Giari các bpt sau
a,\(\left(x+2\right)^2+3\left(x+1\right)^2\ge4x^2-4\)
b,\(\dfrac{x-1}{2}-\dfrac{x-2}{3}\le x-\dfrac{x-3}{4}\)
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a: \(\Leftrightarrow4\left(5x^2-3\right)+5\left(3x-1\right)< 10x\left(2x+3\right)-100\)
\(\Leftrightarrow20x^2-12x+15x-5< 20x^2+30x-100\)
=>3x-5<=30x-100
=>30x-100>3x-5
=>27x>95
hay x>95/27
b: \(\Leftrightarrow4\left(5x-2\right)-6\left(2x^2-x\right)< 4x\left(1-3x\right)-15x\)
\(\Leftrightarrow20x-8-12x^2+6x< 4x-12x^2-15x\)
=>26x-8<-11x
=>37x<8
hay x<8/37
a: \(\Leftrightarrow15\left(x-1\right)-2\left(7x+3\right)\le10\left(2x+1\right)+6\left(3-2x\right)\)
\(\Leftrightarrow15x-15-14x-6\le20x+10+18-12x\)
=>x-21<=8x+28
=>-7x<=49
hay x>=-7
b: \(\Leftrightarrow20\left(2x+1\right)-15\left(2x^2+3\right)< 10x\left(5-3x\right)-12\left(4x+1\right)\)
\(\Leftrightarrow40x+20-30x^2-45< 50x-30x^2-48x-12\)
=>40x-25<2x-12
=>38x<13
hay x<13/38
\(a,\dfrac{x-1}{2}-\dfrac{7x+3}{15}\le\dfrac{2x+1}{3}+\dfrac{3-2x}{5}\\ \Leftrightarrow\dfrac{15\left(x-1\right)}{30}-\dfrac{2\left(7x+3\right)}{30}\le\dfrac{10\left(2x+1\right)}{30}+\dfrac{6\left(3-2x\right)}{30}\\ \Leftrightarrow15x-15-14x-6\le20x+10+18-12x\\ \Leftrightarrow x-21\le8x+28\\ \Leftrightarrow7x+49\ge0\\ \Leftrightarrow x\ge-7\)
\(b,\dfrac{2x+1}{-3}-\dfrac{2x^2+3}{-4}>\dfrac{x\left(5-3x\right)}{-6}-\dfrac{4x+1}{-5}\\ \Leftrightarrow\dfrac{20\left(2x+1\right)}{-60}-\dfrac{15\left(2x^2+3\right)}{-60}>\dfrac{10x\left(5-3x\right)}{-60}-\dfrac{12\left(4x+1\right)}{-60}\\ \Leftrightarrow40x+20-30x^2-45>50x-30x^2-48x-12\\ \Leftrightarrow38x-13>0\\ \Leftrightarrow x>\dfrac{13}{38}\)
1:
c: =>1/3x+2/3-x+1>x+3
=>-2/3x+5/3-x-3>0
=>-5/3x-4/3>0
=>-5x-4>0
=>x<-4/5
d: =>3/2x+5/2-1<=1/3x+2/3+x
=>3/2x+3/2<=4/3x+2/3
=>1/6x<=2/3-3/2=-5/6
=>x<=-5
2:
a: \(\Leftrightarrow x^2+4x+4+3x^2+6x+3>=4x^2-4\)
=>10x+7>=-4
=>10x>=-11
hay x>=-11/10
b: \(\Leftrightarrow6\left(x-1\right)-4\left(x-2\right)\le12x-3\left(x-3\right)\)
=>6x-6-4x+8<=12x-3x+9
=>2x+2<=9x+9
=>-7x<=7
hay x>=-1
a: ⇔x2+4x+4+3x2+6x+3>=4x2−4⇔x2+4x+4+3x2+6x+3>=4x2−4
=>10x+7>=-4
=>10x>=-11
hay x>=-11/10
b: ⇔6(x−1)−4(x−2)≤12x−3(x−3)⇔6(x−1)−4(x−2)≤12x−3(x−3)
=>6x-6-4x+8<=12x-3x+9
=>2x+2<=9x+9
=>-7x<=7
hay x>=-1