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Ta có :
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\frac{a+b}{c+d}\right)^2\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}-\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\left(\frac{a-b}{c-d}\right)^2\left(2\right)\)
Từ ( 1 ) và ( 2 ) suy ra : \(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{a-b}{c-d}\right)^2\)
TH1 : \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)+\left(a-b\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2a}{2c}=\frac{a}{b}\left(3\right)\)
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)-\left(a-b\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2b}{2d}=\frac{b}{d}\left(4\right)\)
từ ( 3 ) và ( 4 ) suy ra : \(\frac{a}{c}=\frac{b}{d}\text{ hay }\frac{a}{b}=\frac{c}{d}\)
TH2 : \(\frac{a+b}{c+d}=\frac{b-a}{c-d}=\frac{\left(a+b\right)+\left(b-a\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2b}{2c}=\frac{b}{c}\left(5\right)\)
\(\frac{a+b}{c+d}=\frac{b-a}{c-d}=\frac{\left(a+b\right)-\left(b-a\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2a}{2d}=\frac{a}{d}\left(6\right)\)
Từ ( 5 ) và ( 6 ) suy ra : \(\frac{b}{c}=\frac{a}{d}\text{ hay }\frac{a}{b}=\frac{d}{c}\)
Vậy : \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\text{ thì }\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\)
kinh quá
Ta có : \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
\(\Rightarrow\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
=> \(\frac{a}{c}=\frac{b}{d}\)
=> \(\frac{a}{b}=\frac{c}{d}\) nếu khố hiểu thì bạn chứng mình kiểu này :
Ta có : \(\frac{a}{b}=\frac{c}{d}\)
=> \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
Mặt khác \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
=> \(\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
Vậy \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=\)\(1\)
\(\hept{\begin{cases}\hept{\begin{cases}\frac{a}{b}=1\\\frac{b}{c}=1\end{cases}}\\\frac{c}{d}=1\\\frac{d}{a}=1\end{cases}\Rightarrow\hept{\begin{cases}\hept{\begin{cases}a=b\\b=c\end{cases}}\\c=d\\d=a\end{cases}\Rightarrow}a=b=c=d}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Theo tính chất dãy tỉ số bằng nhau có:
\(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
ta có a+b/a-b=c+d/c-d
suy ra (a+b)(c-d)=(a-b)(c+d)
ac-ad+bc-bd=ac+ad-bc-bd
ac-ac+bc+bc-bd+bd=ad+ad
2bc=2ad
nen bc=ad=a/b=c/d
vay tu a/b=c/d ta co the suy ra a+b/a-b=c+d/c-d
Ta có : \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd+b^2cd=abc^2+abd^2\)
\(\Leftrightarrow\left(a^2cd-abd^2\right)+\left(b^2cd-abc^2\right)=0\)
\(\Leftrightarrow ad\left(ac-bd\right)-bc\left(ac-bd\right)=0\)
\(\Leftrightarrow\left(ac-bd\right)\left(ad-bc\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}ac=bd\\ad=bc\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\) (đpcm)
\(\frac{a}{b}< \frac{c}{d}\)
\(\Leftrightarrow ad< bc\)
\(\Leftrightarrow ad+ab< bc+ab\)
\(\Leftrightarrow a\left(b+d\right)< b\left(a+c\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\) (1)
\(\frac{a}{b}< \frac{c}{d}\)
\(\Leftrightarrow ad< bc\)
\(\Leftrightarrow ad+cd< bc+cd\)
\(\Leftrightarrow d\left(a+c\right)< c\left(b+d\right)\)
\(\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\) (2)
Từ (1) ; (2) \(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\) (đpcm)
Ta có :
\(\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}+\frac{d}{a+b+c}\)\(< \frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}=\frac{a+b+c+d}{a+b+c+d}=1\)\(\left(1\right)\)
Ta lại có :
\(\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}+\frac{d}{a+b+c}\)\(< \frac{2a}{a+b+c+d}+\frac{2b}{a+b+c+d}+\frac{2c}{a+b+c+d}+\frac{2d}{a+b+c+d}=\frac{2a+2b+2c+2d}{a+b+c+d}=2\)\(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)suy ra \(1< S< 2\)
Vậy \(S\)không là số tự nhiên
Ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\left(\frac{a-b}{c-d}\right)^2\left(2\right)\)
Từ (1) và (2) \(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{a-b}{c-d}\right)^2\)
TH1: \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b+a-b}{c+d+c-d}=\frac{2a}{2c}=\frac{a}{c}\left(3\right)\)
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b-a+b}{c+d-c+d}=\frac{2b}{2d}=\frac{b}{d}\left(4\right)\)
Từ (3) và (4) => \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\)
TH2: \(\frac{a+b}{c+d}=\frac{b-a}{d-c}=\frac{a+b+b-a}{c+d+d-c}=\frac{2b}{2d}=\frac{b}{d}\left(5\right)\)
\(\frac{a+b}{c+d}=\frac{b-a}{d-c}=\frac{a+b-b+a}{c+d-d+c}=\frac{2a}{2c}=\frac{a}{c}\left(6\right)\)
Từ (5) và (6) => \(\frac{b}{c}=\frac{a}{d}\Rightarrow\frac{a}{b}=\frac{d}{c}\)
Vậy nếu \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\) thì \(\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\)
Đề phải thêm là \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\) nhé.
Ta có: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}.\)
\(\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a+b+c}{b+c+d}.\frac{a+b+c}{b+c+d}.\frac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}\)
\(\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}.\)
\(\Rightarrow\frac{a}{d}=\frac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\left(đpcm\right).\)
Chúc bạn học tốt!
