
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


\(2x+\frac{1}{2}=\frac{-5}{3}\)
\(2x=\frac{-5}{3}-\frac{1}{2}\)
\(2x=\frac{-10}{6}-\frac{3}{6}\)
\(2x=\frac{-13}{6}\)
\(x=\frac{-13}{6}:2\)
\(x=\frac{-13}{12}\)

a/ \(\left|x-3\right|=\left|4-x\right|\) \(\Rightarrow\left[\begin{matrix}x-3=4-x\\x-3=x-4\end{matrix}\right.\) => x = \(\frac{7}{2}\)
b/ \(\left|2x-1\right|+5=5\Rightarrow\left|2x-1\right|=0\Rightarrow x=\frac{1}{2}\)
c/ \(\left|x-1\right|.\left(-\frac{1}{2}\right)^3=\left(-\frac{1}{2}\right)^5\)
\(\Rightarrow\left|x-1\right|=\left(-\frac{1}{2}\right)^5:\left(-\frac{1}{2}\right)^3\)
\(\Rightarrow\left|x-1\right|=\frac{1}{4}\)
\(\Rightarrow x-1=\pm\frac{1}{4}\)
\(\Rightarrow\left[\begin{matrix}x=\frac{5}{4}\\x=\frac{3}{4}\end{matrix}\right.\)
d/ \(3x-\left|2x+1\right|=2\)
\(\Rightarrow\left[\begin{matrix}3x-\left(2x+1\right)=2\\3x-\left(-2x-1\right)=2\end{matrix}\right.\)
\(\Rightarrow x=3\) (loại x = 1/5)
e/ \(\left|x\right|+\left|-x\right|=3-x\)
Xét với x = 0 không thỏa mãn
Xét với x < 0 thì : (-x) + (-x) = 3 - x => x = -3
Xét với x > 0 thì : x + x = 3 - x => x = 1
Vậy x = 1 hoặc x = -3

nhìu dữ
a)3/2
b)-1/3
c)-5/6
d)0
e)-1/2
Bài 2
a=3
b=1/2
c=-1/3
d=0
e=9
f=-2/3

Bài 2:
a: \(A=-\left(x-2\right)^2+3\le3\)
Dấu '=' xảy ra khi x=2
b: \(B=-\left(2x-3\right)^2+\dfrac{1}{2}\le\dfrac{1}{2}\)
Dấu '=' xảy ra khi x=3/2
c: \(C=-\left(2x-\dfrac{3}{5}\right)^2-\dfrac{1}{3}\le-\dfrac{1}{3}\)
Dấu '=' xảy ra khi x=3/10
d: \(D=-\left|\dfrac{5}{3}-x\right|\le0\)
Dấu '=' xảy ra khi x=5/3
e: \(E=-\left|x-\dfrac{1}{10}\right|+9\le9\)
Dấu '=' xảy ra khi x=1/10
f: \(F=-\left|\dfrac{1}{2}x+1\right|-\dfrac{2}{3}\le-\dfrac{2}{3}\)
Dấu '=' xảy ra khi x=-2

\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự