DC
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K
Khách
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PP
2
HN
4 tháng 7 2018
Mình làm câu c trước để bạn hình dung ra nhé, câu a tương tự:
c) \(7\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(=\left(8-1\right)\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(=\left[\left(2^3-1\right)\left(2^3+1\right)\right]\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(=\left(2^6-1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(=\left(2^{12}-1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(=\left(2^{12}-1\right)\left(2^{24}+1\right)\)
\(=2^{36}-1\)
b) \(\left(x^2-x+4\right)\left(x^2+x+1\right)\left(x^2-1\right)\)
\(=\left(x^2.x^2.x^2\right).\left(-x+4+x+1+\left(-1\right)\right)\)
\(=x^8.\left(-4\right)\)
4 tháng 7 2018
\(a,\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\)
\(=2^{16}-1\)
HP
1
5 tháng 9 2015
dặt a=(7^3+2)(7^6+4)(7^12+16)(7^24+256)
(7^3-2)a=(7^3-2)(7^3+2)...........................
=>341a=[(7^3)^2-2^2](7^6+4)......
=>341a=(7^6-4)(7^6+4)(7^12+16)(7^24+256)
=>341a=[(7^6)^2-4^2](7^12+16)(7^24+256)
=>341a=(7^12-16)(7^12+16)(7^24+256)
=>341a=[(7^12)^2-16^2](7^24+256)
=>341a=(7^24-256)(7^24+256)
=>341a=(7^24)^2-256^2
làm đến đây bạn tự lấy máy tính tinh nha
28 tháng 11 2016
Rút gọn các phân thức:
a) \(\frac{\left(3x+2\right)^2-\left(x+2\right)^2}{x^3-x^2}=\frac{9x^2+12x+4-x^2-4x-4}{x^3-x^2}=\frac{8x^2+8x}{x^3-x^2}=\frac{8x\left(x+1\right)}{x^2\left(x-1\right)}=\frac{8\left(x+1\right)}{x-1}\)
b) \(\frac{x^4-1}{x^3+2x^2-x-2}=\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x^3-x\right)+\left(2x^2-2\right)}=\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x+2\right)\left(x^2-1\right)}=\frac{x^2+1}{x+2}\)
c) \(\frac{x^2+7x+12}{x^2+5x+6}=\frac{\left(x^2+3x\right)+\left(4x+12\right)}{\left(x^2+3x\right)+\left(2x+6\right)}=\frac{\left(x+3\right)\left(x+4\right)}{\left(x++3\right)\left(x+2\right)}=\frac{x+4}{x+2}\)
d) \(\frac{x^{10}-x^8+x^6-x^4+x^2-1}{x^4-1}=\frac{\left(x^{10}-x^8\right)+\left(x^6-x^4\right)+\left(x^2-1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\frac{\left(x^2-1\right)\left(x^8+x^4+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\frac{x^8+x^4+1}{x^2+1}\)
NT
1
G
19 tháng 3 2018
\(M=\frac{1}{a^2-5a+6}+\frac{1}{a^2-7a+12}+\frac{1}{a^2-9a+20}+\frac{1}{a^2-11a+30}\)
\(M=\frac{1}{\left(a-2\right)\left(a-3\right)}+\frac{1}{\left(a-3\right)\left(a-4\right)}+\frac{1}{\left(a-4\right)\left(a-5\right)}+\frac{1}{\left(a-5\right)\left(a-6\right)}\)
\(M=\frac{1}{a-2}-\frac{1}{a-3}+\frac{1}{a-3}-\frac{1}{a-4}+\frac{1}{a-4}-\frac{1}{a-5}+\frac{1}{a-5}-\frac{1}{a-6}\)
\(M=\frac{1}{a-2}-\frac{1}{a-6}\)
\(A=7.\left(2^3+1\right).\left(2^6+1\right).\left(2^{12}+1\right).\left(2^{24}+1\right)\)
\(A=\left(2^3-1\right).\left(2^3+1\right).\left(2^6+1\right).\left(2^{12}+1\right).\left(2^{24}+1\right)\)
\(A=\left(2^6-1\right).\left(2^6+1\right).\left(2^{12}+1\right).\left(2^{24}+1\right)\)
\(A=\left(2^{12}-1\right).\left(2^{12}+1\right).\left(2^{24}+1\right)\)
\(A=\left(2^{24}-1\right).\left(2^{24}+1\right)\)
\(A=2^{48}-1.\)
Chúc bạn học tốt!