\(P=\left(\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)....\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right).....\left(23^4+4\right)}\right)\). Rút gọn biểu thức
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rút gọn biểu thức
P=\(\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
rút gọn biểu thức P=\(\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+1\right)...\left(21^4+1\right)}{\left(3^4+1\right)\left(7^4+1\right)\left(11^4+1\right)...\left(23^4+1\right)}\)
Rút gọn biểu thức P=\(\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+1\right)\left(7^4+1\right)\left(11^4+1\right)...\left(23^4+1\right)}\)
Rút gọn phân thức P=\(\frac{\left(1^4+4\right)\left(5^4 +4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
\(P=\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)\(=\frac{\left(1+4\right)\left(4^2+1\right)\left(6^2+1\right)\left(8^2+1\right)\left(10^2+1\right)...\left(20^2+1\right)\left(\cdot22^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)\left(6^2+1\right)\left(8^2+1\right)\left(10^2+1\right)\left(12^2+1\right)...\left(22^2+1\right)\left(24^2+1\right)}\)
\(=\frac{1+4}{\left(2^2+1\right)\left(24^2+1\right)}=\frac{5}{5\left(24^2+1\right)}=\frac{1}{24^2+1}=\frac{1}{577}\)
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cái bước tách ra bn nhân lại là có kết quả y chang, VD:
\(\left(5^4+4\right)=\left(4^2+1\right)\left(6^2+1\right)=629\)
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rút gọn biểu thức:
\(P=\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right).....\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)....\left(23^4+4\right)}\)
rút gọn bieu thuc P=\(\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+1\right)...\left(21^4+1\right)}{\left(3^4+1\right)\left(7^4+1\right)\left(11^4+1\right)...\left(23^4+1\right)}\)
rút gọn các biểu thức ( viết kết quả dưới dạng phân số )
\(B=\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
Rút gọn biểu thức :
\(P=\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)....\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)....\left(23^4+4\right)^{ }}\)
ta co dang tong quat cho tu so la : n^4+4=(n^2+2)^2=(n^2+2n+2)(n^2-2n+2)=[(n-1)^2+1][(n+1)^2+1]
Nen A=(0+1)(2^2+1)/(2^2+1)(4^2+1) . (4^2+1)(6^2+1)/(6^2+1)(8^2+1) .........(20^2+1)(22^2+1)/(22^2+1)(24^2+1) = 1/24^2+1=1/577
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Rút gọn \(C=\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)