\(\frac{a}{3}=\frac{b}{4}=\frac{c}{5}\Rightarrow\frac{3a}{9}=\frac{2b}{8}=\frac{c}{5}\)

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

\(\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=\frac{3a-2b+c}{9-8+5}=\frac{36}{6}=6\)

=> a = 18; b =24; c = 30

Ap dung tinh chat day ti so bang nhau ta co 

\(\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=\frac{3a-2b+c}{9-8+5}=\frac{36}{6}=6\)

\(a=18;b=24;c=30\)

Ta có :\(|x-2|-|1-2x|=0\)

\(\Rightarrow|x-2|=|1-2x|\)

\(\Rightarrow\orbr{\begin{cases}x-2=1-2x\\x-2=2x-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=3\\x=-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Vậy \(x=1\)hoặc \(x=-1\)

\(\left|x-2\right|-\left|1-2x\right|=0\)

\(\Leftrightarrow\left|x-2\right|=\left|1-2x\right|\)Xet tung TH :

TH1 :  \(x-2=1-2x\Leftrightarrow3x=3\Leftrightarrow x=1\)

TH2 : \(x-2=-1+2x\Leftrightarrow-x=1\Leftrightarrow x=-1\)

Vay \(x=\left\{\pm1\right\}\)

a) Ta có : \(f\left(0\right)=2.0^2-10=-10\)

               \(f\left(1\right)=2.1^2-10=-8\)

               \(f\left(-1\frac{1}{2}\right)=f\left(\frac{-3}{2}\right)=2.\left(\frac{-3}{2}\right)^2-10=2.\frac{9}{4}-10=\frac{9}{2}-10=\frac{-11}{2}\)

b)Vì \(f\left(x\right)=2\)

\(\Rightarrow2x^2-10=-2\)

\(\Rightarrow2x^2=8\)

\(\Rightarrow x^2=4\)

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

Vậy \(x=2\)hoặc \(x=-2\)

a, \(f\left(0\right)=2.0^2-10=-10\)

\(f\left(1\right)=2.1^2-10=2-10=-8\)

Ta co \(-1\frac{1}{2}=-\frac{3}{2}\)

\(f\left(-\frac{3}{2}\right)=2.\left(-\frac{3}{2}\right)^2-10=2.\frac{9}{4}-10=\frac{18}{4}-\frac{40}{4}=-\frac{22}{4}=-\frac{11}{2}\)

b, Ta co : \(f\left(x\right)=-2\)hay \(2x^2-10=-2\Leftrightarrow2x^2=8\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)

ko thấy rõ đề bn