a,\(x-\dfrac{3}{5}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}+\dfrac{3}{5}\)
\(x=\dfrac{6}{5}\)
b,\(\left|x\right|-\dfrac{4}{5}=\dfrac{2}{5}\)
\(\left|x\right|=\dfrac{2}{5}+\dfrac{4}{5}\)
\(\left|x\right|=\dfrac{6}{5}\)
\(\Rightarrow x=\pm\dfrac{6}{5}\)
c,\(\dfrac{x}{-5}=\dfrac{24}{15}\)
\(x=\dfrac{-5.24}{15}\)
\(x=\dfrac{-24}{5}\)
d,Áp dụng tc dãy TSBN, ta có:
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x-y}{4-5}=\dfrac{21}{-1}=-21\)
+\(\dfrac{x}{4}=-21\Rightarrow x=-21.4=-84\)
+\(\dfrac{y}{5}=-21\Rightarrow y=-21.5=-105\)
Vậy x=-84 ; y=-105
a/ \(x-\dfrac{3}{5}=\dfrac{3}{5}\)
\(\Leftrightarrow x=\dfrac{3}{5}+\dfrac{3}{5}\)
\(\Leftrightarrow x=\dfrac{6}{5}\)
Vậy...
b/ \(\left|x\right|-\dfrac{4}{5}=\dfrac{2}{5}\)
\(\Leftrightarrow\left|x\right|=\dfrac{2}{5}+\dfrac{4}{5}\)
\(\Leftrightarrow\left|x\right|=\dfrac{6}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=-\dfrac{6}{5}\end{matrix}\right.\)
Vậy...
c/ \(\dfrac{x}{-5}=\dfrac{24}{15}\)
\(\Leftrightarrow15x=-120\)
\(\Leftrightarrow x=-8\)
Vậy...
c/ Theo t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x-y}{4-5}=\dfrac{21}{-1}=-21\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=-21\\\dfrac{y}{5}=-21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-84\\y=-105\end{matrix}\right.\)
Vậy..
a, \(x-\dfrac{3}{5}=\dfrac{3}{5}\)
\(x\) \(=\dfrac{3}{5}+\dfrac{3}{5}=\dfrac{6}{5}\)
b, \(\left|x\right|-\dfrac{4}{5}=\dfrac{2}{5}\)
\(\left|x\right|\) \(=\dfrac{2}{5}+\dfrac{4}{5}=\dfrac{6}{5}\)
=> \(x=\pm\dfrac{6}{5}\)
c, \(\dfrac{x}{-5}=\dfrac{24}{15}\)
=> \(x:\left(-5\right)=24:15\)
=> \(x\) \(=\left(24:15\right).\left(-5\right)=-8\)
d, \(\dfrac{x}{4}=\dfrac{y}{5}\) và \(x-y=21\)
Theo tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x-y}{4-5}=\dfrac{21}{-1}=-21\)
Suy ra:
\(\dfrac{x}{4}=-21\) => \(\left(-21\right).4=-84\)
\(\dfrac{y}{5}=-21\) => \(\left(-21\right).5=-105\)
Vậy x = -84 và y = -105.
a \(\dfrac{6}{5}\)
b \(\dfrac{6}{5}\) và - \(\dfrac{6}{5}\)
c, -8
d,x=-84
y= -105
