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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
\(\Leftrightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{2ab}{2cd}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{2ab}{2cd}=\frac{a^2+2ab+b^2}{c^2+2cd+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{2ab}{2cd}=\frac{a^2-2ab+b^2}{c^2-2cd+d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\left(2\right)\)
Từ điều (1) và (2)
\(\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(\Rightarrow\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
\(\Rightarrow\left(a+b\right)\left(c-d\right)=\left(a-b\right)\left(c+d\right)\)
\(\Rightarrow c\left(a+b\right)-d\left(a+b\right)=c\left(a-b\right)+d\left(a-b\right)\)
\(\Rightarrow ac+bc-ad-bd=ac-bc+ad-bd\)
\(\Rightarrow bc-ad=-bc+ad\)
\(\Rightarrow2bc=2ad\)
\(\Rightarrow bc=ad\)
\(\Rightarrow\left[\begin{matrix}\frac{a}{b}=\frac{c}{d}\\\frac{b}{a}=\frac{d}{c}\end{matrix}\right.\) ( đpcm )
đề sai phải là CMR \(\frac{a}{b}=\frac{c}{d}\) hoặc \(\frac{b}{a}=\frac{d}{c}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}\)
Áp dụng tính chất của dãy tỉ số bằng nhau,ta có :
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\frac{a+b}{c+d}\right)^2\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\left(\frac{a-b}{c-d}\right)^2\left(2\right)\)
Từ ( 1 ) và ( 2 ) suy ra : \(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{a-b}{c-d}\right)^2\)
xét 2 TH :
TH1 : \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)-\left(a-b\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2a}{2c}=\frac{a}{c}\left(3\right)\)
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)+\left(a-b\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2b}{2d}=\frac{b}{d}\left(4\right)\)
Từ ( 3 ) và ( 4 ) suy ra : \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\)
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 : \(\frac{b}{c}=\frac{a}{d}\Rightarrow\frac{a}{b}=\frac{d}{c}\)
Từ hai trường hợp trên , nếu \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)thì \(\frac{a}{b}=\frac{c}{d}\text{ hay }\frac{a}{b}=\frac{d}{c}\)
ta có \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\left(a,b,c,d\ne0;c\ne\pm d\right)\)
\(\Rightarrow\)cd(a2+b2)=ab(c2+d2)\(\Rightarrow\)a2cd+b2cd=abc2+abd2
\(\Rightarrow\)a2cd-abc2=abd2-b2cd \(\Rightarrow\)ac(ad-bc)=bd(ad-bc)
\(\Rightarrow\)(ad-bc) (ac-bd)=0\(\Rightarrow\orbr{\begin{cases}ad-bc=0\\ac-bd=0\end{cases}}\Rightarrow\orbr{\begin{cases}ad=bc\\ac=bd\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\)(DPCM)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{a}{b}=\frac{c}{d}\)
\(\Leftrightarrow\frac{b}{a}=\frac{d}{c}\Leftrightarrow\frac{b}{a}+1=\frac{d}{c}+1\Leftrightarrow\frac{a+b}{a}=\frac{c+d}{c}\) (1)
\(\Rightarrow\frac{a}{a+b}=\frac{c}{c+d}\)
\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{b}{a}=\frac{d}{c}\Leftrightarrow1-\frac{b}{a}=1-\frac{d}{c}\)
\(\Leftrightarrow\frac{a-b}{a}=\frac{c-d}{c}\Leftrightarrow\frac{a}{a-b}=\frac{c}{c-d}\) (2)
Nhân vế (1) và (2) lại ta được:
\(\frac{a+b}{a}\cdot\frac{a}{a-b}=\frac{c+d}{c}\cdot\frac{c}{c-d}\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right)\cdot cd=\left(c^2+d^2\right)\cdot ab\)
\(\Rightarrow a^2\cdot cd+b^2\cdot cd=c^2\cdot ab+d^2\cdot ab\)
\(\Rightarrow a^2\cdot cd+b^2\cdot cd-c^2\cdot ab-d^2\cdot ab=0\)
\(\Rightarrow\left(a^2\cdot cd-c^2\cdot ab\right)+\left(b^2\cdot cd-d^2\cdot ab\right)=0\)
\(\Rightarrow ac\cdot\left(ad-bc\right)+bd\cdot\left(bc-ad\right)=0\)
\(\Rightarrow ac\cdot\left(ad-bc\right)-bd\cdot\left(ad-bc\right)=0\)
\(\Rightarrow\left(ac-bd\right)\cdot\left(ad-bc\right)=0\)
Tự làm tiếp nhé.......
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a}{c}.\frac{a}{c}=\frac{b}{d}.\frac{b}{d}=\frac{ab}{cd}=\left(\frac{a-b}{c-d}\right)^2\)
\(\Leftrightarrow\frac{ab}{cd}=\left(\frac{a-b}{c-d}\right)^2\)
đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)
a) Ta có:
\(\frac{a}{a+b}=\frac{bk}{bk+b}=\frac{bk}{b\left(k+1\right)}=\frac{k}{k+1}\) (1)
\(\frac{c}{c+d}=\frac{dk}{dk+d}=\frac{dk}{d\left(k+1\right)}=\frac{k}{k+1}\) (2)
Từ (1) và (2) suy ra \(\frac{a}{a+b}=\frac{c}{c+d}\)
b) Ta có:
\(\frac{a}{a-b}=\frac{bk}{bk-b}=\frac{bk}{b\left(k-1\right)}=\frac{k}{k-1}\) (1)
\(\frac{c}{c-d}=\frac{dk}{dk-d}=\frac{dk}{d\left(k-1\right)}=\frac{k}{k-1}\) (2)
Từ (1) và (2) suy ra \(\frac{a}{a-b}=\frac{c}{c-d}\)
c) Ta có:
\(\frac{a+b}{a-b}=\frac{bk+b}{bk-b}=\frac{b\left(k+1\right)}{b\left(k-1\right)}=\frac{k+1}{k-1}\) (1)
\(\frac{c+d}{c-d}=\frac{dk+d}{dk-d}=\frac{d\left(k+1\right)}{d\left(k-1\right)}=\frac{k+1}{k-1}\) (2)
Từ (1) và (2) suy ra \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}=\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}\)
\(\Rightarrow\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\Rightarrow\frac{a^2-b^2}{ab}=\frac{c^2-d^2}{cd}\left(đpcm\right)\)
Đặt : \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Khi đó : \(\frac{\left(bk\right)^2-b^2}{kb^2}=\frac{\left(dk\right)^2-d^2}{kd^2}\)
\(\Rightarrow\frac{b^2.k^2-b^2}{kb^2}=\frac{d^2.k^2-d^2}{kd^2}\)
\(\Rightarrow\frac{b^2\left(k^2-1\right)}{kb^2}=\frac{d^2\left(k^2-1\right)}{kd^2}\)
\(\Rightarrow\frac{k^2-1}{k}=\frac{k^2-1}{k}\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ giả thiết: \(\frac{a}{b}=\frac{c}{d}\)=>ad=bc (1)
Ta có: ab(c2-d2)=abc2-abd2=acbc-adbd (2)
cd(a2-b2)=a2cd-b2cd=acad-bcbd (3)
Từ (1) ,(2),(3)=> ab(c2-d2)=cd(a2-b2)=>\(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\) (đpcm)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
\(\Rightarrow\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd+b^2cd=abc^2+abd^2\)
\(\Leftrightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)
\(\Leftrightarrow\left(ac-bd\right)\left(ad-bc\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}ac=bd\\ad=bc\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{a}{b}=\frac{d}{c}\\\frac{a}{b}=\frac{c}{d}\end{cases}}\).