Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x-\dfrac{1}{3}=\dfrac{2}{3}.\dfrac{9}{14}+\dfrac{3}{7}\)
\(x-\dfrac{1}{3}=\dfrac{1}{7}+\dfrac{3}{7}\)
\(x-\dfrac{1}{3}=\dfrac{4}{7}\)
\(x=\dfrac{19}{21}\)
\(\left|2x-3\right|=3-2x\)
\(ĐK:x\le\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3-2x\\3-2x=3-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\left(đúng\right)\end{matrix}\right.\)
Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)
\(\dfrac{1}{2}-\dfrac{5}{12}x=\dfrac{2}{3}\)
\(\dfrac{5}{12}x=\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{3}{6}-\dfrac{4}{6}\)
\(\dfrac{5}{12}x=\dfrac{-1}{6}\)
\(x=\dfrac{-1}{6}:\dfrac{5}{12}=\dfrac{-1}{6}.\dfrac{12}{5}\)
\(x=\dfrac{-2}{5}\)
\(3\left|4x-1\right|-2=19\)
\(3\left|4x-1\right|=21\)
\(\left|4x-1\right|=7\)
⇔\(\left[{}\begin{matrix}4x-1=7\\4x-1=-7\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left|4x-1\right|=21:3=7\\ \Rightarrow\left[{}\begin{matrix}4x-1=7\\4x-1=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
a/\(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x=1+\dfrac{5}{7}+\dfrac{13}{14}\)
\(x=\dfrac{14}{14}+\dfrac{10}{14}+\dfrac{13}{14}\)
\(x=\dfrac{37}{14}\)
Vậy \(x=\dfrac{37}{14}\)
b/\(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{3}{5}+\dfrac{6}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{55}{15}-\dfrac{27}{15}\)
\(x=\dfrac{28}{15}\)
Vậy \(x=\dfrac{28}{15}\)
#kễnh
a) \(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x-\dfrac{23}{14}=1\)
\(x=1+\dfrac{23}{14}\)
\(x=\dfrac{37}{14}\)
b) \(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+1+\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{28}{15}\)
a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)
=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)
=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)
b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)
=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)
=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)
Với $x=1$ ta có :
$-7.(x+3)^3 .|2x-1|+42$
$=-7.(-1+3)^3.|2.(-1)-1|+42$
$=-7.2^3.|-3|+42$
$=-7.8.3 + 42$
$=-126$
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}+\dfrac{x+y-3}{z}\\ =\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=\dfrac{2\left(z+y+x\right)}{x+y+z}=2\\ \to\left\{{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\end{matrix}\right.\to\left\{{}\begin{matrix}x+y+z=3x-1\\x+y+z=3y-2\\x+y+z=3z+3\end{matrix}\right.\)
Mặt khác \(\dfrac{1}{x+y+z}=2\to x+y+z=\dfrac{1}{2}\)
\(\to\left\{{}\begin{matrix}3x-1=\dfrac{1}{2}\\3y-2=\dfrac{1}{2}\\3z+3=\dfrac{1}{2}\end{matrix}\right.\to\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=-\dfrac{5}{6}\end{matrix}\right.\)
Ta có : \(\frac{x-1}{3}=\frac{y-4}{6}=\frac{y-4-\left(x-1\right)}{6-3}\) \(=\frac{y-4-x+1}{3}=\frac{y-x-3}{3}=\frac{1}{3}\)
Nên : \(\frac{x-1}{3}=\frac{1}{3}\Rightarrow x-1=1\Rightarrow x=2\)
\(\frac{y-4}{6}=\frac{1}{3}\Rightarrow y-4=2\Rightarrow y=6\)
Vậy x = 2 ; y = 6