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a: \(\Leftrightarrow\left|\dfrac{5}{3}x\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{5}{3}=\dfrac{1}{6}\\x\cdot\dfrac{5}{3}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}:\dfrac{5}{3}=\dfrac{3}{30}=\dfrac{1}{10}\\x=-\dfrac{1}{10}\end{matrix}\right.\)
b: \(\Leftrightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x-1\right|=\dfrac{3}{2}:\dfrac{3}{4}=2\)
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
c: \(\Leftrightarrow\left|x+\dfrac{3}{5}\right|=\left|x-\dfrac{7}{3}\right|\)
\(\Leftrightarrow x+\dfrac{3}{5}=\dfrac{7}{3}-x\)
=>2x=44/15
hay x=22/15
a) \(\left|x-17\right|=2,3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-17=2,3\\x-17=-2,3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=19,3\\x=14,7\end{matrix}\right.\)
b) \(\left|x+\dfrac{3}{4}\right|=0\)
\(\Leftrightarrow x+\dfrac{3}{4}=0\Leftrightarrow x=-\dfrac{3}{4}\)
c) \(\left|x+\dfrac{3}{4}\right|+\dfrac{1}{3}=0\)
\(\Leftrightarrow\left|x+\dfrac{3}{4}\right|=-\dfrac{1}{3}\)( vô lý do \(\left|x+\dfrac{3}{4}\right|\ge0\forall x\))
Vậy \(S=\varnothing\)
A, \(x\cdot x+2x-3=0\)
\(x^2+2x-3=0\)
\(x^2+3x-x-3=0\)
\(x\left(x+3\right)-\left(x+3\right)=0\)
\(\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow x+3=0\) => x=-3
\(\Leftrightarrow x-1=0\)=> x=1
b,
\(2x^2+3x+1=0\)
\(2x^2+2x+x+1=0\)
\(2x\left(x+1\right)+\left(x+1\right)=0\)
\(\left(x+1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow x+1=0\)=> x=-1
\(\Leftrightarrow\)\(2x+1=0\)=> x=\(\frac{-1}{2}\)
\(\left(x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-4\right)=0\)
\(\left(x-4\right)\left(x+3+x+5\right)=0\)
\(\left(x-4\right)\left(2x+8\right)=0\)
\(\left[{}\begin{matrix}x-4=0\Leftrightarrow x=4\\2x+8=0\Leftrightarrow x=-4\end{matrix}\right.\)
1)\(x^2-x=x\left(x-1\right)=0\)
\(\orbr{\begin{cases}x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
B1: Đk: 5x ≥ 0 => x ≥ 0
Vì |x + 1| ≥ 0 => |x + 1| = x + 1
|x + 2| ≥ 0 => |x + 2| = x + 2
|x + 3| ≥ 0 => |x + 3| = x + 3
|x + 4| ≥ 0 => |x + 4| = x + 4
=> |x + 1| + |x + 2| + |x + 3| + |x + 4| = 5x
=> x + 1 + x + 2 + x + 3 + x + 4 = 5x
=> 4x + 10 = 5x
=> x = 10
B2: Ta có: |x - 2018| = |2018 - x|
=> A=|x + 2000| + |2018 - x| ≥ |x + 2000 + 2018 - x| = |4018| = 4018
Dấu " = " xảy ra <=> (x + 2000)(x - 2018) ≥ 0
Th1: \(\hept{\begin{cases}x+2000\ge0\\x-2018\ge0\end{cases}\Rightarrow}\hept{\begin{cases}x\ge-2018\\x\le2018\end{cases}}\Rightarrow-2018\le x\le2018\)
Th2: \(\hept{\begin{cases}x+2000\le0\\x-2018\le0\end{cases}\Rightarrow}\hept{\begin{cases}x\le-2018\\x\ge2018\end{cases}}\)(vô lý)
Vậy GTNN của A = 4018 khi -2018 ≤ x ≤ 2018
B3:
a, Vì |x + 1| ≥ 0 ; |2y - 4| ≥ 0
=> |x + 1| + |2y - 4| ≥ 0
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+1=0\\2y-4=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\y=2\end{cases}}\)
Vậy...
b, Vì |x - y + 1| ≥ 0 ; (y - 3)2 ≥ 0
=> |x - y + 1| + (y - 3)2 ≥ 0
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x-y+1=0\\y-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-y=-1\\y=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x-3=-1\\y=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=3\end{cases}}\)
Vậy...
c, Vì |x + y| ≥ 0 ; |x - z| ≥ 0 ; |2x - 1| ≥ 0
=> |x + y| + |x - z| + |2x - 1| ≥ 0
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+y=0\\x-z=0\\2x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=0\\x=z\\x=\frac{1}{2}\end{cases}\Leftrightarrow}}\hept{\begin{cases}\frac{1}{2}+y=0\\x=z=\frac{1}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}y=\frac{-1}{2}\\x=z=\frac{1}{2}\end{cases}}\)
Ta có : \(\dfrac{\left(x-3\right)\left(x+2\right)\left(x+1\right)}{\left(x+3\right)\left(x-4\right)}>0\)
- Đặt \(f\left(x\right)=\dfrac{\left(x-3\right)\left(x+2\right)\left(x+1\right)}{\left(x+3\right)\left(x-4\right)}\)
- Lập bảng xét dấu :
- Từ bảng xét dấu : - Để f(x) > 0
\(\Leftrightarrow\left[{}\begin{matrix}-3< x< -2\\-1< x< 3\\x>4\end{matrix}\right.\)
Vậy ...
1) \(|5x-3|=|7-x|\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=7-x\\5x-3=x-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x=10\\4x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)
Vậy...
2) \(2.|3x-1|-3x=7\)
\(\Leftrightarrow2.|3x-1|=7+3x\)
\(\Leftrightarrow\orbr{\begin{cases}2.\left(3x-1\right)=7+3x\\2.\left(3x-1\right)=-7-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x-2=7+3x\\6x-2=-7-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=9\\9x=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{9}\end{cases}}\)
Vậy...
\(\left(x-3\right)\left(x+\frac{3}{4}\right)>0\)<=> \(x-3\) và \(x+\frac{3}{4}\) khác dấu
\(\left(+\right)\hept{\begin{cases}x-3>0\\x+\frac{3}{4}>0\end{cases}=>\hept{\begin{cases}x>3\\x>-\frac{3}{4}\end{cases}=>x>3}}\)
\(\left(+\right)\hept{\begin{cases}x-3< 0\\x+\frac{3}{4}< 0\end{cases}=>\hept{\begin{cases}x< 3\\x< -\frac{3}{4}\end{cases}=>x< -\frac{3}{4}}}\)
Vậy \(x>3\) hoặc \(x< -\frac{3}{4}\) thì \(\left(x-3\right)\left(x+\frac{3}{4}\right)>0\)