??Đề kiểu j đây??

\(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{2c+3d}\Rightarrow\dfrac{2a-3d}{2c-3d}=\dfrac{2a+3b}{2c-3d}\)

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

vì a/b = c/d

theo dãy tỉ số bằng nhau ta có

a/b =c/d = a+c/b+d = a-c/b-d (đỗi vị trí)

⇒  2a-2b/2a+3b = 2c-3d/2c-3d

\(ab\cdot\sqrt{\dfrac{a}{3b}}-a^2\sqrt{\dfrac{3b}{a}}\)

\(=a\sqrt{ab}-a^2\cdot\dfrac{\sqrt{3b}}{\sqrt{a}}\)

\(=a\sqrt{ab}-a\sqrt{a}\cdot\sqrt{3b}\)

\(=a\sqrt{ab}\left(1-\sqrt{3}\right)\)

\(\Leftrightarrow m=\dfrac{a\sqrt{ab}\left(1-\sqrt{3}\right)}{\sqrt{3ab}}=\dfrac{a\left(\sqrt{3}-3\right)}{3}\)

b.\(ĐK:x;y\in Z^+;x;y\ne0\)

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{5}{x}+\dfrac{5}{y}=1\)

\(\Leftrightarrow\dfrac{5}{x}=1-\dfrac{5}{y}\)

\(\Leftrightarrow\dfrac{5}{x}=\dfrac{y-5}{y}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{y-5}\)

\(\Leftrightarrow x=\dfrac{5y}{y-5}\)

\(\Leftrightarrow x=5+\dfrac{25}{y-5}\) ( bạn chia \(5y\) cho \(y-5\) ý )

Để x;y là số nguyên dương thì \(25⋮y-5\) hay \(y-5\in U\left(25\right)=\left\{\pm1;\pm5;\pm25\right\}\)

TH1: 

\(y-5=1\) 

\(\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=30\end{matrix}\right.\) ( tm )   ( bạn thế y=6 vào \(x=5+\dfrac{25}{y+5}\) nhé )

Xét tương tự, ta ra được nghiệm nguyên dương của phương trình:

\(\left\{{}\begin{matrix}x=30\\y=6\end{matrix}\right.\)  \(\left\{{}\begin{matrix}x=10\\y=10\end{matrix}\right.\)  \(\left\{{}\begin{matrix}x=6\\y=30\end{matrix}\right.\)

Bài này chắc phải giải theo kiểu lớp 7

Giải:

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

\(\dfrac{2a}{3b}=\dfrac{3b}{4c}=\dfrac{4c}{5d}=\dfrac{5d}{2a}=\dfrac{2a+3b+4c+5d}{3b+4c+5d+2a}=1\)

\(\Rightarrow\left\{{}\begin{matrix}2a=3b\\3b=4c\\4c=5d\\5d=2a\end{matrix}\right.\)\(\Rightarrow2a=3b=4c=5d\)

\(\Rightarrow C=\dfrac{2a}{3b}+\dfrac{3b}{4c}+\dfrac{4c}{5d}+\dfrac{5d}{2a}\)

\(=\dfrac{2a}{2a}+\dfrac{2a}{2a}+\dfrac{2a}{2a}+\dfrac{2a}{2a}\)

\(=1+1+1+1\)

\(=4\)

Vậy \(C=4\)

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

\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\left(1\right)\)

\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\left(2\right)\)

Từ (1) và (2) suy ra \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3b}\left(=\dfrac{2k+3}{2k-3}\right)\)

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

\(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=>\dfrac{2a}{2c}=\dfrac{3b}{3d}=>\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3d}{2c-3d}=>\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\left(đpcm\right)\)

\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Leftrightarrow\left\{{}\begin{matrix}2b+c-a=2a\\2c-b+a=2b\\2a+b-c=2c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a-2b=c\\3b-2c=a\\3c-2a=b\end{matrix}\right.\text{ và }\left\{{}\begin{matrix}3a-c=2b\\3b-a=2c\\3c-b=2a\end{matrix}\right.\\ \Leftrightarrow P=\dfrac{a\cdot b\cdot c}{2a\cdot2b\cdot3c}=\dfrac{1}{8}\)