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a) \(\left|4x-1\right|-\left|3x-\dfrac{1}{2}\right|=0\\ \Leftrightarrow\left|4x-1\right|=\left|3x-\dfrac{1}{2}\right|\\ \Leftrightarrow\left[{}\begin{matrix}4x-1=3x-\dfrac{1}{2}\\4x-1=\dfrac{1}{2}-3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x-3x=1-\dfrac{1}{2}\\4x+3x=\dfrac{1}{2}+1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{3}{14}\right\}\) là nghiệm của pt.
b) \(\left|x-1\right|-2x=\dfrac{1}{2}\\ \Leftrightarrow\left|x-1\right|=2x+\dfrac{1}{2}\left(ĐK:x\ge\dfrac{-1}{4}\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x+\dfrac{1}{2}\\x-1=-2x-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-2x=1+\dfrac{1}{2}\\x+2x=1-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{3}{2}\\3x=\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\left(ktmđk\right)\\x=\dfrac{1}{6}\left(tmđk\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{6}\) là nghiệm của pt.
Lời giải:
a.
$|4x-1|-|3x-\frac{1}{2}|=0$
$\Leftrightarrow |4x-1|=|3x-\frac{1}{2}$
\(\Leftrightarrow \left[\begin{matrix} 4x-1=3x-\frac{1}{2}\\ 4x-1=\frac{1}{2}-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=\frac{3}{14}\end{matrix}\right.\)
b. Nếu $x\geq 1$ thì:
$|x-1|-2x=\frac{1}{2}$
$\Leftrightarrow x-1-2x=\frac{1}{2}$
$\Leftrightarrow -x-1=\frac{1}{2}$
$\Leftrightarrow x=\frac{-3}{2}$ (vô lý vì $x\geq 1$)
Nếu $x< 1$ thì:
$1-x-2x=\frac{1}{2}$
$\Leftrightarrow x=\frac{1}{6}$ (tm)
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\(a.\)
\(\left(x-8\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
\(S=\left\{8,-2\right\}\)
\(b.\)
\(\left(4x-3\right)-\left(x+5\right)=3\cdot\left(10-x\right)\)
\(\Leftrightarrow4x-3-x-5-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{38}{6}\)
\(S=\left\{\dfrac{38}{6}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x - 8 )( x3 + 8) = 0
\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x^3=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
b)(4x - 3) – ( x + 5) = 3(10 - x)
\(\Leftrightarrow4x-3-x-5=30-3x\)
\(\Leftrightarrow3x-8=30-3x\)
\(\Leftrightarrow3x-8-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{19}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
d: Ta có: (-4,6)+x=-3,5
nên x=-3,5+4,6
hay x=1,1
a: Ta có: \(1.5-\left|x-0.3\right|=0\)
\(\Leftrightarrow\left|x-0.3\right|=1.5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-0.3=1.5\\x-0.3=-1.5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.8\\x=-1.2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow-x^2-3x+x+3+x^2-6x=11\)
=>-8x+3=11
=>-8x=8
hay x=-1
b: \(\Leftrightarrow3x^2-15x+x-5-3x^2+3x=5\)
=>-11x=10
hay x=-10/11
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2
P(x) + Q(x) = x3 – 6x + 2 + 2x2 - 4x3 + x - 5 = - 3x3 + 2x2 – 5x - 3
P(x) - Q(x) = x3 – 6x + 2 - 2x2 + 4x3 - x + 5 = 5x3 − 2x2 − 7x+7
![](https://rs.olm.vn/images/avt/0.png?1311)
a: 2x-1=0
nên 2x=1
hay x=1/2
b: 4x2-16=0
=>(x-2)(x+2)=0
=>x=2 hoặc x=-2
c: x2-2x=0
=>x(x-2)=0
=>x=0 hoặc x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\Leftrightarrow2\left|3x-1\right|=\dfrac{4}{5}\)
\(\Leftrightarrow\left|3x-1\right|=\dfrac{2}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=\dfrac{2}{5}\\3x-1=-\dfrac{2}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{15}\\x=\dfrac{1}{5}\end{matrix}\right.\)
b)TH1: \(x\ge3\)
\(\Leftrightarrow x+5+x-3=9\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\left(tm\right)\)
TH2: \(-5\le x< 3\)
\(\Leftrightarrow x+5-x+3=9\Leftrightarrow8=9\left(VLý\right)\)
TH3: \(x< -5\)
\(\Leftrightarrow-x-5-x+3=9\Leftrightarrow2x=-11\Leftrightarrow x=-\dfrac{11}{2}\left(tm\right)\)
\(a,2.|3x-1|-\dfrac{3}{4}=\dfrac{1}{20}\)
\(2.|3x-1|=\dfrac{1}{20}+\dfrac{3}{4}\)
\(2.|3x-1|=\dfrac{4}{5}\)
\(|3x-1|=\dfrac{4}{5}:2\)
\(|3x-1|=\dfrac{2}{5}\)
\(\Rightarrow3x-1=\pm\dfrac{2}{5}\)
\(3x-1=\dfrac{2}{5}\)
\(3x=\dfrac{2}{5}+1\)
\(3x=\dfrac{7}{5}\)
\(x=\dfrac{7}{5}:3\)
\(x=\dfrac{7}{15}\)
\(3x-1=-\dfrac{2}{5}\)
\(3x=-\dfrac{2}{5}+1\)
\(3x=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:3\)
\(x=\dfrac{1}{5}\)
Đặt x^2+1=t => pt có dạng t^2+3tx+2x^2=0
<=> (t+x)(t+2x)=0
TH1: t+x=0
<=> x^2+1+x=0
=> pt vô nghjệm
TH2: t+2x=0
<=> x^2+1+2x=0
<=> (x+1)^2=0
<=> x=-1
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