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4 tháng 3 2022

\(a,\left(x-1\right)^2+x^2\le\left(x+1\right)^2+\left(x+2\right)^2\\ \Leftrightarrow x^2-2x+1+x^2\le x^2+2x+1+x^2+4x+4\\ \Leftrightarrow2x^2-2x+1\le2x^2+6x+5\\ \Leftrightarrow-8x-6\le0\\ \Leftrightarrow x\ge\dfrac{3}{4}\)

\(b,\left(x^2+1\right)\left(x-6\right)\le\left(x-2\right)^3\\ \Leftrightarrow x^3+x-6x^2-6\le x^3-6x^2+12x-8\\ \Leftrightarrow11x-2\ge0\\ \Leftrightarrow x\ge\dfrac{2}{11}\)

a: \(\Leftrightarrow x^2-2x+1+x^2< =x^2+2x+1+x^2+4x+4\)

=>-2x+1<=6x+5

=>-7x<=4

hay x>=-4/7

b: \(\Leftrightarrow x^3-6x^2+x-6-x^3+6x^2-12x+8< =0\)

=>-11x+2<=0

=>-11x<=-2

hay x>=2/11

a: \(\Leftrightarrow2x^2-2-3>-5x+\left(2x+1\right)\left(x-3\right)\)

\(\Leftrightarrow2x^2-5>-5x+2x^2-6x+x-3\)

\(\Leftrightarrow2x^2-5>2x^2-10x-3\)

=>-5>-10x-3

=>5<10x+3

=>10x+3>5

=>10x>2

hay x>1/5

b: \(\Leftrightarrow x^2-6x+9+8-4x>x+7\)

\(\Leftrightarrow x^2-10x+17-x-7>0\)

\(\Leftrightarrow x^2-11x+10>0\)

=>x>10 hoặc x<1

5 tháng 3 2022

a: ⇔2x2−2−3>−5x+(2x+1)(x−3)⇔2x2−2−3>−5x+(2x+1)(x−3)

⇔2x2−5>−5x+2x2−6x+x−3⇔2x2−5>−5x+2x2−6x+x−3

⇔2x2−5>2x2−10x−3⇔2x2−5>2x2−10x−3

=>-5>-10x-3

=>5<10x+3

=>10x+3>5

=>10x>2

hay x>1/5

b: ⇔x2−6x+9+8−4x>x+7⇔x2−6x+9+8−4x>x+7

⇔x2−10x+17−x−7>0⇔x2−10x+17−x−7>0

⇔x2−11x+10>0⇔x2−11x+10>0

=>x>10 hoặc x<1

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

5 tháng 3 2022

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

a: (2x+3)(x+1)<0

=>2x+3 và x+1 khác dấu

=>x>-1 hoặc x<-3/2

b: (4-x)(x+2)>0

=>(x-4)(x+2)<0

=>-2<x<4

5 tháng 3 2022

a: (2x+3)(x+1)<0

=>2x+3 và x+1 khác dấu

=>x>-1 hoặc x<-3/2

b: (4-x)(x+2)>0

=>(x-4)(x+2)<0

=>-2<x<4

10 tháng 7 2017

\(\left(2x+1\right)^2+\left(1-x\right)3x\le\left(x+2\right)^2\)

\(\Leftrightarrow4x^2+4x+1+3x-3x^2\le x^2+4x+4\)

\(\Leftrightarrow4x^2+4x+3x-3x^2-x^2-4x\le4-1\)

\(\Leftrightarrow3x\le3\Leftrightarrow x\le1\) vậy \(x\le1\)

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

4 tháng 3 2022

\(a,\dfrac{x-1}{2}-\dfrac{7x+3}{15}\le\dfrac{2x+1}{3}+\dfrac{3-2x}{5}\\ \Leftrightarrow\dfrac{15\left(x-1\right)}{30}-\dfrac{2\left(7x+3\right)}{30}\le\dfrac{10\left(2x+1\right)}{30}+\dfrac{6\left(3-2x\right)}{30}\\ \Leftrightarrow15x-15-14x-6\le20x+10+18-12x\\ \Leftrightarrow x-21\le8x+28\\ \Leftrightarrow7x+49\ge0\\ \Leftrightarrow x\ge-7\)

\(b,\dfrac{2x+1}{-3}-\dfrac{2x^2+3}{-4}>\dfrac{x\left(5-3x\right)}{-6}-\dfrac{4x+1}{-5}\\ \Leftrightarrow\dfrac{20\left(2x+1\right)}{-60}-\dfrac{15\left(2x^2+3\right)}{-60}>\dfrac{10x\left(5-3x\right)}{-60}-\dfrac{12\left(4x+1\right)}{-60}\\ \Leftrightarrow40x+20-30x^2-45>50x-30x^2-48x-12\\ \Leftrightarrow38x-13>0\\ \Leftrightarrow x>\dfrac{13}{38}\)

15 tháng 10 2023

a: \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)

\(=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)

\(=\left(2x+1+2x-1\right)^2=\left(4x\right)^2=16x^2\)

b: \(\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3+2x^2-x-2-x^3+8\)

\(=2x^2-x+6\)

15 tháng 10 2023

a) \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)

\(=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)

\(=\left[\left(2x+1\right)+\left(2x-1\right)\right]^2\)

\(=\left(2x+1+2x-1\right)^2\)

\(=\left(4x\right)^2\)

\(=16x^2\)

b) \(\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=\left(x^3+2x^2-x-2\right)-\left(x^3-8\right)\)

\(=x^3+2x^2-x-2-x^3+8\)

\(=2x^2-x+6\)