Đặt :

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

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

Lời giải:

Đặt $\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk$.

Khi đó:

$\frac{5a+3b}{5a-3b}=\frac{5bk+3bk}{5bk-3bk}=\frac{8bk}{2bk}=4(1)$

$\frac{5c+3d}{5c-3d}=\frac{5dk+3dk}{5dk-3dk}=\frac{8dk}{2dk}=4(2)$

Từ $(1); (2)$ suy ra điều phải chứng minh.

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3b}{3d}=\dfrac{5a}{5c}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\\ \Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)

a) \(\dfrac{a}{b}=\dfrac{c}{d}\left(a;b;c;d\ne0\right)\)

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

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

\(\Rightarrow dpcm\)

b) \(\dfrac{a}{b}=\dfrac{c}{d}\)

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

\(\Rightarrow\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)

\(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)

\(\Rightarrow dpcm\)

Thanks

Lời giải:

Đặt $\frac{a}{b}=\frac{c}{d}=k$

$\Rightarrow a=bk; c=dk$. Khi đó:

1.

$\frac{a+b}{b}=\frac{bk+b}{b}=\frac{b(k+1)}{b}=k+1(1)$

$\frac{c+d}{d}=\frac{dk+d}{d}=\frac{d(k+1)}{d}=k+1(2)$
Từ $(1); (2)\Rightarrow \frac{a+b}{b}=\frac{c+d}{d}$

2.

$\frac{5a+3b}{5a-3b}=\frac{5bk+3b}{5bk-3b}=\frac{b(5k+3)}{b(5k-3)}=\frac{5k+3}{5k-3}(3)$

$\frac{5c+3d}{5c-3d}=\frac{5dk+3d}{5dk-3d}=\frac{d(5k+3)}{d(5k-3)}=\frac{5k+3}{5k-3}(4)$

Từ $(3); (4)\Rightarrow \frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}$ (đpcm)

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

nên \(\dfrac{5a}{3b}=\dfrac{5c}{3d}\)

hay \(\dfrac{5a}{5c}=\dfrac{3b}{3d}\)

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

\(\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)

\(\Leftrightarrow\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)

hay \(\dfrac{5a+3n}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)(đpcm)

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

Ta có: \(\left\{{}\begin{matrix}\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\\\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\end{matrix}\right.\Rightarrowđpcm\)

b) \(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\)

\(\Rightarrow\left(\dfrac{x-1}{2017}-1\right)+\left(\dfrac{x-2}{2016}-1\right)=\left(\dfrac{x-3}{2015}-1\right)+\left(\dfrac{x-4}{2014}-1\right)\)\(\Rightarrow\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}=\dfrac{x-2018}{2015}+\dfrac{x-2018}{2014}\)

\(\Rightarrow\left(x-2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)

vì \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\) nên \(x-2018=0\Leftrightarrow x=2018\)

Mk chỉ làm 1 câu thôi mấy câu sau tương tự theo cách đó nhoa:v

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

\(\Rightarrow\left(\dfrac{a-b}{c-d}\right)^4=\left(\dfrac{bk-b}{dk-d}\right)^4=\left[\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right]^4=\dfrac{b^4}{d^4}\)

\(\dfrac{a^4+b^4}{c^4+d^4}=\dfrac{bk^4+b^4}{dk^4+d^4}=\dfrac{b^4\left(k^4+1\right)}{d^4\left(k^4+1\right)}=\dfrac{b^4}{d^4}\)

\(\Rightarrow\left(\dfrac{a-b}{c-d}\right)^4=\dfrac{a^4+b^4}{c^4+d^4}\Rightarrowđpcm\)

Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a^4}{c^4}\)=\(\dfrac{b^4}{d^4}\)=\(\dfrac{a^4+b^4}{c^4+d^4}\)(1)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a-b}{c-d}\)=\(\left(\dfrac{a-b}{c-d}\right)^4\)(2)
Từ (1) và (2)suy ra:
\(\left(\dfrac{a-b}{c-d}\right)^4\)=\(\dfrac{a^4+b^4}{c^4+d^4}\)(đpcm)
b) Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{5a}{5c}\)=\(\dfrac{3b}{3d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{5a}{5c}\)=\(\dfrac{3b}{3d}\)=\(\dfrac{5a+3b}{5c+3d}\)(1)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{5a}{5b}\)=\(\dfrac{3b}{3d}\)=\(\dfrac{5a-3b}{5c-3d}\)(2)
Từ (1) và (2) suy ra:
\(\dfrac{5a+3b}{5c+3d}\)=\(\dfrac{5a-3b}{5c-3d}\)=\(\dfrac{5a+3b}{5a-3b}\)=\(\dfrac{5c+3d}{5c-3d}\) (đpcm)
c) Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)
Do đó: \(\dfrac{a}{c}\).\(\dfrac{b}{d}\)=\(\left(\dfrac{a}{c}\right)^2\)và \(\dfrac{a}{c}\).\(\dfrac{b}{d}\)=\(\left(\dfrac{b}{d}\right)^2\)
=>\(\dfrac{ab}{cd}\)=\(\dfrac{a^2}{c^2}\) và \(\dfrac{ab}{cd}\)=\(\dfrac{b^2}{d^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{ab}{cd}\)=\(\dfrac{a^2}{c^2}\)=\(\dfrac{b^2}{d^2}\)=\(\dfrac{7a^2}{7c^2}\)=\(\dfrac{8b^2}{8d^2}\)=\(\dfrac{3ab}{3cd}\)=\(\dfrac{7a^2+3ab}{7c^2+3cd}\)(1)
Ta có: \(\dfrac{a^2}{c^2}\)=\(\dfrac{b^2}{d^2}\)=> \(\dfrac{11a^2}{11c^2}\)=\(\dfrac{8b^2}{8d^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a^2}{c^2}\)=\(\dfrac{b^2}{d^2}\)=\(\dfrac{11a^2}{11c^2}\)=\(\dfrac{8b^2}{8d^2}\)=\(\dfrac{11a^2-8b^2}{11c^2-8d^2}\)(2)
Từ (1) và (2) suy ra:
\(\dfrac{7a^2+3ab}{7c^2+3cd}\)=\(\dfrac{11a^2-8b^2}{11c^2-8d^2}\)=\(\dfrac{7a^2+3ab}{11a^2-8b^2}\)=\(\dfrac{7c^2+3cd}{11c^2-8d^2}\)

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

a, Ta có: \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\)\(\Rightarrow\dfrac{\left(bk-b\right)^2}{\left(ck-c\right)^2}=\dfrac{bk.b}{dk.d}\)

\(\Rightarrow\dfrac{\left[b.\left(k-1\right)\right]^2}{\left[d.\left(k-1\right)\right]^2}=\dfrac{b^2}{d^2}\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2}{d^2}\)

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

b, Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{5a}{5c}=\dfrac{3b}{3d}\)

\(\Rightarrow\dfrac{5a+3b}{5c+3d}=\dfrac{5a+3b}{5c+3d}\)

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

a) \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2}{d^2}\)\(=\dfrac{\dfrac{a}{k}.b}{\dfrac{c}{k}.d}=\dfrac{ab}{cd}=VT\)

Vậy...

b) \(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{5k+3}{5k-3}\)

\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{5k+3}{5k-3}\)

Suy ra \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)

c) \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7\left(bk\right)^2+3\left(bk\right).b}{11\left(bk\right)^2-8b^2}\)\(=\dfrac{7k^2+3k}{11k^2-8}\)

\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7\left(dk\right)^2+3\left(dk\right).d}{11\left(dk\right)^2-8d^2}=\dfrac{7k^2+3k}{11k^2-8}\)

Suy ra \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)

a) Có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

=> \(ad=bc\)

=> \(\dfrac{a}{c}=\dfrac{b}{d}\) => \(\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2=\dfrac{ab}{cd}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\)

(theo tính chất dãy tỉ số bằng nhau)

=> (đpcm)

b) Có: \(\dfrac{a}{b}=\dfrac{c}{d}\) => \(\dfrac{a}{c}=\dfrac{b}{d}\)

=> \(\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)(theo tính chất dãy tỉ số bằng nhau)

=> \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\) (đpcm)

c) Có: \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

=> \(\dfrac{a^2}{c^2}=\dfrac{ab}{cd}=\dfrac{b^2}{d^2}\)          => \(\dfrac{7a^2}{7c^2}=\dfrac{3ab}{3cd}=\dfrac{11a^2}{11c^2}=\dfrac{8b^2}{8d^2}\)

=> \(\dfrac{7a^2+3ab}{7c^2+3cd}=\dfrac{11a^2-8b^2}{11c^2-8d^2}\) (theo tính chất dãy tỉ số bằng nhau)

=> \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)(đpcm)

#Ayumu