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

=>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=

ko biết làm

mới học lớp 6

thôi

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

=>4(3a²-b²)=3(a²+b²)

=>12a²-4b²=3a²+3b²

=>12a²-3a²=4b²+3b²

=>9a²=7b²

=>\(\frac{a^2}{b^2}\)=\(\frac{7}{9}\)

=>\(\frac{a}{b}\)=\(\sqrt{\frac{7}{9}}\)

a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)

Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)

b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)

\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)

\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)

\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)

\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)

Do đó:  +)  \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)

+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)

+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)

\(\frac{3a^2-b^2}{a^2+b^2}=\frac{3\left(\frac{a}{b}\right)^2-1}{\left(\frac{a}{b}\right)^2+1}=\frac{3}{4}\Leftrightarrow4.3\left(\frac{a}{b}\right)^2-4=3.\left(\frac{a}{b}\right)^2+3\)

\(\Leftrightarrow9.\left(\frac{a}{b}\right)^2=7\Leftrightarrow\left(\frac{a}{b}\right)^2=\frac{7}{9}\Leftrightarrow\int^{\frac{a}{b}=\frac{\sqrt{7}}{3}}_{\frac{a}{b}=-\frac{\sqrt{7}}{3}}\)

Suy ra:

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

=> \(12a^2-4b^2=3a^2+3b^2\)

=> \(12a^2-3a^2=3b^2+4b^2\)

=> \(9a^2=7b^2=>\frac{a^2}{b^2}=\left(\frac{a}{b}\right)^2=\frac{7}{9}=>\frac{a}{b}=\sqrt{\frac{7}{9}}\)

Vay: \(\frac{a}{b}=\sqrt{\frac{7}{9}}\)

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