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b) \(\left|4-7x\right|-\dfrac{3}{2}:5=\left|-1\dfrac{1}{3}\right|\)
\(\left|4-7x\right|-\dfrac{3}{10}=\dfrac{4}{3}\)
\(\left|4-7x\right|=\dfrac{49}{30}\) (*)
+) Nếu 4 - 7x \(\ge\) 0 \(\Rightarrow x\le\dfrac{4}{7}\)
PT (*) \(\Leftrightarrow4-7x=\dfrac{49}{30}\)
\(-7x=-\dfrac{71}{30}\)
x = \(\dfrac{71}{210}\) (t/m)
+) Nếu \(4-7x< 0\Rightarrow x>\dfrac{4}{7}\)
Pt (*) \(\Leftrightarrow-4+7x=\dfrac{49}{30}\)
x = \(\dfrac{169}{210}\) (t/m)
Vậy x=\(\dfrac{71}{210}\) hoặc x = \(\dfrac{169}{210}\)
c, \(\left(7-3x\right)\left(2x+1\right)=0\)
=> \(7-3x=0\) hoặc \(2x+1=0\)
\(3x=7-0\) hoặc \(2x=0-1\)
\(3x=7\) hoặc \(2x=-1\)
\(x=7:3\) hoặc \(x=-1:2\)
\(x=\dfrac{7}{3}\) hoặc \(x=-0,5\)
Vậy, \(x\in\left\{\dfrac{7}{3};-0,5\right\}\)
\(d,x-5\sqrt{x}=0\)
\(ĐKXĐ:x\ge0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\\sqrt{x}=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}\)(Thỏa mãn ĐKXĐ)
Vậy...
b: 2x-3<0
=>2x<3
hay x<3/2
c: \(\left(2x-4\right)\left(9-3x\right)>0\)
=>(x-2)(x-3)<0
=>2<x<3
d: \(\dfrac{2}{3}x-\dfrac{3}{4}>0\)
=>2/3x>3/4
hay x>9/8
1) |x|=x+2
=> \(\left[{}\begin{matrix}x=x+2\\x=-x-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0=2\left(voli\right)\\2x=-2\Rightarrow x=-1\end{matrix}\right.\)
vậy x=-1
c;b tương tự
2) \(\left|x-\dfrac{3}{2}\right|=\left|\dfrac{5}{2}-x\right|\)
=> \(\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{5}{2}-x\\x-\dfrac{3}{2}=x-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\Rightarrow x=2\\0=-1\left(voli\right)\end{matrix}\right.\)
vậy x=2
\(\text{a) }3-2\left|4x-5\right|=\dfrac{2}{6}\\ \Leftrightarrow2\left|4x-5\right|=\dfrac{8}{3}\\ \Leftrightarrow\left|4x-5\right|=\dfrac{4}{3}\\ \Leftrightarrow4x-5=-\dfrac{4}{3}\text{ hoặc :}\\ 4x-5=-\dfrac{4}{3}\\ \Leftrightarrow\left[{}\begin{matrix}4x-5=-\dfrac{4}{3}\\4x-5=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{11}{3}\\4x=\dfrac{19}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=\dfrac{19}{12}\end{matrix}\right.\\ \text{Vậy }x=\dfrac{11}{12}\text{ hoặc }x=\dfrac{19}{12}\)
sao có mỗi ý a vậy bạn ?