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a)
$|3x-2|=2x\Rightarrow x\geq 0$.
Xét 2 TH:
TH1: $x\geq \frac{2}{3}$ thì pt trở thành:
$3x-2=2x\Leftrightarrow x=2$ (thỏa mãn)
TH2: $0\leq x< \frac{2}{3}$ thì pt trở thành:
$2-3x=2x\Leftrightarrow x=\frac{2}{5}$ (thỏa mãn)
b)
PT $\Rightarrow x\geq 0$
$\Rightarrow |4+2x|=4+2x$. PT trở thành:
$4+2x=4x\Leftrightarrow x=2$ (thỏa mãn)
c)
Xét các TH sau:
TH1: $x\geq \frac{3}{2}$. Khi đó, pt trở thành:
$2x-3=-x+21$
$\Leftrightarrow x=8$ (thỏa mãn)
TH2: $x< \frac{3}{2}$. Khi đó, pt trở thành:
$3-2x=-x+21$
$\Leftrightarrow x=-18$ (thỏa mãn)
d)
Từ PT suy ra $x-2\geq 0\Leftrightarrow x\geq 2(*)$
Khi đó: $|3x-1|=3x-1$. PT trở thành:
$3x-1=x-2$
$\Leftrightarrow 2x=-1<0\Rightarrow x<0$ (mâu thuẫn với $(*)$)
Vậy PT vô nghiệm.
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1) \(\Rightarrow16x^2+24x+9+9x^2-24x+16+4-25x^2=x\)
\(\Rightarrow x=29\)
2)
a) \(=x^2-9-x^2+6x-9=6x-18\)
b) \(=\left(3x-1+2x+1\right)^2=\left(5x\right)^2=25x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>2x^2+2x-3x-3+x^2+2x=3x^2+12x+12
=>12x+12=x-3
=>11x=-15
=>x=-15/11
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)|2x+1|=5`
`<=>` \(\left[ \begin{array}{l}2x+1=5\\2x+1=-5\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\)
`b)|2x+1|=0`
`<=>2x+1=0`
`<=>2x=-1`
`<=>x=-1/2`
`c)|2x+1|=7`
`<=>` \(\left[ \begin{array}{l}2x+1=7\\2x+1=-7\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=6\\2x=-8\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\)
`d)|2x+5|=|3x-7|`
`<=>` \(\left[ \begin{array}{l}2x+5=3x-7\\2x+5=7-3x\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=12\\5x=2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=12\\x=\dfrac25\end{array} \right.\)
`e)|2x+7|=1`
`<=>` \(\left[ \begin{array}{l}2x+7=1\\2x+7=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=-6\\2x=-8\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=3\\x=-4\end{array} \right.\)
`g)|x-2|+|2x-3|=2`
Nếu `x>=2=>|x-2|=x-2,|2x-3|=2x-3`
`pt<=>x-2+2x-3=2`
`<=>3x-5=2`
`<=>3x=7`
`<=>x=7/3(tm)`
Nếu `x<=3/2=>|x-2|=2-x,|2x-3|=3-2x`
`pt<=>2-x+3-2x=2`
`<=>5-3x=2`
`<=>3x=3`
`<=>x=1(tm)`
Nếu `3/2<=x<=2=>|x-2|=2-x,|2x-3|=2x-3`
`pt<=>2-x+2x-3=2`
`<=>x-1=2`
`<=>x=3(l)`
`h)|x+2|+|1-x|=3x+2`
Vì `VT>=0=>3x+2>=0=>x>=-2/3`
`=>|x+2|=x+2`
`pt<=>x+2+|1-x|=3x+2`
`<=>|1-x|=2x(x>=0)`
`<=>` \(\left[ \begin{array}{l}2x=1-x\\2x=x-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}3x=1\\x=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac13(TM)\\x=-1(KTM)\end{array} \right.\)
a.
$|2x+1|=5$
\(\Leftrightarrow \left[\begin{matrix}
2x+1=5\\
2x+1=-5\end{matrix}\right.\Leftrightarrow \left[\begin{matrix}
x=2\\
x=-3\end{matrix}\right.\)
b.
$|2x+1|=0$
$\Leftrightarrow 2x+1=0$
$\Leftrightarrow x=-\frac{1}{2}$
c.
$|2x+1|=7$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=7\\ 2x+1=-7\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=3\\ x=-4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+3\right)^2+\left(x-2\right)^2=2x^2\\ \Leftrightarrow x^2+6x+9+x^2-4x+4=2x^2\\ \Leftrightarrow2x=-13\Leftrightarrow x=-\dfrac{13}{2}\)
<=> x2+6x+9+x2+4x+4-2x2 =0
<=>10x+13=0
<=> x= -13/10
Chúc bạn học tốt nha!
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Mk xin phép ko vt lại đề nx
\(\Rightarrow A=\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x^2-1\right)\right]\div x+1\)
\(\Rightarrow A=3x-2-\left(2x-5\right)\left(x-1\right)\)
\(\Rightarrow x=\dfrac{1}{2}\)
\(\Rightarrow A=\dfrac{3}{2}-2-\left(1-5\right)\left(\dfrac{1}{2}-1\right)=-\dfrac{5}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b, ( 5/2 - x ) ^2
=25/4-4/5x+x^2
c,( xy/2 - x/3 ) ( xy/2 + x/3)
=(xy/2)^2-(x/3)^2
c: \(\left(\dfrac{xy}{2}-\dfrac{x}{3}\right)\left(\dfrac{xy}{2}+\dfrac{x}{3}\right)=\dfrac{x^2y^2}{4}-\dfrac{x^2}{9}\)
e: \(\left(2x+3y\right)^2=4x^2+12xy+9y^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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a: \(=\dfrac{x-2x-1}{x+1}=\dfrac{-\left(x+1\right)}{x+1}=-1\)
b: \(=\dfrac{2+2x}{x\left(x+1\right)}=\dfrac{2\left(x+1\right)}{x\left(x+1\right)}=\dfrac{2}{x}\)
c: \(=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{18x^2-12x+2}{2\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x-2)3+6(x+1)2-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6x2+12x+6-x3+12=0
\(\Rightarrow\)24x+10=0
\(\Rightarrow\)24x=-10
\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)
b)(x-5)(x+5)-(x+3)2+3(x-2)2=(x+1)2-(x-4)(x+4)+3x2
\(\Rightarrow\)x2-25-(x2+6x+9)+3(x2-4x+4)=x2+2x+1-(x2-16)+3x2
\(\Rightarrow\)x2-25-x2-6x-9+3x2-12x+12=x2+2x+1-x2+16+3x2
\(\Rightarrow\)3x2-18x-22=3x2+2x+17
\(\Rightarrow\)3x2-18x-22-3x2-2x-17=0
\(\Rightarrow\)-20x-39=0
\(\Rightarrow\)-20x=39
\(\Rightarrow\)x=\(-\dfrac{39}{20}\)
(2x - 3)(x + 1) + x(x - 2) = 3(x + 2)2
<=> 2x2 - x - 3 + x2 - 2x = 3(x2 + 4x + 4)
<=> 3x2 - 3x - 3 = 3x2 + 12x + 12
<=> 15x = 15
<=> x = 1
Vậy x = 1 là nghiệm của phương trình
( 2x - 3 )( x + 1 ) + x( x - 2 ) = 3( x + 2 )2
<=> 2x2 - x - 3 + x2 - 2x = 3( x2 + 4x + 4 )
<=> 3x2 - 3x - 3 - 3x2 - 12x - 12 = 0
<=> -15x - 15 = 0
<=> x = -1
Vậy phương trình có một nghiệm x = -1