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Ta có: \(5x\left(12x+7\right)-3x\left(20x-5\right)=-100\)
\(\Leftrightarrow60x^2+35x-60x^2+15x=-100\)
\(\Leftrightarrow50x=-100\)
hay x=-2
\(5x.\left(12x+7\right)-3x.\left(20x-5\right)=-100\)
\(5x.12x+5x.7+\left(-3x\right).20x+\left(-3x\right).\left(-5\right)=-100\)
\(60x^2+35x-60x^2+15x=-100\)
\(\left(60x^2-60x^2\right)+\left(35x+15x\right)=-100\)
\(50x=-100\)
\(x=\left(-100\right):50\)
\(x=-2\)
Vậy:\(x=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(60x^2+35x-60x^2+15x=100\)
35x+15x=100
50x=100 =>x=2
b)\(10x^2-35x+16x-10x^2=5\)
-35x+16x=5
-19x=5 =>x=-5/19
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>1,5x^2-2x+1,2x-1,5x^2>=0,32
=>-0,8x>=0,32
=>x<=4
b: =>60x^2+35x-60x^2+15x<=-100
=>50x<=-100
=>x<=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(0,5x\left(3x-4\right)+1,5x\left(0,7-x\right)\ge0,32\)
\(\Leftrightarrow1,5x^2-2+1,05-1,5x^2\ge0,32\)
\(\Leftrightarrow1,5x^2-1,5x^2-2+1,05\ge0,32\)
\(\Leftrightarrow-0,95\ge0,32\)(vô lí)
Vậy bất phương trình vô nghiệm
b)\(5x\left(12x+7\right)-3x\left(20x-5\right)\le-100\)
\(\Leftrightarrow60x^2+35x-60x^2+15x\le-100\)
\(\Leftrightarrow50x\le-100\)
\(\Leftrightarrow x\le-2\)
Vậy bất phương trình có nghiệm là \(S=\left\{xIx\le-2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 5x(12x + 7) - 3x(20x - 5) = -100
<=> 60x2 + 35x - 60x2 + 15x = -100
<=> 50x = -100
<=> x = -2
b) 0,6x(x - 0,5) - 0,3x(2x + 1,3) = 0,138
<=> 0,6x2 - 0,3x - 0,6x2 - 0,39x = 0,138
<=> -0,69x = 0,138
<=> x = -0,2
c) 4x(3x - 7) - 6(2x2 - 5x + 1) = 12
<=> 12x2 - 28x - 12x2 + 30x - 6 = 12
<=> 2x - 6 = 12
<=> 2x = 18
<=> x = 9
\(5x\left(12x+7\right)-3x\left(20x-5\right)=-100\)
\(\Leftrightarrow60x^2+35x-60x^2+15x=-100\)
\(\Leftrightarrow50x=-100\)
hay x=-2