Ta có: \(\frac{a+3b}{a-3b}=\frac{c+3d}{c-3d}\)

\(\rightarrow\left(a+3b\right)\left(c-3d\right)=\left(a-3b\right)\left(c+3d\right)\)

\(\rightarrow ac+3bc-3ad-9bd=ac-3bc+3ad-9bd\)

\(\rightarrow3bc-3ad=3ad-3bc\)

\(\rightarrow6bc=6ad\)

\(\rightarrow bc=ad\rightarrow\frac{a}{c}=\frac{b}{d}\left(đpcm\right)\)

Chúc bn học tốt

Vì 3a=12b=>\(\frac{a}{12}=\frac{b}{3}\)

=>\(\frac{a}{60}=\frac{b}{15}\)

Vì 7b=5c=>\(\frac{b}{5}=\frac{c}{7}\)

=>\(\frac{b}{15}=\frac{c}{21}\)

=>\(\frac{a}{60}=\frac{b}{15}=\frac{c}{21}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

=>\(\frac{a}{60}=\frac{b}{15}=\frac{c}{21}=\frac{a-b+c}{60-15+21}=\frac{16}{33}\)

=>\(\frac{a}{60}=\frac{16}{33}=>a=16.60:33=\frac{320}{11}\)

=>\(\frac{b}{15}=\frac{16}{33}=>b=15.16:33=\frac{80}{11}\)

=>\(\frac{c}{21}=\frac{16}{33}=>c=16.21:33=\frac{112}{11}\)

Vậy a=\(\frac{320}{11}\)

       b=\(\frac{80}{11}\)

       c=\(\frac{112}{11}\)

Ta có: \(2|5x-3|-2x=14\)

\(\Leftrightarrow2|5x-3|=14+2x\)

\(\Leftrightarrow|5x-3|=7+x\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=10\\6x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{-2}{3}\end{cases}}\)

Vậy \(x\in\left\{\frac{5}{2};\frac{-2}{3}\right\}\)

\(2\left|5x-3\right|-2x=14\)

\(\Rightarrow2\left|5x-3\right|=14+2x\)

\(\Rightarrow\left|5x-3\right|=\left(14+2x\right):2=7+x\)

\(\Rightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}4x=10\\6x=-4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{-2}{3}\end{cases}}\)

Vậy \(x\in\left\{\frac{5}{2};\frac{-2}{3}\right\}\)

Chúc em học tốt nhé!

Ta có : \(\left|x-2011\right|\ge0;\left|x-200\right|\ge0\)

            =>|x-2011|+|x-200|\(\ge0\)

            =>A\(\ge0\)

Dấu bằng xảy ra <=> x-2011=0<=>x=2011

                                  x-200=0<=>x=200

Vậy Amin=0<=>x\(\in\left\{2011;200\right\}\)

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

 \(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}\) =>\(\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)

=> \(\hept{\begin{cases}\frac{x}{10}=2\\\frac{y}{6}=2\\\frac{z}{21}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.10=20\\y=2.6=12\\z=2.21=42\end{cases}}\)

Vậy ...

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