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Bai 1: 5.2x+3.2x+2 = 5.25 + 3.27
=> 5.2x + 3.22.2x = 5.25 + 3.27
=> 2x .(5+3.22) = 25 . (5+3.22)
=> 2x = 25
=> x=5
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{5k+3}{5k-3}\)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{5k+3}{5k-3}\)
Do đó: \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
Bài 1:
Dễ rồi nhé bn tự lm đi nha
Bài 2:
\(a)\dfrac{5a+3b}{5a-3b}=\dfrac{5a+3d}{5c-3d}\Leftrightarrow\dfrac{5bk+3b}{5bk-3b}=\dfrac{5dk+3d}{5dk-3d}\)
Xét VT \(\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-2}\left(1\right)\)
Xét VP \(\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k-3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\rightarrowđpcm\)
\(b)\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
\(\left\{{}\begin{matrix}b,d\ne0\\11a^2\ne8b^2\\11c^2\ne8d^2\end{matrix}\right.\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\left(\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\right)\Rightarrow\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7.\dfrac{a^2}{b^2}+3\dfrac{a}{b}}{11.\dfrac{a^2}{b^2}-8}=\dfrac{7.\dfrac{c^2}{d^2}+3\dfrac{c}{d}}{11.\dfrac{c^2}{d^2}-8}=\dfrac{7c^2+3cd}{11c^2-8d^2}=VP\)
Chúc bạn học tốt!
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{7a-4b}{3a+5b}=\dfrac{7bk-4b}{3bk+5b}=\dfrac{7k-4}{3k+5}\)
\(\dfrac{7c-4d}{3c+5d}=\dfrac{7dk-4d}{3dk+5d}=\dfrac{7k-4}{3k+5}\)
Do đó: \(\dfrac{7a-4b}{3a+5b}=\dfrac{7c-4d}{3c+5d}\)
b: \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2\)
Do đó: \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)