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Ta có:
\(\dfrac{a}{3}=\dfrac{b}{5}\Leftrightarrow a=\dfrac{3b}{5}\)
Khi đó:
\(b^2-a^2=36\Leftrightarrow b^2-\dfrac{9b^2}{25}=36\\ \Leftrightarrow\dfrac{16b^2}{25}=36\Leftrightarrow b^2=\dfrac{225}{4}\Leftrightarrow b=\dfrac{\pm15}{2}\)
Với \(b=\dfrac{15}{2}\) suy ra: \(a=\dfrac{3b}{5}=\dfrac{3}{5}.\dfrac{15}{2}=\dfrac{9}{2}\)
Với \(b=\dfrac{-15}{2}\) suy ra: \(a=\dfrac{3b}{5}=\dfrac{3}{5}.\dfrac{-15}{2}=\dfrac{-9}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{a}{5}=\dfrac{b}{4}\Rightarrow\dfrac{a^2}{25}=\dfrac{b^2}{16}\)
Áp dụng tính chất DTSBN :
\(\dfrac{a^2}{25}=\dfrac{b^2}{16}=\dfrac{a^2-b^2}{25-16}=\dfrac{1}{9}\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{1}{9}\cdot25=\dfrac{25}{9}\\b^2=\dfrac{1}{9}\cdot16=\dfrac{16}{9}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\dfrac{5}{3};b=\dfrac{4}{3}\\a=\dfrac{-5}{3};b=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(\left(a;b\right)\in\left\{\left(\dfrac{5}{3};\dfrac{4}{3}\right);\left(-\dfrac{5}{3};-\dfrac{4}{3}\right)\right\}\)
b) \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)
Áp dụng tính chất DTSBN :
\(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}=\dfrac{2c^2}{32}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=4.4=16\\b^2=4.9=36\\c^2=4,16=64\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=4;=6;c=8\\a=-4;b=-6;c=-8\end{matrix}\right.\)
Vậy (a;b;c) \(\in\left\{\left(4;6;8\right);\left(-4;-6;-8\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\)
b) U1 + V1= 180o (kề bù)
V1= 180o -U1 = 180o - 36o= 144o
U2 = V1 (đồng vị)
=> U2= 144o
Vậy V1= U2= 144o
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Sửa \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)
Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\Rightarrow a=2k;b=3k;c=4k\)
\(a^2-b^2+2c^2=108\\ \Rightarrow4k^2-9k^2+32k^2=108\\ \Rightarrow27k^2=108\Rightarrow k^2=4\\ \Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4;y=6;z=8\\x=-4;y=-6;z=-8\end{matrix}\right.\)
Ta có:
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{a^2}{2^2}=\dfrac{b^2}{3^2}=\dfrac{2c^2}{2.4^2}=\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{2c^2}{32}\)
Áp dụng tcdtsbn , ta có:
\(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{2c^2}{32}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4\)
\(\Rightarrow\left\{{}\begin{matrix}a=8\\b=12\\c=16\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{3a^2-b^2}{a^2+b^2}=\dfrac{3}{4}\)
\(\Leftrightarrow4\cdot\left(3a^2-b^2\right)=3\left(a^2+b^2\right)\)
\(\Leftrightarrow12a^2-4b^2=3a^2+3b^2\)
\(\Leftrightarrow12a^2-3a^2=3b^2+4b^2\)
\(\Leftrightarrow9a^2=7b^2\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{7}{9}\)
hay \(\dfrac{a}{b}=\pm\dfrac{\sqrt{7}}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(=\dfrac{\left(3^3\right)^{15}\left(3^2\right)^{20}}{\left(3^4\right)^{12}\cdot3^{36}}=\dfrac{3^{45}\cdot3^{40}}{3^{48}\cdot3^{36}}=3\)
ta kó: a:b=3:4 => \(\frac{a}{3}=\frac{b}{4}\)=>\(\frac{a^2}{9}=\frac{b^2}{16}\)=\(\frac{a^2+b^2}{9+16}=\frac{36}{25}\)\(=\left(\frac{6}{5}\right)^2\)
=>a2/9 =(6/5)2 =>(a/3)2=(6/5)2 =>a/3=6/5 hoặc a/3=-6/5
a/3=6/5 =>5a=18 =>a=18/5
a/3=-6/5 =>5a=-18=>a=-18/5
=>b2/16=(6/5)2 =>(b/4)2=(6/5)2=>b/4=6/5 hoặc b/4=-6/5
b/4=6/5 =>5b=24=>b=24/5
b/4=-6/5 =>5a=-24=>a=-24/5
vậy a=18/5 ; b=24/5
hoặc a=-18/5; b=-24/5