Cho \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với ( với a, b, c, d khác 0, và c \(\ne\pm d\) ). Chứng minh rằng hoặc \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\) ?

ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow1+\dfrac{a}{b}=1+\dfrac{c}{d}\Leftrightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\left(đpcm\right)\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=b\cdot k\) ;\(c=d\cdot k\)

=>\(\dfrac{a+b}{b}=\dfrac{b\cdot k+b}{b}=\dfrac{b\cdot\left(k+1\right)}{b}=k+1\) (1)

=>\(\dfrac{c+d}{d}=\dfrac{d\cdot k+d}{d}=\dfrac{d\cdot\left(k+1\right)}{d}=k+1\) (2)

Từ (1) và (2) => \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)

Từ \(\dfrac{a+b}{c+d}=\dfrac{b+c}{d+a}\Rightarrow\left(a+b\right)\left(d+a\right)=\left(b+c\right)\left(c+d\right)\)

\(\Rightarrow ad+a^2+bd+ba=bc+bd+c^2+cd\)

\(\Rightarrow a^2+a\left(b+d\right)=c^2+c\left(b+d\right)\)

Vì đt trên bằng nhau : \(\Rightarrow a\left(b+d\right)=c\left(b+d\right)\Leftrightarrow a=c\)

Bài giải:

Với \(a,b,c,d\ne0\) ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{b}+1=\dfrac{c}{d}+1\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\Rightarrow\dfrac{a+b}{c+d}=\dfrac{b}{d}\left(1\right)\)

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a-b}{b}=\dfrac{c-d}{d}\Rightarrow\dfrac{a-b}{c-d}=\dfrac{b}{d}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\left(ĐPCM\right)\)

Đặt:

\(\dfrac{a}{b}=\dfrac{c}{d}=t\Leftrightarrow\left\{{}\begin{matrix}a=bt\\c=dt\end{matrix}\right.\)

Khi đó:

\(\dfrac{a+b}{a-b}=\dfrac{bt+b}{bt-b}=\dfrac{b\left(t+1\right)}{b\left(t-1\right)}=\dfrac{t+1}{t-1}\)

\(\dfrac{c+d}{c-d}=\dfrac{dt+d}{dt-d}=\dfrac{d\left(t+1\right)}{d\left(t-1\right)}=\dfrac{t+1}{t-1}\)

Ta có đpcm

Đặt \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=k

=>a=bk;c=dk

Ta có:

\(\dfrac{a+b}{a-b}\)=\(\dfrac{bk+b}{bk-b}\)=\(\dfrac{b\left(k+1\right)}{b\left(k-1\right)}\)=\(\dfrac{k+1}{k-1}\)(1)

\(\dfrac{c+d}{c-d}\)=\(\dfrac{dk+d}{dk-d}\)=\(\dfrac{d\left(k+1\right)}{d\left(k-1\right)}\)=\(\dfrac{k+1}{k-1}\)(2)

Từ (1) và (2) =>\(\dfrac{a+b}{a-b}\)=\(\dfrac{c+d}{c-d}\)

Vậy từ tỉ lệ thức\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)ta có thể suy ra tỉ lệ thức\(\dfrac{a+b}{a-b}\)=\(\dfrac{c+d}{c-d}\)(đpcm)

Ta có: \(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\)

\(\Leftrightarrow\left(a+d\right)^2-\left(b+c\right)^2=\left(a-d\right)^2-\left(b-c\right)^2\)

\(\Leftrightarrow\left(a+d-a+d\right)\left(a+d+a-d\right)=\left(b+c-b+c\right)\left(b+c+b-c\right)\)

\(\Leftrightarrow2d\cdot2a=2c\cdot2b\)

\(\Leftrightarrow ad=bc\)

hay \(\dfrac{a}{c}=\dfrac{b}{d}\)

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

Áp dụng t/c dtsbn:

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}=\dfrac{a+b}{c+d}\)

\(\Rightarrow\dfrac{a-b}{c-d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)

Đặt :

\(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(VT=\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\left(1\right)\)

\(VP=\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)

Đề bài sai

Ví dụ: với \(a=1;b=2;c=3,d=4\) thì \(x=\dfrac{1}{2}\) ; \(y=\dfrac{3}{4}\) ; \(z=\dfrac{2}{3}\)

Khi đó  \(x< y\) nhưng \(z< y\)

\(\text{Vì }\dfrac{a}{b}< \dfrac{c}{d}\text{ nên }ad< bc\left(1\right)\)

\(\text{Xét tích}:a\left(b+d\right)=ab+ad\left(2\right)\)

                \(b\left(a+c\right)=ba+bc\left(3\right)\)

\(\text{Từ(1);(2);(3)}\Rightarrow a\left(b+d\right)< b\left(a+c\right)\text{ do đó }\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(4\right)\)

\(\text{Tương tự ta có:}\dfrac{a+c}{b+d}< \dfrac{c}{d}\left(5\right)\)

\(\text{Từ (4);(5) ta được }\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)

\(\Rightarrow x< y< z\)