theo tinh chat cua day ti so bang nhau ta co:

a/b=b/c=c/a =a+b+c/b+c+a=1

suy ra: a/b=1

b/c=1

c/a=1

vay a=b=c=

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

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)

\(\Rightarrow a=b=c\)

b, Ta có: \(a^2=bc\Rightarrow\dfrac{a}{c}=\dfrac{b}{a}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

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

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

\(\Rightarrowđpcm\)

a) $\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1$

(tính chất dãy tỉ số bằng nhau)

$\dfrac{a}{b}=1=>a=b$

$\dfrac{b}{c}=1=>b=c$

$\dfrac{c}{a}=1=>c=a$

Vậy a = b = c.

b) Ta có : $a^2=bc=>\dfrac{a}{c}=\dfrac{b}{a}=\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}$(tính chất dãy tỉ số bằng nhau)

$=>\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}$

$=>\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}$

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)

\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)

\(\Rightarrow a=b=c\)

\(a^2=bc\Rightarrow\dfrac{a}{c}=\dfrac{b}{a}\)

Đặt:

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

\(\Rightarrow\left\{{}\begin{matrix}a=ck\\b=ak\end{matrix}\right.\)

\(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{ck+ak}{ck-ak}=\dfrac{k\left(c+a\right)}{k\left(c-a\right)}=\dfrac{c+a}{c-a}\)

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

cái này chắc k ai làm đâu. mệt lắm

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

\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\frac{a+b}{c+d}\right)^2=\frac{ab}{cd}\)

Vậy \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)và \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)

\(\frac{a}{b}=\frac{c}{d}\\ \Rightarrow\frac{a}{c}=\frac{b}{d}\\ \Rightarrow\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}\)

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

\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\\ \Rightarrow\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=\left(\frac{a-b}{c-d}\right)^{2013}\left(1\right)\)

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

\(\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow\left(\frac{a-b}{c-d}\right)^{2013}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

\(\Rightarrow VT=\frac{bk}{bk-b}=\frac{bk}{b\left(k-1\right)}=\frac{k}{k-1}\left(1\right)\)

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

Từ (1) và (2) =>Đpcm

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có

\(VT:\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{b^{2018}\cdot k^{2018}+d^{2018}\cdot k^{2018}}{b^{2018}+d^{2018}}=\frac{k^{2018}\left(b^{2018}+d^{2018}\right)}{b^{2018}+d^{2018}}=k^{2018}\)

\(VP:\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}=\frac{\left(bk+dk\right)^{2018}}{\left(b+d\right)^{2018}}=\frac{k^{2018}\cdot\left(b+d\right)^{2018}}{\left(b+d\right)^{2018}}=k^{2018}\)

\(\Rightarrow VT=VP\)

Hay \(\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}\left(đpcm\right)\)

Ta có: 

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

\(\Rightarrow ad=bc\)

\(\Rightarrow ac-ad=ac-bc\)

\(\Rightarrow a\left(c-d\right)=c\left(a-b\right)\)

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

Vậy \(\frac{a}{a-b}=\frac{c}{c-d}\)

Ta có :

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

\(\Rightarrow ad=bc\)

\(\Rightarrow ac-ad=ac-bc\)

\(\Rightarrow a\left(c-d\right)=c\left(a-b\right)\)

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

\(KL:\frac{a}{a-b}=\frac{c}{c-d}\)

\(\text{Ta có : }\frac{a}{b}=\frac{c}{d}\left(a,b,c\ne0;a\ne b\ne c\ne d\right)\)

\(\Rightarrow ad=cb\left(\text{tính chất tỉ lệ thức}\right)\)

\(\Rightarrow ac-ad=ac-cb\left(\text{tính chất của đẳng thức}\right)\)

\(\Rightarrow a\left(c-d\right)=c\left(a-b\right)\)

\(\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\left(đpcm\right)\)