bài 2 : a) \(\dfrac{a-1}{2}=\dfrac{b+3}{4}=\dfrac{c-5}{6}\)

áp dụng dảy tỉ số bằng nhau

ta có : \(\dfrac{5\left(a-1\right)-3\left(b+3\right)-4\left(c-5\right)}{5.2-3.4-4.6}\)

\(=\dfrac{5a-5-3b-9-4c+20}{10-12-24}=\dfrac{\left(5a-3b-4c\right)-5-9+20}{-26}\)

\(=\dfrac{46+6}{-26}=\dfrac{52}{-26}=-2\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a-1}{2}=-2\\\dfrac{b+3}{4}=-2\\\dfrac{c-5}{6}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a-1=-4\\b+3=-8\\c-5=-12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=-11\\c=-7\end{matrix}\right.\)

vậy \(a=-3;b=-11;c=-7\)

b) ta có : \(3a=2b\Leftrightarrow6a=4b=5c\Leftrightarrow\dfrac{6a}{2}=\dfrac{4b}{2}=\dfrac{5c}{2}\)

áp dụng dảy tỉ số bằng nhau

ta có \(\dfrac{-60a-60b+60c}{-10.2-15.2+12.2}=\dfrac{60\left(-a-b+c\right)}{-20-30+24}\)

\(=\dfrac{60\left(-52\right)}{-26}=\dfrac{-3120}{-26}=120\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{6a}{2}=120\\\dfrac{4b}{2}=120\\\dfrac{5c}{2}=120\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6a=240\\4b=240\\5c=240\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=40\\b=60\\c=48\end{matrix}\right.\)

vậy \(a=40;b=60;c=48\)

\(\dfrac{3a+4b}{5a-6b}=\dfrac{3c+4d}{5c-6d}\)

=> \(\dfrac{3a+4b}{3c+4d}=\dfrac{5a-6b}{5c-6d}\)

ta có

\(\dfrac{3a+4b}{3c+4d}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{a}{c}=\dfrac{b}{d}=>\dfrac{a}{b}=\dfrac{c}{d}\)(đpcm)

Ta có:

\(\dfrac{3a+4b}{5a-6b}=\dfrac{3c+4d}{5c-6d}\)

\(\Leftrightarrow\left(3a+4b\right)\left(5c-6d\right)=\left(3c+4d\right)\left(5a-6b\right)\)

\(\Rightarrow15ac-18ad+20bc-24bd=15ac-18bc+20ad-24bd\)

\(\Rightarrow15ac-15ac-18ad-20ad=-24bd+24bd-18bc-20bc\)

\(\Rightarrow-38ad=-38bc\)

\(\Rightarrow ad=bc\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)

\(\frac{\left(5a-4b\right)6}{36}=\frac{\left(6a-4c\right)5}{25}=\frac{\left(6b-5c\right)4}{16}=\frac{\left(5a-4b\right)6-\left(6a-4c\right)5+\left(6b-5c\right)4}{36-25+16}=\frac{0}{27}\)

\(\Rightarrow5a=4b\Leftrightarrow\frac{a}{4}=\frac{b}{5}\)

\(\Rightarrow6a=4c\Leftrightarrow\frac{a}{4}=\frac{c}{6}\)

\(\Rightarrow\frac{a}{4}=\frac{b}{5}=\frac{c}{6}\)

ờ, vậy chúc hai n` giải toán zui zẻ

 Mk săpp thi rồi nên hơi nhiều bài mong mn giúp mk !!!

\(1,\\ a,3^{2^3}=3^8>3^6=\left(3^2\right)^3\\ b,\left(-8\right)^9=\left(-2\right)^{27}< \left(-2\right)^{25}=\left(-32\right)^5\\ c,2^{21}=8^7< 9^7=3^{14}\\ 2,\)

\(a,\) Áp dụng tcdtsbn:

\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)

\(b,\) Sửa: \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow a=bk;c=dk\)

\(\Leftrightarrow\dfrac{ab}{cd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2};\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2}{d^2}\\ \LeftrightarrowĐpcm\)

b) Ta có : \(\dfrac{2a}{3}=\dfrac{3b}{4}=\dfrac{4c}{5}\)

\(\Leftrightarrow\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}=\dfrac{a+b+c}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

Khi đó \(a=12.\dfrac{3}{2}=18;b=12.\dfrac{4}{3}=16;c=12.\dfrac{5}{4}=15\)

Vậy (a,b,c) = (18,16,15) 

Ta có : a - 1 / 2 = b + 3 / 4 = c - 5 / 6

<=> 5a - 5 / 10 = 3b + 9 / 12 = 4c - 20 / 24

Áp dụng t/c dãy tỉ số bằng nhau , ta có :

5a - 5 / 10 = 3b + 9 / 12 = 4c - 20 / 24 = ( 5a - 3b - 4c ) - 5 - 9 + 20 / 10 - 12 - 24 = 52/-26 = -2

=> a - 1 / 2 = -2 <=> a = -3

=> b + 3 / 4 = -2 <=> b = -5

=> c - 5 / 6 = -2 <=> c = -7

Vậy a = -3 ; b = -5 ; c = -7