nếu a/b=c/d thì (a-b/c-d)^2003=a^2005+b^2005/c^2005+d^2005
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cho hỏi chút
\(\frac{a}{b}=\frac{c}{d}\)
trong đó
\(a=c\) hay \(a\ne c\)
\(b=d\) hay \(b\ne d\)
( bài có thiếu điều kiện ko vậy )
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Lời giải:
a)
\(2006.2005^{2003}> 2005.2005^{2003}=2005^{1+2003}=2005^{2004}\)
Vậy \(2006.2005^{2003}> 2005^{2004}\)
b)
\(2005^{2004}+2005^{2003}=2005^{2003}(2005+1)=2005^{2003}.2006< 2006^{2003}.2006\)
hay \(2005^{2004}+2005^{2003}< 2006^{2004}\)
c) Thiếu đề
d)
\(72^{27}-72^{26}=72^{26}(72-1)=71.72^{26}\)
\(72^{28}-72^{27}=72^{27}(72-1)=71.72^{27}> 71.72^{26}\)
\(\Rightarrow 72^{28}-72^{27}> 72^{27}-72^{26}\)
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Thư viện Đề thi & Kiểm tra
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Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\)
a) \(\frac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\frac{\left(kb\right)^{2004}-b^{2004}}{\left(kb\right)^{2004}+b^{2004}}=\frac{k^{2004}b^{2004}-b^{2004}}{k^{2004}b^{2004}+b^{2004}}=\frac{b^{2004}\left(k^{2004}-1\right)}{b^{2004}\left(k^{2004}+1\right)}=\frac{k^{2004}-1}{k^{2004}+1}\)(1)
\(\frac{c^{2004}-d^{2004}}{d^{2004}+d^{2004}}=\frac{\left(kd\right)^{2004}-d^{2004}}{\left(kd\right)^{2004}+d^{2004}}=\frac{k^{2004}d^{2004}-d^{2004}}{k^{2004}d^{2004}+d^{2004}}=\frac{d^{2004}\left(k^{2004}-1\right)}{d^{2004}\left(k^{2004}+1\right)}=\frac{k^{2004}-1}{k^{2004}+1}\)(2)
Từ (1) và (2) => đpcm
b) \(\frac{a^{2005}}{b^{2005}}=\frac{\left(kb\right)^{2005}}{b^{2005}}=\frac{k^{2005}b^{2005}}{b^{2005}}=k^{2005}\)(1)
\(\frac{\left(a-c\right)^{2005}}{\left(b-d\right)^{2005}}=\frac{\left(kb-kd\right)^{2005}}{\left(b-d\right)^{2005}}=\frac{\left[k\left(b-d\right)\right]^{2005}}{\left(b-d\right)^{2005}}=\frac{k^{2005}\left(b-d\right)^{2005}}{\left(b-d\right)^{2005}}=k^{2005}\)(2)
Từ (1) và (2) => đpcm
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\(\hept{\begin{cases}a+b=c+d\Rightarrow\left(a+b\right)^2=\left(c+d\right)^2\Rightarrow a^2+2ab+b^2=c^2+2cd+d^2\\a^2+b^2=c^2+d^2\end{cases}}\)
\(\Rightarrow2ab=2cd\Rightarrow ab=cd\Rightarrow\frac{a}{d}=\frac{b}{c}=k\Rightarrow\hept{\begin{cases}a=dk\\b=ck\end{cases}}\)
Xét \(a^2+b^2=c^2+d^2\Leftrightarrow\left(dk\right)^2+b^2=\left(ck\right)^2+d^2\Leftrightarrow d^2\left(k^2-1\right)=b^2\left(k^2-1\right)\)
\(\Leftrightarrow\left(d^2-b^2\right)\left(k^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}d^2-b^2=0\\k^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}d=\pm b\\k=\pm1\end{cases}}\Rightarrow\orbr{\begin{cases}a=\pm c\\a=\pm d;c=\pm b\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}d^{2005}=b^{2005};a^{2005}=c^{2005}\\a^{2005}=d^{2005};c^{2005}=b^{2005}\end{cases}\Rightarrow\orbr{\begin{cases}a^{2005}+b^{2005}=c^{2005}+d^{2005}\\a^{2005}+b^{2005}=c^{2005}+d^{2005}\end{cases}}}\)
\(\Rightarrow a^{2005}+b^{2005}=c^{2005}+d^{2005}\left(đpcm\right)\)
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ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\left(\frac{a}{c}\right)^{2003}=\left(\frac{b}{d}\right)^{2003}\)(1)
áp dụng t/c dãy tỉ số = nhau
\(\left(\frac{a}{c}\right)^{2003}=\left(\frac{b}{d}\right)^{2003}=\left(\frac{a-b}{c-d}\right)^{2003}\)(2)
ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^{2005}}{c^{2005}}=\frac{b^{2005}}{d^{2005}}\)(3)
áp dụng t/c dãy tỉ số = nhau
\(\frac{a^{2005}}{c^{2005}}=\frac{b^{2005}}{d^{2005}}=\frac{a^{2005}+b^{2005}}{c^{2005}+d^{2005}}\)(4)
từ (1) ;(2) ; (3) và (4)
ta suy ra
\(\left(\frac{a-b}{c-d}\right)^{2003}=\frac{a^{2005}+b^{2005}}{c^{2005}+d^{2005}}\)