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Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
Ta có: \(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{b+a+d}=\frac{d}{c+b+a}\)
\(\Rightarrow\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{b+a+d}+1=\frac{d}{c+b+a}+1\)
\(\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{b+a+d}=\frac{a+b+c+d}{c+b+a}\)
Mà a+b+c+d khác 0
=> b+c+d = a+c+d = b+a+d = c+b+a
=> b = a = c = d
Ta có:
\(P=\frac{2a+5b}{3c+4d}-\frac{2b+5c}{3d+4a}-\frac{2c+5d}{3a+4b}-\frac{2d+5a}{3c+4b}\)
\(P=\frac{2a+5a}{3a+4a}-\frac{2b+5b}{3b+4b}-\frac{2c+5d}{3c+4c}-\frac{2d+5d}{3d+4d}\)
\(P=\frac{7a}{7a}-\frac{7b}{7b}-\frac{7c}{7c}-\frac{7d}{7d}\)
\(P=1-1-1-1=-2\)
Thiếu đề <3