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d: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{3\cdot\left(dk\right)^2+5\cdot\left(bk\right)^2}{3d^2+5b^2}=k^2\)
\(\dfrac{c^2}{d^2}=\dfrac{\left(dk\right)^2}{d^2}=k^2\)
Do đó: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{c^2}{d^2}\)
Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
a: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a}{b}-1=\dfrac{c}{d}-1\)
hay \(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b}{c}=2\)
\(\Rightarrow P=2+2+2=6\)
\(TH1:a+b+c+d\ne0\)
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
\(\Rightarrow a=b=c=d\)
\(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\)
\(=1+1+1+1\)
\(=4\)
\(TH2:a+b+c+d=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\)
\(=-\dfrac{c+d}{c+d}-\dfrac{d+a}{d+a}-\dfrac{a+b}{a+b}-\dfrac{b+c}{b+c}\)
\(=-1-1-1-1\)
\(=-4\)
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=k\)(k<>0)
=>\(\left\{{}\begin{matrix}d=a\cdot k\\c=d\cdot k=a\cdot k\cdot k=ak^2\\b=ck=ak^3\\a=bk=ak^4\end{matrix}\right.\)
\(a=ak^4\)
=>\(ak^4-a=0\)
=>\(a\left(k^4-1\right)=0\)
=>\(k^4-1=0\)
=>\(\left[{}\begin{matrix}k=1\\k=-1\end{matrix}\right.\)
\(M=\dfrac{3a+b}{5b-d}+\dfrac{2b+3c}{2d-b}\)
\(=\dfrac{3\cdot ak^4+ak^3}{5\cdot ak^3-ak}+\dfrac{2\cdot ak^3+3\cdot ak^2}{2\cdot ak-ak^3}\)
\(=\dfrac{ak^3\left(3k+1\right)}{ak\left(5k^2-1\right)}+\dfrac{ak^2\left(2k+3\right)}{ak\left(2-k^2\right)}\)
\(=\dfrac{k^2\left(3k+1\right)}{5k^2-1}+\dfrac{k\left(2k+3\right)}{2-k^2}\)
TH1: k=1
=>\(M=\dfrac{1^2\left(3\cdot1+1\right)}{5\cdot1^2-1}+\dfrac{1\left(2\cdot1+3\right)}{2-1^2}=\dfrac{4}{4}+\dfrac{5}{1}=6\)
TH2: k=-1
=>\(M=\dfrac{\left(-1\right)^2\cdot\left(-3+1\right)}{5\cdot\left(-1\right)^2-1}+\dfrac{\left(-1\right)\left(2\cdot\left(-1\right)+3\right)}{2-\left(-1\right)^2}\)
\(=\dfrac{-2}{4}+\dfrac{1}{1}=-\dfrac{1}{2}+1=\dfrac{1}{2}\)
cíu tuii
ghép câu thành có nghĩa: H/ồ/g/.../B/a/o/n