Giải các phương trình :
a) \(\left|2x\right|=3x-2\)
b) \(\left|-3,5x\right|=1,5x+5\)
c) \(\left|x+15\right|=3x-1\)
d) \(\left|2-x\right|=0,5x-4\)
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17 tháng 5 2022
a: \(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(\dfrac{1}{2}x\right)^2-\left(2x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(\dfrac{1}{2}x-2x+3\right)\left(\dfrac{1}{2}x+2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(3-\dfrac{3}{2}x\right)\left(\dfrac{5}{2}x-3\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{\dfrac{6}{5}\right\}\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{4}{3}\\\left(3x+4\right)^2-\left(2x\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{4}{3}\\\left(5x+4\right)\left(x+4\right)=0\end{matrix}\right.\)
\(\Leftrightarrow x=-\dfrac{4}{5}\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=12\\\left(5x-x+12\right)\left(5x+x-12\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=12\\\left(4x+12\right)\left(6x-12\right)=0\end{matrix}\right.\)
hay \(x\in\varnothing\)
d: \(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{10}{3}\\\left(2,5x-1,5x-5\right)\left(2,5x+1,5x+5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{10}{3}\\\left(x-5\right)\left(4x+5\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{5}{4};5\right\}\)
2 tháng 6 2023
a: =>x+3=x-2 hoặc x+3=2-x
=>2x=-1
=>x=-1/2
b: =>3x+7=x-2 hoặc 3x+7=-x+2
=>2x=-9 hoặc 4x=-5
=>x=-5/4 hoặc x=-9/2
c: =>|3x-4|=|2x-5|
=>3x-4=2x-5 hoặc 3x-4=-2x+5
=>x=-1 hoặc x=9/5
16 tháng 2 2021
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
H
1 tháng 6 2023
\(\left|x\right|=x+1\)
Ta có : \(\left\{{}\begin{matrix}x\ge0\\x< 0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=x+1\\-x=x+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0=1\\-2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0=1\left(ktm\right)\\x=-\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{-\dfrac{1}{2}\right\}\)
__
\(\left|3x\right|=x-2\)
Ta có : \(\left\{{}\begin{matrix}3x\ge0\Leftrightarrow x\ge0\\3x< 0\Leftrightarrow x< 0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=x-2\\-3x=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-2\\-4x=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\left(ktm\right)\)
Vâỵ phương trình vô nghiệm
__
\(\left|-2x\right|=3x-4\)
Ta có : \(\left\{{}\begin{matrix}-2x\ge0\Leftrightarrow x\ge0\\-2x< 0\Leftrightarrow x< 0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=3x-4\\-\left(-2x\right)=3x-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-5x=-4\\2x=3x-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\-x=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\left(tm\right)\\x=4\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{4\right\}\)
14 tháng 3 2020
e sẽ cố gắng !!!
\(3x-15=2x\left(x-5\right)\)
\(3x-15=2x^2-10x\)
\(3x-15-2x^2+10x=0\)
\(13x-15-2x^2=0\)
\(x\left(13-2x\right)-15=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\13-2x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\-2-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\2x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
14 tháng 3 2020
\(f,x\left(2x-7\right)-4x+14=0\)
\(2x^2-7x-4x+14=0\)
\(2x^2-11x+14=0\)
\(x\left(2x-11\right)=-14\)
\(\Rightarrow\orbr{\begin{cases}x=-14\\2x-11=-14\end{cases}\Rightarrow\orbr{\begin{cases}x=-14\\2x=-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-14\\x=-\frac{3}{2}\end{cases}}}\)
LD
31 tháng 3 2022
\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)
\(\Leftrightarrow-28x+37\ge12\)
\(\Leftrightarrow-28x\ge12-37\)
\(\Leftrightarrow-28x\ge-25\)
\(\Leftrightarrow x\le\dfrac{25}{28}\)
Vậy \(S=\left\{x\left|x\le\dfrac{25}{28}\right|\right\}\)
b, \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)
\(\Leftrightarrow x^2-x^2-6x\ge9+5+16\)
\(\Leftrightarrow-6x\ge30\)
\(\Leftrightarrow x\le-5\)
Vậy \(S=\left\{x\left|x\le-5\right|\right\}\)
\(c,\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)
\(\Leftrightarrow9x^2-6x-1-9x^2+36< 5x\)
\(\Leftrightarrow9x^2-9x^2-6x-5x+36+1< 0\)
\(\Leftrightarrow-11x+37< 0\)
\(\Leftrightarrow-11x< -37\)
\(\Leftrightarrow x>\dfrac{37}{11}\)
vậy \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)
21 tháng 3 2021
a)(2x+1)(3x-2)=(5x-8)(2x+1)
⇔(2x+1)(3x-2)-(5x-8)(2x+1)=0
⇔(2x+1)(3x-2-5x+8)=0
⇔(2x+1)(-2x+6)=0
⇔2x+1=0 hoặc -2x+6=0
1.2x+1=0⇔2x=-1⇔x=-1/2
2.-2x+6=0⇔-2x=-6⇔x=3
phương trình có 2 nghiệm x=-1/2 và x=3
23 tháng 5 2022
a: =>|x-7|=3-2x
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)
b: =>|2x-3|=4x+9
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)
c: =>3x+5=2-5x hoặc 3x+5=5x-2
=>8x=-3 hoặc -2x=-7
=>x=-3/8 hoặc x=7/2
a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2-2x\right)\left(3x-2+2x\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(x-2\right)\left(5x-2\right)=0\end{matrix}\right.\)
hay x=2
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{10}{3}\\\left(-3,5x-1,5x-5\right)\left(-3,5x+1,5x+5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{10}{3}\\\left(-5x-5\right)\left(-2x+5\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-1;\dfrac{5}{2}\right\}\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(3x-1-x-15\right)\left(3x-1+x+15\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(2x-16\right)\left(4x+14\right)=0\end{matrix}\right.\Leftrightarrow x=8\)
d: \(\Leftrightarrow\left|x-2\right|=0,5x-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=8\\\left(0,5x-4-x+2\right)\left(0,5x-4+x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=8\\\left(-0,5x-2\right)\left(1,5x-6\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)