a) \(\left(2\dfrac{3}{4}-1\dfrac{4}{5}\right)\cdot x=1\)
\(\left(\dfrac{11}{4}-\dfrac{9}{5}\right)\cdot x=1\)
\(\dfrac{19}{20}x=1\)
\(x=\dfrac{20}{19}\)
Vậy \(x=\dfrac{20}{19}\)
b) \(\left(x^2-9\right)\left(3-5x\right)=0\)
TH1:
\(x^2-9=0\)
\(x^2=9\)
\(x^2=3^2=\left(-3\right)^2\)
=>\(x\in\left\{3;-3\right\}\)
TH2:
\(3-5x=0\)
\(5x=3\)
\(x=\dfrac{3}{5}\)
Vậy \(x\in\left\{3;-3;\dfrac{3}{5}\right\}\)
\(a,\dfrac{19}{20}x=1\)
x=\(\dfrac{21}{19}\)
b,\(\left(x^2-9\right)=0hoac\left(3-5x\right)=0\)
x^2=9 hoac 5x=3
x=3 hoac x= \(\dfrac{3}{5}\)
c./3x-1 /+\(\dfrac{11}{4}\)= \(\dfrac{49}{16}\)
/3x-1/= \(\dfrac{5}{16}\)
3x =\(\dfrac{21}{16}\)
x=\(\dfrac{7}{16}\)
c) \(\left|3x-1\right|+2\dfrac{3}{4}=3\dfrac{1}{16}\)
\(\left|3x-1\right|=3\dfrac{1}{16}-2\dfrac{3}{4}\)
\(\left|3x-1\right|=\left(3-2\right)+\left(\dfrac{1}{16}-\dfrac{3}{4}\right)\)
\(\left|3x-1\right|=1+\dfrac{-11}{16}\)
\(\left|3x-1\right|=\dfrac{5}{16}\)
=> \(3x-1\in\left\{\dfrac{5}{16};\dfrac{-5}{16}\right\}\)
TH1:
\(3x-1=\dfrac{5}{16}\)
\(3x=\dfrac{21}{16}\)
\(x=\dfrac{7}{16}\)
TH2:
\(3x-1=\dfrac{-5}{16}\)
\(3x=\dfrac{11}{16}\)
\(x=\dfrac{11}{48}\)
Vậy \(x\in\left\{\dfrac{11}{48};\dfrac{7}{16}\right\}\)
a) Ta có: \(\left(2\dfrac{3}{4}-1\dfrac{4}{5}\right)x=1\)
\(\Leftrightarrow x\cdot\left(\dfrac{11}{4}-\dfrac{9}{5}\right)=1\)
\(\Leftrightarrow x\cdot\dfrac{55-36}{20}=1\)
\(\Leftrightarrow x=\dfrac{20}{19}\)
c) Ta có: \(\left|3x-1\right|+2\dfrac{3}{4}=3\dfrac{1}{16}\)
\(\Leftrightarrow\left|3x-1\right|=\dfrac{49}{16}-\dfrac{11}{4}=\dfrac{49}{16}-\dfrac{44}{16}=\dfrac{5}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=\dfrac{5}{16}\\3x-1=\dfrac{-5}{16}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{21}{16}\\3x=\dfrac{11}{16}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=\dfrac{11}{48}\end{matrix}\right.\)
