so sánh
\(\left(\frac{2006-2005}{2006+2005}\right)^2va\frac{2006^2-2005^2}{2006^2+2005^2}\)
Những câu hỏi liên quan
so sánh: \(\left(\frac{2006-2005}{2006+2005}\right)^2\) và \(\frac{2006^2-2005^2}{2006^2+2005^2}\)
\(\left(\frac{2006-2005}{2006+2005}\right)^2=\frac{2006^2-2005^2}{2006^2+2005^2}.\)
Vì \(\frac{2006^2-2005^2}{2006^2+2005^2}=\frac{2006^2+2005^2}{2006^2+2005^2}\)nên => \(\left(\frac{2006-2005}{2006+2005}\right)^2=\frac{2006^2-2005^2}{2006^2+2005^2}.\)
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So sánh: \(\left(\frac{2006-2005}{2006+2005}\right)^2\)
và \(\frac{2006^2-2005^2}{2006^2+2005^2}\)
Ta có :
\(\left(\frac{2006-2005}{2006+2005}\right)^2=\frac{\left(2006-2005\right)^2}{\left(2006+2005\right)^2}=\frac{2006^2-2.2006.2005+2005^2}{2006^2+2.2006.2005+2005^2}=\frac{2006^2-2005^2}{2006^2+2005^2}\)
Vậy \(\left(\frac{2006-2005}{2006+2005}\right)^2=\frac{2006^2-2005^2}{2006^2+2005^2}\)
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SO SÁNH : \(A=\left(\frac{2006-2005}{2006+2005}\right)^2\)và \(B=\frac{2006^2-2005^2}{2006^2+2005^2}\)
Giúp tui zới
ai đúng cho 3 tick
TA CÓ A= \(\left(\frac{2006-2005}{2006+2005}\right)^2\)=\(\frac{1}{4011^2}\)
B=\(\frac{2006^2-2005^2}{2006^2+2005^2}\) = \(\frac{\left(2006-2005\right)\left(2006+2005\right)}{\left(2006+2005\right)^2-2.2005.2006}\) = \(\frac{4011}{4011^2-2.2006.2005}\)
VÌ 1.(\(4011^2\)-2.200.2005)<\(4011^2\).4011 (DO \(4011^2\)>\(4011^2\)-2.2006.2005)
\(\Rightarrow\)\(\frac{1}{4011^2}\)< \(\frac{4011}{4011^2-2.2005.2006}\) .HAY A<B
VẬY A<B
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Cho frac{a}{b}frac{c}{d}. Chứng minh:a) frac{left(a-bright)^3}{left(c-dright)^3}frac{3a^2+2b^2}{3c^2+2d^2}b)frac{4a^4+5b^4}{4c^4+5d^4}frac{a^2b^2}{c^2d^2}c)left(frac{a-b}{c-d}right)^{2005}frac{2a^{2005}-b^{2005}}{2c^{2005}-d^{2005}}d)frac{2a^{2005}+5b^{2005}}{2c^{2005}+5d^{2005}}frac{left(a+bright)^{2005}}{left(c+dright)^{2005}}e)frac{left(20a^{2006}+11b^{2006}right)^{2007}}{left(20a^{2007}-11b^{2007}right)^{2006}}frac{left(20c^{2006}+11d^{2006}right)^{2007}}{left(20c^{2007}-11d^{2007}right)^{20...
Đọc tiếp
Cho \(\frac{a}{b}=\frac{c}{d}\). Chứng minh:
a) \(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{3a^2+2b^2}{3c^2+2d^2}\)
b)\(\frac{4a^4+5b^4}{4c^4+5d^4}=\frac{a^2b^2}{c^2d^2}\)
c)\(\left(\frac{a-b}{c-d}\right)^{2005}=\frac{2a^{2005}-b^{2005}}{2c^{2005}-d^{2005}}\)
d)\(\frac{2a^{2005}+5b^{2005}}{2c^{2005}+5d^{2005}}=\frac{\left(a+b\right)^{2005}}{\left(c+d\right)^{2005}}\)
e)\(\frac{\left(20a^{2006}+11b^{2006}\right)^{2007}}{\left(20a^{2007}-11b^{2007}\right)^{2006}}=\frac{\left(20c^{2006}+11d^{2006}\right)^{2007}}{\left(20c^{2007}-11d^{2007}\right)^{2006}}\)
f)\(\frac{\left(20a^{2007}-11c^{2007}\right)^{2006}}{\left(20a^{2006}+11c^{2006}\right)^{2007}}=\frac{\left(20b^{2007}-11d^{2007}\right)^{2006}}{\left(20b^{2006}+11d^{2006}\right)^{2007}}\)
ừ, bạn bik làm thì giúp mình nha ^^
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Số nào lớn hơn
\( \left(\frac{2006-2005}{2006+2005}\right)^2hay\frac{2006^2-2005^2}{2006^2+2005^2}\)
Số nào lớn hơn:
\(\left(\frac{2006-2005}{2006+2005}\right)^2hay\frac{2006^2-2005^2}{2006^2+2005^2}\)
Theo tính chất của phân thức ta có:
\(\left(\frac{2006-2005}{2006+2005}\right)^2=\frac{2006-2005}{2006+2005}.\frac{2006-2005}{2006+2005}< \frac{2006^2-2005^2}{\left(2006+2005\right)^2}\)
\(=\frac{2006^2-2005^2}{2006^2+2.2006.2005+2005^2}< \frac{2006^2-2005^2}{2006^2+2005^2}\)
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số nào lớn hơn \(\left(\frac{2006-2005}{2006+2005}\right)^2hay\frac{2006^2-2005^2}{2006^2+2005^2}\)
tôi lm đc rồi nếu ai lm giống tôi tôi sẽ tick cho
\(\left(\frac{2006-2005}{2006+2005}\right)^2=\frac{1}{\left(2006+2005\right)^2}<\frac{4011}{2006^2+2005^2}=\frac{2006^2-2005^2}{2006^2+2005^2}\)
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Cho \(S=\frac{2}{2005+1}+\frac{2^2}{2005^2+1}+...+\frac{2^{n+1}}{2005^{^{2^n}}+1}+...+\frac{2^{2006}}{2006^{2^{2005}}+1}\). So sánh S với \(\frac{1}{1002}\)
\(\frac{1}{2}\left(2004^{2005}+2005^{2006}+2006^{2007}\right)=?\)