(100^3-1).(100^3+1).
giúp mìn vs nhá các bn!!!! còn nhiều bài nữa mk sẽ gửi lên lun.
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt biểu thức đã cho là A.
Ta có: 2A = (3 - 1) * (3 + 1) * (3^2 + 1) * .... * (3^64 + 1)
= (3^2 - 1) * (3^2 + 1) * ... * (3^64 + 1) (hằng đẳng thức a^2 - b^ 2 = (a+b)(a-b))
Rút gọn triệt tiêu ta được 2A=3^64 - 1
=> A = (3^64 - 1)/2
\(A=1+3+3^2+3^3+...+3^{99}+3^{100}\\ \Rightarrow3A=3+3^2+3^3+...+3^{100}+3^{101}\\ \Rightarrow3A-A=3^{101}-1\\ \Rightarrow2A=3^{101}-1\\ \Rightarrow A=\left(3^{101}-1\right).\dfrac{1}{2}\\ \Rightarrow\dfrac{3^{101}}{2}-\dfrac{1}{2}.\)
\(A=1+3+3^2+3^3+...+3^{99}+3^{100}\)
Ta có: \(3A=3+3^2+3^3+...+3^{99}+3^{100}\)
Khi đó: \(3A-A=3+3^2+3^3+...+3^{99}+3^{100}+3^{101}-\left(1+3+3^2+3^3+...+3^{99}+3^{100}\right)\)
\(=3^{101}-1\)
\(\Leftrightarrow2A=3^{101}-1\)
Vậy \(A=\left(3^{101}-1\right):2\)
\(A=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right)\div\left(1-\frac{1}{x+3}\right)\)
\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right)\div\left(\frac{x+3}{x+3}-\frac{1}{x+3}\right)\)
\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{x^2-x-12}{\left(x-3\right)\left(x+3\right)}-\frac{x^2-4x+3}{\left(x-3\right)\left(x+3\right)}\right)\div\frac{x+2}{x+3}\)
\(=\frac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}\times\frac{x+3}{x+2}\)
\(=\frac{3x+6}{x-3}\times\frac{1}{x+2}=\frac{3\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}=\frac{3}{x-3}\)
\(A=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)
\(=\left(\frac{21}{x^2-9}+\frac{x-4}{x-3}-\frac{x-1}{x+3}\right):\left(\frac{x+2}{x+3}\right)\)
\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{x+2}{x+3}\right)\)
\(=\left(\frac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{x+2}{x+3}\right)\)
\(=\frac{6+3x}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}=\frac{3\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}=\frac{3}{x-3}\)
\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)
\(3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)\)
\(2C=1-\frac{1}{3^{99}}< 1\)
\(\Rightarrow C=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
1.
B = 3100 - 399 + 398 - 397 + ... + 32 - 3 + 1
3B = 3101 - 3100 + 399 - 398 + ... + 33 - 32 + 3
3B + B = ( 3101 - 3100 + 399 - 398 + ... + 33 - 32 + 3 ) + ( 3100 - 399 + 398 - 397 + ... + 32 - 3 + 1 )
4B = 3101 + 1
B = \(\frac{3^{101}+1}{4}\)
a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012
2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013
3M=2^0+2^2013
M=(2^0+2^2013)÷3
Vậy.......
b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012
3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013
4N=3-3^2013
N=(3-3^2013)÷4
Vậy........
K tao nhé ko lên lớp tao đánh m😈😈😈
a) Ta có: \(\dfrac{7\cdot25}{14\cdot10}\)
\(=\dfrac{7\cdot5\cdot5}{7\cdot2\cdot2\cdot5}\)
\(=\dfrac{5}{4}\)
b) Ta có: \(\dfrac{24\cdot15-14\cdot9}{36\cdot12}\)
\(=\dfrac{9\cdot8\cdot5-14\cdot9}{36\cdot12}\)
\(=\dfrac{9\cdot\left(8\cdot5-14\right)}{9\cdot4\cdot12}\)
\(=\dfrac{40-14}{4\cdot12}\)
\(=\dfrac{13}{24}\)