Tìm x, biết :
a) / 3x - 5 / = 3x - 5
b) / 7 - x / = x - 7
c) / 2x - 3 / = 3 - 2x
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18 tháng 5 2021
a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)
\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)
\(\Leftrightarrow-9x=18\)
hay x=-2
Vậy: S={-2}
b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)
\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
\(\Leftrightarrow14x=7\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)
\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)
\(\Leftrightarrow5.2x=-6.5\)
hay \(x=-\dfrac{5}{4}\)
Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)
18 tháng 5 2021
d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x+16=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
Vậy: S={-5}
e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
Vậy: S={0}
29 tháng 10 2021
1.
a) \(=x^2-6x+9+3x^2-15x=4x^2-21x+9\)
b) \(=9x^2+12x+4-x^2+9=8x^2+12x+13\)
2.
a) \(\Leftrightarrow x^2+8x+16-x^2+4-5=0\\ \Leftrightarrow8x=-15\\ \Leftrightarrow x=-\dfrac{15}{8}\)
b) \(\Leftrightarrow9x^2-6x+1-8x^2+12x-2x+3-5-x^2=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)
11 tháng 7 2021
`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)`
`<=>` \(\left[ \begin{array}{l}2x=1-x\\2x=x-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}3x=1\\x=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac13(TM)\\x=-1(KTM)\end{array} \right.\)
AH
Akai Haruma
Giáo viên
11 tháng 7 2021
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.
$|2x+1|=7$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=7\\ 2x+1=-7\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=3\\ x=-4\end{matrix}\right.\)
VN
0
21 tháng 7 2018
a) ( 2x - 3 ) - ( x - 5 ) = ( x + 7 ) - ( x + 2 )
<=> 2x - 3 - x + 5 = x + 7 - x - 2
<=> x = 3
b)(7x-5)-(6x+4)=(2x+3)-(2x+1)
<=> 7x - 5 - 6x - 4 = 2x + 3 - 2x - 1
<=> x = 11
c)(9x-3)-(8x+5)=(3x+2)
<=> 9x - 3 - 8x - 5 = 3x + 2
<=> -2x = 10
<=> x = -5
d)(x+7)-(2x+3)=(3x+5)-(2x+4)
<=> x + 7 - 2x - 3 = 3x + 5 - 2x - 4
<=> -2x = -3
<=> x = 3/2
2 tháng 2
ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠᅠ ᅠ
CW
18 tháng 6 2016
f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)
\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)
\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)
\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)
\(-x^3=27\)
\(x=-3\)
31 tháng 10 2021
\(A=x^2+6x+9-4x-1-2x-x^2=9\\ B=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ C=\left(3x+5-3x+5\right)^2=100\)
31 tháng 10 2021
a: \(A=x^2+6x+9-4x-1-2x-x^2=8\)
b: \(B=2x^2+3x-10x-15-2x^2+6x+x+7=-8\)
20 tháng 9 2021
\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)
\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)
a) |3x - 5| = 3x - 5 => 3x - 5 > 0 => 3x > 5 => x > 5/3
b) |7 - x| = x - 7 => 7 - x < 0 => - x < - 7 => x > 7
c) |2x - 3| = 3 - 2x => 2x - 3 < 0 => 2x < 3 => x < 3/2