giúp mình vớiiii

\(\Leftrightarrow12a^2-4b^2=3a^2+3b^2\)

\(\Leftrightarrow9a^2=7b^2\)

\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{7}{9}\)

hay \(\dfrac{a}{b}\in\left\{\dfrac{\sqrt{7}}{3};-\dfrac{\sqrt{7}}{3}\right\}\)

\(\dfrac{3a^2-b^2}{a^2+b^2}=\dfrac{3}{4}\)

\(\Leftrightarrow4.\left(3a^2-b^2\right)=3\left(a^2+b^2\right)\)

\(\Leftrightarrow12a^2-4b^2=3a^2+3b^2\)

\(\Leftrightarrow12a^2-3a^2=3b^2+4b^2\)

\(\Leftrightarrow9a^2=7b^2\)

\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{7}{9}\)

\(\text{hoặc }\dfrac{a}{b}=\pm\dfrac{\sqrt{7}}{3}\)

làm ơn trả lời hộ mk với ah mai mk phải nộp bài r

Ta có công thức tổng quát như sau:

\(A=n^k+n^{k+1}+n^{k+2}+...+n^{k+x}\Rightarrow A=\dfrac{n^{k+x+1}-n^k}{n-1}\)

Áp dụng ta có:

\(A=1+4+4^2+...+4^6=\dfrac{4^7-1}{3}\) 

\(\Rightarrow B-3A=4^7-3\cdot\dfrac{4^7-1}{3}=1\)

______

\(A=2^0+2^1+...+2^{2008}=2^{2009}-1\)

\(\Rightarrow B-A=2^{2009}-2^{2009}+1=1\)

_____

\(A=1+3+3^2+....+3^{2006}=\dfrac{3^{2007}-1}{2}\)

\(\Rightarrow B-2A=3^{2007}-2\cdot\dfrac{3^{2007}-1}{2}=1\)

a: \(A=-2y^3-3y^4+2\)

\(B=7y^4+2y^3+3y^2-3y-3\)

b: \(A+B=4y^4+3y^2-3y-1\)

\(A-B=-10y^4-4y^3-3y^2+3y+5\)