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\(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)
\(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)
\(\Rightarrow\frac{a^2+b^2}{c^2+d^2}.\frac{a^2+c^2}{b^2+d^2}=\frac{a^2}{c^2}.\frac{c^2}{d^2}=\frac{a^2}{d^2}\)
Đặt \(S=\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}\)
Ta có: \(\frac{a}{a+b+c}< \frac{a}{a+c}\)
\(\frac{b}{b+c+d}< \frac{b}{b+d}\)
\(\frac{c}{c+d+a}< \frac{c}{a+c}\)
\(\frac{d}{d+a+b}< \frac{d}{d+b}\)
\(\Rightarrow S< \left(\frac{a}{a+c}+\frac{c}{a+c}\right)+\left(\frac{b}{b+d}+\frac{d}{d+b}\right)\)
\(\Rightarrow S< 2\left(1\right)\)
Lại có: \(\frac{a}{a+b+c}>\frac{a}{a+b+c+d}\)
\(\frac{b}{b+c+d}>\frac{b}{b+c+a+d}\)
\(\frac{c}{c+d+a}>\frac{c}{a+b+c+d}\)
\(\frac{d}{d+a+b}>\frac{d}{a+b+c+d}\)
\(\Rightarrow S>1\left(2\right)\)
Từ (1) và (2) \(\Rightarrowđpcm\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
Suy ra \(\begin{cases}a=bk\\c=dk\end{cases}\)
\(VT=\frac{a^2+ac}{c^2-ac}=\frac{\left(bk\right)^2+bkdk}{\left(dk\right)^2-bkdk}=\frac{b^2k^2+bdk^2}{d^2k^2-bdk^2}=\frac{k^2\left(b^2+bd\right)}{k^2\left(d^2-bd\right)}=\frac{b^2+bd}{d^2-bd}=VP\)
=>Đpcm
Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk,c=dk\)
Ta có:
\(\frac{a^2+ac}{c^2-ac}=\frac{\left(bk\right)^2+bkdk}{\left(dk\right)^2-bkdk}=\frac{b^2.k^2+b.d.k^2}{d^2.k^2-b.d.k^2}=\frac{b.k^2\left(b+d\right)}{d.k^2\left(d-b\right)}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\) (1)
\(\frac{b^2+bd}{d^2-bd}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\) (2)
Từ (1) và (2) suy ra \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\) ( đpcm )
\(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}.Đặt:a=ck;b=dk\)
\(\Rightarrow\frac{a^2+ac}{c^2-ac}=\frac{c^2k^2+c^2k}{c^2-kc^2}=\frac{c^2\left(k^2+k\right)}{c^2\left(1-k\right)}=\frac{k^2+k}{1-k}\)
\(\frac{b^2+bd}{d^2-bd}=\frac{d^2k^2+kd^2}{d^2-kd^2}=\frac{d^2\left(k^2+k\right)}{d^2\left(1-k\right)}=\frac{k^2+k}{1-k}\)
\(\Rightarrow\frac{b^2+bd}{d^2-bd}=\frac{a^2+ac}{c^2-ac}\left(\text{đpcm}\right)\)
Ta có \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc\)
\(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\Leftrightarrow ad\left(a+c\right)\left(d-b\right)=bc\left(b+d\right)\left(c-a\right)\)
Rút gọn ad với bc \(\Rightarrow\left(a+c\right)\left(d-b\right)=\left(b+d\right)\left(c-a\right)\)
\(\Leftrightarrow ad+cd-ab-bc=bc+cd-ab-ad\)
Rút gọn 2 vế ta đc 0=0
vì 0=0 luôn đúng nên cái phương trình trên luôn đúng
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2\)
Do đó: \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
\(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{2c^2}{2d^2}=\frac{ac}{bd}\)
Ta có : \(\frac{2c^2}{2d^2}=\frac{ac}{bd}=\frac{2c^2-ac}{2d^2-bd}\)
Vậy \(\frac{a^2}{b^2}=\frac{2c^2-ac}{2d^2-bd}\)
Có:a2/b2=c2/d2=ac/bd=>a2+ac/b2+bd=c2-ac/b2-bd=>a2+ac/c2-ac=b2+bd/d2-bd
Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk,c=dk\)
Ta có:
\(\frac{ac}{bd}=\frac{bkdk}{bd}=k^2\) (1)
\(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\frac{b^2.k^2+d^2.k^2}{b^2+d^2}=\frac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (2)
Từ (1) và (2) suy ra \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)
mk cũng định làm thế nhưng ko rảnh