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\(\frac{3a^2-b^2}{a^2+b^2}\)=\(\frac{3}{4}\)
=>4.(3a2-b=2)=3.(a2+b2)
4.3a2-4.b2=3.a2+3.b2
12a2-4b2=3a2+3b2
12a2-3a2=4b2+3b2
9a2=7b2
\(\frac{a^2}{b^2}\)=\(\frac{7}{9}\)
=>\(\frac{a}{b}\)=\(\sqrt{\frac{7}{9}}\)
đặt a/b=t
chia cả tử mấu cho b^2
\(\Leftrightarrow\frac{3t^2-1}{t^2+1}=\frac{3}{4}\)\(12t^2-4=3t^2+3\Rightarrow9t^2=7\Rightarrow\orbr{\begin{cases}\frac{a}{b}=t=\frac{\sqrt{7}}{3}\\\frac{a}{b}=t=\frac{-\sqrt{7}}{3}\end{cases}}\)
a/b=
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Ta có \(\frac{2a+b+c}{b+c}=\frac{2b+c+a}{c+a}=\frac{2c+a+b}{a+b}\Rightarrow\frac{2a}{b+c}+1=\frac{2b}{a+c}+1=\frac{2c}{a+b}+1\)
=> \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{3}{2}\)
^_^
Bài 1: Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=k\)
\(\Rightarrow\hept{\begin{cases}a=2016k\\b=2017k\\c=2018k\end{cases}}\).Thay vào M,ta có:
\(M=4\left(2016k-2017k\right)\left(2017k-2018k\right)-\left(2018k-2016k\right)^2\)
\(=4.\left(-1k\right)\left(-1k\right)-\left(2k\right)^2\)
\(=4k^2-4k^2=0\)
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\(a^2+b^2+c^2=\frac{b^2-c^2}{a^2+3}+\frac{c^2-a^2}{b^2+4}+\frac{a^2-b^2}{c^2+5}\)
<=>\(a^2-\frac{a^2-b^2}{c^2+5}+b^2-\frac{b^2-c^2}{a^2+3}+c^2-\frac{c^2-a^2}{b^2+4}=0\)
<=>\(\frac{ac^2+4a^2+b^2}{c^2+5}+\frac{ba^2+4b^2+c^2}{a^2+3}+\frac{ab^2+4c^2+a^2}{b^2+4}=0\)
Vì \(VT\ge0\) nên dấu "=" xảy ra khi a=b=c=0 => S = 2017 + bc + 20c=2017+0.0+20.0=2017
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Ta có:
\(a^2+b^2+c^2=\frac{b^2-c^2}{3+a^2}+\frac{c^2-a^2}{4+b^2}+\frac{a^2-b^2}{5+c^2}\)
\(\Leftrightarrow a^2+\frac{a^2}{4+b^2}-\frac{a^2}{5+c^2}+b^2+\frac{b^2}{5+c^2}-\frac{b^2}{3+a^2}+c^2+\frac{c^2}{3+a^2}-\frac{c^2}{4+b^2}=0\)
\(\Leftrightarrow a^2.\frac{b^2c^2+4b^2+5c^2+21}{\left(4+b^2\right)\left(5+c^2\right)}+b^2.\frac{a^2c^2+6a^2+2c^2+13}{\left(3+a^2\right)\left(5+c^2\right)}+c^2.\frac{a^2b^2+3a^2+4b^2+13}{\left(3+a^2\right)\left(4+b^2\right)}=0\)
Dấu = xảy ra khi \(a=b=c=0\)
Thế vô ta có: \(S=2016ab+bc+20c=0\)
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\(\frac{3a^2-b^2}{a^2+b^2}\)=\(\frac{3}{4}\)
=>4(3a2-b2)=3(a2+b2)
=>12a2-4b2=3a2+3b2
=>12a2-3a2=4b2+3b2
=>9a2=7b2
=>\(\frac{a^2}{b^2}\)=\(\frac{7}{9}\)
=>\(\frac{a}{b}\)=\(\sqrt{\frac{7}{9}}\)
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\(\frac{a}{b}=\frac{3}{4}\) \(< =>\frac{a}{3}=\frac{b}{4}\)
Đặt \(\frac{a}{3}=\frac{b}{4}=k=>a=3k;b=4k\)
Ta có: \(A=\frac{a^2+b^2}{a^2-b^2}=\frac{\left(3k\right)^2+\left(4k\right)^2}{\left(3k\right)^2-\left(4k\right)^2}=\frac{9k^2+16k^2}{9k^2-16k^2}=\frac{\left(9+16\right).k^2}{\left(9-16\right).k^2}=\frac{28k^2}{-7k^2}=\frac{28}{-7}=-4\)
Vậy A=-4
\(\frac{a}{b}=\frac{3}{4}\Rightarrow4a=3b\Rightarrow\frac{a}{3}=\frac{b}{4}\)
\(\Rightarrow\frac{a}{3}=\frac{b}{4}=\frac{a^2}{9}=\frac{b^2}{16}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a^2}{9}=\frac{b^2}{16}=\frac{a^2+b^2}{9+16}=\frac{a^2-b^2}{9-16}\)
\(\Rightarrow\left(a^2+b^2\right)\left(9-16\right)=\left(a^2-b^2\right)\left(9+16\right)\)
\(\Rightarrow\left(a^2+b^2\right)\left(-7\right)=\left(a^2-b^2\right)25\)
\(\Rightarrow\frac{a^2+b^2}{a^2-b^2}=\frac{25}{-7}\)