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\(\Leftrightarrow12a^2-4b^2=3a^2+3b^2\)
\(\Leftrightarrow9a^2=7b^2\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{7}{9}\)
hay \(\dfrac{a}{b}\in\left\{\dfrac{\sqrt{7}}{3};-\dfrac{\sqrt{7}}{3}\right\}\)
\(\dfrac{3a^2-b^2}{a^2+b^2}=\dfrac{3}{4}\)
\(\Leftrightarrow4.\left(3a^2-b^2\right)=3\left(a^2+b^2\right)\)
\(\Leftrightarrow12a^2-4b^2=3a^2+3b^2\)
\(\Leftrightarrow12a^2-3a^2=3b^2+4b^2\)
\(\Leftrightarrow9a^2=7b^2\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{7}{9}\)
\(\text{hoặc }\dfrac{a}{b}=\pm\dfrac{\sqrt{7}}{3}\)
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a: \(A=-2y^3-3y^4+2\)
\(B=7y^4+2y^3+3y^2-3y-3\)
b: \(A+B=4y^4+3y^2-3y-1\)
\(A-B=-10y^4-4y^3-3y^2+3y+5\)
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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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theo bài ra ta có:
\(\frac{3a^2-b^2}{a^2+b^2}=\frac{3}{4}\)
=> 4( 3a2 - b2) = 3( a2+ b2)
=> 12a2- 4b2 = 3a2 + 3b2
=> 12a2- 3a2 = 3b2 + 4b2
=> 9a2= 7b2
=> \(\frac{7}{9}=\frac{a^2}{b^2}\Rightarrow\sqrt{\frac{7}{9}}=\frac{a}{b}\)
vậy \(\sqrt{\frac{7}{9}}=\frac{a}{b}\)