a) 2x - 3 = -5
b) 13 - ( 3x - 4 ) = -7
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`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.
$|2x+1|=7$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=7\\ 2x+1=-7\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=3\\ x=-4\end{matrix}\right.\)
\(A-B=\left(3x^4+x^3-x^2+3x-5\right)-\left(7x^4-x^3+2x^2-x+4\right)\)
\(=3x^4+x^3-x^2+3x-5-7x^4+x^3-2x^2+x-4\)
\(=\left(3x^4-7x^4\right)+\left(x^3+x^3\right)-\left(x^2+2x^2\right)+\left(3x+x\right)-\left(5+4\right)\)
\(=-4x^4+2x^3-3x^2+4x-9\)
a: 2x-3/2+3/4=-4
=>2x-3/4=-4
=>2x=-13/4
hay x=-13/8
b: \(\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)\cdot\left(\dfrac{-3}{2}-\dfrac{10}{3}\right)=\dfrac{2}{5}\)
\(\Leftrightarrow-\dfrac{2}{3}x-\dfrac{3}{5}=\dfrac{2}{5}:\dfrac{-29}{6}=\dfrac{-2}{5}\cdot\dfrac{6}{29}=\dfrac{-12}{145}\)
=>2/3x+3/5=12/145
=>2/3x=-15/29
hay x=-45/58
c: \(\dfrac{x}{2}-\left(\dfrac{3}{5}x-\dfrac{13}{5}\right)=-\left(\dfrac{7}{10}x+\dfrac{7}{5}\right)\)
=>1/2x-3/5x+13/5=-7/10x-7/5
=>-1/10x+7/10x=-7/5-13/5
=>3/5x=-2
hay x=-2:3/5=-10/3
a: \(\left\{{}\begin{matrix}3x-2y=4\\2x+y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x-2y=4\\4x+2y=10\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x=14\\2x+y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=5-2x=5-2\cdot2=1\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}-x+2y=2\\2x-y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2x+4y=4\\2x-y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y=3\\x-2y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=1\\x=-2+2y=-2+2\cdot1=0\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}2x-y=13\\y-5=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-y=13\\y=-7+5=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=y+13=-2+13=11\\y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{11}{2}\\y=-2\end{matrix}\right.\)
d: \(\left\{{}\begin{matrix}3x+y=8\\2x-3y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}9x+3y=24\\2x-3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=25\\3x+y=8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{25}{11}\\y=8-3x=8-3\cdot\dfrac{25}{11}=8-\dfrac{75}{11}=\dfrac{13}{11}\end{matrix}\right.\)
a; -2\(x\) - 3.(\(x-17\)) = 34 - 2.( - \(x\) + 25)
- 2\(x\) - 3\(x\) + 51 = 34 + 2\(x\) - 50
2\(x\) + 2\(x\) + 3\(x\) = - 34 + 50 + 51
7\(x\) = 67
\(x\) = 67 : 7
\(x\) = \(\dfrac{67}{7}\)
Vậy \(x\) = \(\dfrac{67}{7}\)
b; 17\(x\) + 3.(- 16\(x\) - 37) = 2\(x\) + 43 - 4\(x\)
17\(x\) - 48\(x\) - 111 = 2\(x\) - 4\(x\) + 43
- 31\(x\) - 2\(x\) + 4\(x\) = 111 + 43
- \(x\) x (31 + 2 - 4) = 154
- \(x\) x (33 - 4) = 154
- \(x\) x 29 = 154
- \(x\) = 154 : (-29)
\(x\) = - \(\dfrac{154}{29}\)
Vậy \(x=-\dfrac{154}{29}\)
a) \(2x-4< 5\)
\(\Leftrightarrow\) \(2x< 4+5\)
\(\Leftrightarrow\) \(x< 4,5\)
b) \(4-3x\ge6\)
\(\Leftrightarrow\) \(-3x\ge-4+6\)
\(\Leftrightarrow-3x\ge2\)
\(\Leftrightarrow\) \(x\le-0,6\)
c) \(3x-7< 5x-2\)
\(\Leftrightarrow\) \(3x-5x< 7-2\)
\(\Leftrightarrow\) \(-2,5x< 5\)
\(\Leftrightarrow x>-2,5\)
a: =>2x=-2
hay x=-1
b: =>3x-4=20
=>3x=24
hay x=8
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