\(\frac{a}{b}=\frac{c}{d}\)

\(\Rightarrow\frac{a}{c}=\frac{b}{d}.Đặt:a=ck;b=dk\)

\(\Rightarrow\frac{a^2+ac}{c^2-ac}=\frac{c^2k^2+c^2k}{c^2-kc^2}=\frac{c^2\left(k^2+k\right)}{c^2\left(1-k\right)}=\frac{k^2+k}{1-k}\)

\(\frac{b^2+bd}{d^2-bd}=\frac{d^2k^2+kd^2}{d^2-kd^2}=\frac{d^2\left(k^2+k\right)}{d^2\left(1-k\right)}=\frac{k^2+k}{1-k}\)

\(\Rightarrow\frac{b^2+bd}{d^2-bd}=\frac{a^2+ac}{c^2-ac}\left(\text{đpcm}\right)\)

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

Có:a²/b²=c²/d²=ac/bd=>a²+ac/b²+bd=c²-ac/b²-bd=>a²+ac/c²-ac=b²+bd/d²-bd
 

tính chất dãy tỉ số=nhau

ta có: a/b = c/d = (a + c)/ (b + d) = (c - a)/ (d - b)

điều cần chứng minh là:   

(a²  + ac) / (c² - ac) = (b² + bd) / (d² - bd)     => (a² + ac) / (b² + bd)  = (c² - ac) / (d² - bd) 

                                                               = a (a + c) /  b (b + d)   = c (c - a)  / d (d - b)

mà theo chứng minh trên ta có:

a/b = c/d ; (a + c)/ (b + d) = (c - a)/ (d - b)

từ đó ta  =>   (a² + ac) / (c² - ac) = (b² + bd) / (d² - bd)         (đpcm)

tinh chatday tỉ số:nhau

Theo đề bài ta được:

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

Ta có:

\(\dfrac{a^2+ac}{c^2-ac}=\dfrac{a\left(a+c\right)}{c\left(c-a\right)}=\dfrac{bk\left(bk+dk\right)}{dk\left(dk-bk\right)}=\dfrac{bk\left[k\left(b+d\right)\right]}{dk\left[k\left(d-b\right)\right]}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\left(1\right)\)

\(\dfrac{b^2+bd}{d^2-bd}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\left(2\right)\)

Từ (1) và (2) suy ra:\(\dfrac{a^2+ac}{c^2-ac}=\dfrac{b^2+bd}{d^2-bd}\)

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: \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

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 )

Ta có
\(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}=\frac{ac}{bd}\)|
\(\Rightarrow dpcm\)

đặt \(\frac{a}{b}=\frac{c}{d}=k\) thì \(a=bk\text{ };\text{ }c=dk\text{ }\)

Ta có : \(\frac{ac}{bd}=\frac{bk.dk}{bd}=\frac{bd.k^2}{bd}=k^2\text{ }\left(1\right)\)

\(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\frac{b^2.k^2+d^2.k^2}{b^2+d^2}=\frac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\text{ }\left(1\right)\)

Từ ( 1 ) và ( 2 ) \(\Rightarrow\text{ }\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)

Ta có: \(\frac{a}{b}=\frac{c}{d}\)=> \(\left(\frac{a}{b}\right)^2=\left(\frac{c}{d}\right)^2=\frac{ac}{bd}\)=> \(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{ac}{bd}\)

Áp dụng tính chất dãy tỷ số bằng nhau, ta có:

\(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)

=> \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\) (dpcm)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)\(\Rightarrow a=bk;c=dk\)

1)Xét \(VT=\frac{\left(bk\right)^2+bkdk}{\left(dk\right)^2-bkdk}=\frac{b^2k^2+bdk^2}{d^2k^2-bdk^2}=\frac{k^2\left(b^2+bd\right)}{k^2\left(d^2-bd\right)}=\frac{b^2+bd}{d^2-bd}=VP\)

Suy ra Đpcm

2)Xét \(VT=\frac{3\left(bk\right)^2+\left(dk\right)^2}{3b^2+d^2}=\frac{3b^2k^2+d^2k^2}{3b^2+d^2}=\frac{k^2\left(3b^2+d^2\right)}{3b^2+d^2}=k^2\left(1\right)\)

Xét \(VP=\frac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\frac{k^2\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\left(2\right)\)

Từ (1) và (2) suy ra Đpcm