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 \(x-2018=0\Leftrightarrow x=2018\)

396.99 minh nhanh nhat **** minh nhe!

a) \(AB^2+AC^2=8^2+15^2=17^2=BC^2\)

\(\Rightarrowđpcm\)

b) \(\left\{{}\begin{matrix}S_{ABC}=\dfrac{AB.AC}{2}\\S_{ABC}=\dfrac{AH.BC}{2}\end{matrix}\right.\)

\(\Rightarrow AB.AC=AH.BC\)

\(\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{120}{17}\left(cm\right)\)

xy+3x-2y=11

<=>xy+3x-2y-6=5

<=>x.(y+3)-2.(y+3)=5

<=>(y+3)(x-2)=5

Rồi liệt kê ra,bn tự lm tiếp đc chứ?

la 9 , bai nay trong vong 11 dung ko 

tick nha , minh cung di thi ne