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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 \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) Ta có (a+b+c+d)(a-b-c-d)=(a-b+c-d)(a+b-c-d) với dạng a.d = b.c
\(\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{a+b+c+d}{a+b-c-d}=\frac{a-b-c-d}{a-b+c-d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\left(1\right)\)
\(\frac{a}{b}=\frac{c}{d}=\frac{a-c}{b-d}\left(2\right)\)
Từ (1) và (2) => \(\frac{\left(a+b+c+d\right)=\left(a-b-c-d\right)}{\left(a+b-c-d\right)=\left(a-b+c-d\right)}\Rightarrow\frac{a+b+c+d}{a+b-c-d}=\frac{a-b-c-d}{a-b+c-d}\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a-b}{b-c}=\frac{c-d}{d-a}=\left(a-b\right)\cdot\left(d-a\right)=\left(b-c\right)\cdot\left(c-d\right)=\)
\(\left(a-b\right)\cdot d-\left(a-b\right)\cdot a=\left(b-c\right)\cdot c-\left(b-c\right)\cdot d=\)
\(ad-bd-2a-ab=bc-2c-bd-cd\)
Suy ra a=c
Nhớ cho mình nha!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(K=\frac{a}{a+b+c}+\frac{b}{a+b+d}+\frac{c}{b+c+d}+\frac{d}{a+c+d}\)
Ta có : \(\frac{a}{a+b+c}< \frac{a+d}{a+b+c+d};\frac{b}{a+b+d}< \frac{b+c}{a+b+c+d}\)
\(\frac{c}{c+b+d}< \frac{a+c}{a+b+c+d};\frac{d}{c+a+d}< \frac{b+d}{a+b+c+d}\)
\(\Rightarrow K=\frac{a}{a+b+c}+\frac{b}{a+b+d}+\frac{c}{c+b+d}+\frac{d}{a+c+d}< \frac{a+d}{a+b+c+d}+\frac{b+c}{a+b+c+d}+\frac{c+a}{a+b+c+d}+\frac{d+b}{a+b+c+d}=\frac{1}{2}\)
\(\Rightarrow K^{10}< \left(\frac{1}{2}\right)^{10}=\frac{1}{2^{10}}< 1< 2020\)
Vậy ....
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Ta có:
\(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\left(1\right)\)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\left(đpcm\right)\)
Vậy .....
Bài 2
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Leftrightarrow\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)
\(\Leftrightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\left(đpcm\right)\)
Vậy .....
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
Có ( a-b)/(b-c) = (c-d)/(d-a) = (a-b+c-d)/((b-c+d-a)=-1( tỉ lệ thức)
=> a-b = c-b => a=c
\((a+b+c+d).(a-b-c+d)=(a-b+c-d).(a+b-c-d)\)
\( \Longleftrightarrow (a+d)^2 - (b+c)^2 = (a-d)^2 - (b-c)^2\)
\(\Longleftrightarrow 4ad = 4bc\)
\(\Longleftrightarrow \dfrac{a}{b}=\dfrac{c}{d}\)