Tìm x biết :
a. 3(2x+3)(3x-5)<0 ; b. x2+1>0 ; c. (x2+1)(2x-5)>0 ; d. x2+2x<0
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GH
2
H
12 tháng 12 2021
a) 2x(3x+1) – (2x+3)(3x-2) = 12
\(\Leftrightarrow6x^2+2x-\left(6x^2-4x+9x-6\right)=12\)
\(\Leftrightarrow6x^2+2x-6x^2+4x-9x+6=12\)
\(\Leftrightarrow-3x+6=12\)
\(\Leftrightarrow-3x=6\)
\(\Leftrightarrow x=-2\)
vậy x = -2
b) (x+2)2 – (x-3)(x+3) = 5
\(\Leftrightarrow\left(x+2\right)^2-\left(x^2-9\right)=5\)
\(\Leftrightarrow x^2+4x+4-x^2+9-5=0\)
\(\Leftrightarrow4x+8=0\)
\(\Leftrightarrow4x=-8\)
\(\Leftrightarrow x=-2\)
Vậy x = -2
NT
3
HN
15 tháng 8 2019
a) \(2x\left(x-3\right)+6\left(3x-3\right)=0\)
\(\Leftrightarrow2x^2-6x+18x-18=0\)
\(\Leftrightarrow2x^2+12x-18=0\)
Mà \(2x^2\ge0\)
\(\Rightarrow x\in\varnothing\)
15 tháng 8 2019
a)=>2x^2-6x+18x-18=0 b)=>6x^2-15x-75-30x =????
=>2x^2+12x=0+18
=>2x^2+12x=18
=>x.(2x+12)=18 (tự làm phần còn lai)
ES
3
13 tháng 7 2016
\(a,5\left(3x+5\right)-4\left(2x-3\right)=5x+8\left(2x+12\right)+1\)
\(\Rightarrow5\left(3x+5\right)-4\left(2x-3\right)-5x-8\left(2x+12\right)-1=0\)
\(\Rightarrow15x+25-8x+12-5x-16x-96-1=0\)
\(\Rightarrow-14x-60=0\)
\(\Rightarrow-14x=60\) \(\Rightarrow x=-\frac{60}{14}=\frac{-30}{7}\)
\(b,\left(2x+3\right)\left(x-4\right)-\left(3x-5\right)\left(x-4\right)=\left(5-x\right)\left(x-2\right)\)
\(\Rightarrow2x^2+3x-8x-12-3x^2+5x+12x-20=5x-x^2-10+2x\)
\(\Rightarrow-x^2+12x-32=7x-x^2-10\)
\(\Rightarrow-x^2+12x-32-7x+x^2+10=0\)
\(\Rightarrow5x-22=0\)
\(\Rightarrow5x=22\Rightarrow x=\frac{22}{5}\)
PA
13 tháng 7 2016
a) 5(3x+5)-4(2x-3) = 5x+8(2x+12)+1
15x + 25 - 8x + 12 = 5x + 16x + 96 + 1
15x - 8x - 5x - 16x = 96 + 1 - 25 - 12
-14x = 60
x = \(\frac{60}{-14}\)
x = \(-\frac{30}{7}\)
b) (2x+3)(x-4)-(3x-5)(x-4) = (5-x).(x-2)
(x - 4)(2x + 3 - 3x +5) = 5x - 10 - x2 + 2x
(x - 4)[(2x - 3x) + (3 + 5)] = 5x - 10 - x2 + 2x
(x - 4)(-x + 8) = 5x - 10 - x2 + 2x
-x2 + 8x + 4x - 32 = 5x - 10 - x2 + 2x
(-x2 + x2) + (8x + 4x - 5x - 2x) = -10 + 32
5x = 22
x = \(\frac{22}{5}\)
NT
1
14 tháng 7 2015
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
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
20 tháng 9 2021
\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)
\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)
CW
18 tháng 6 2016
f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)
\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)
\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)
\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)
\(-x^3=27\)
\(x=-3\)