Ôn tập: Bất phương trình bậc nhất một ẩn

<=> 2x - 6 - x - 2 = 4x + 4 
<=> 2x - 6 - x - 2 - 4x - 4 = 0
<=> - 3x - 12 = 0 
<=> - 3x        = 12
<=> x            = \(\dfrac{12}{-3}\)
<=> x               = -4
Vậy S = \(\left\{-4\right\}\)

=>4x+4=2x-6-x-2=x-8

=>3x=-12

hay x=-4

2(x-3) - (x+2) = 4(x+1)

⇔2x-6-x-2=4x+4

⇔2x-x-4x=4+2+6

⇔-3x=12

⇔x=-4

\(\Leftrightarrow20x-6\left(x+1\right)=60+15\left(x-3\right)\)

=>20x-6x-6=60+15x-45

=>15x+15=14x-6

=>x=-21

`<=> (20x)/30 - (6x + 6)/30 = 60/30 + (30x - 90)/30`

`<=> (14x - 6)/30 = (30x - 30)/30`

`=> (7x - 3)/10 = (10x-10)/10`

`=> 7x - 3 = 10x - 10`

`=> 13 = 3x`

`=> x = 13/3`

\(\Leftrightarrow\dfrac{20x}{30}-\dfrac{6\left(x+1\right)}{30}=\dfrac{60}{30}+\dfrac{15\left(x-3\right)}{30}\)

\(\Leftrightarrow20x-6x-6-60-15x+45=0\)

<=>-9 -x = 0

<=> x = -9 - 0 = -9

=> x = - 9 

\(1-\dfrac{x-\dfrac{1+x}{3}}{3}=\dfrac{x}{2}-\dfrac{2x-\dfrac{10-7x}{3}}{2}\\ \Leftrightarrow1-\dfrac{\dfrac{3x-1-x}{3}}{3}=\dfrac{x}{2}-\dfrac{\dfrac{6x-10+7x}{3}}{2}\\ \Leftrightarrow1-\dfrac{2x-1}{9}=\dfrac{x}{2}-\dfrac{13x-10}{6}\\ \Leftrightarrow\dfrac{9-2x+1}{9}=\dfrac{3x-13x+10}{6}\\ \Leftrightarrow\dfrac{10-2x}{9}=\dfrac{-10x+10}{6}\\ \Leftrightarrow6\left(10-2x\right)=9\left(-10x+10\right)\\ \Leftrightarrow60-12x=-90x+90\\ \Leftrightarrow-90x+90+12x-60=0\\ \Leftrightarrow-78x+30=0\\ \Leftrightarrow x=\dfrac{5}{13}\)

Vậy `S={5/13}`

a: \(A=x^2+20x+100-91=\left(x+10\right)^2-91\ge-91\)

Dấu '=' xảy ra khi x=-10

b: \(B=3\left(x^2-\dfrac{4}{3}x+\dfrac{5}{3}\right)\)

\(=3\left(x^2-2\cdot x\cdot\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{11}{9}\right)\)

\(=3\left(x-\dfrac{2}{3}\right)^2+\dfrac{11}{3}\ge\dfrac{11}{3}\)

Dấu '=' xảy ra khi x=2/3

\(A=x^2+20x+9=\left(x+10\right)^2-91\ge-91\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+10\right)^2=0\Leftrightarrow x=-10\)

Vậy \(MinA=-91\)

\(B=3x^2-4x+5=3\left(x^2-\dfrac{4}{3}x\right)+5=3\left[\left(x-\dfrac{2}{3}\right)^2-\dfrac{4}{9}\right]+5=3\left(x-\dfrac{2}{3}\right)^2-\dfrac{4}{3}+5=\left(x-\dfrac{2}{3}\right)^2+\dfrac{11}{3}\ge\dfrac{11}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-\dfrac{2}{3}\right)^2=0\Leftrightarrow x=\dfrac{2}{3}\)

Vậy \(MinB=\dfrac{11}{3}\)

cần bài nào ne?

1d  2b  3a  4a  5a  6d

a: A=B

=>x^3-1-2x=x(x^2-1)

=>x^3-2x-1=x^3-x

=>-2x-1=-x

=>-x=1

=>x=-1

b: A=B

=>x^2-4+3x^2=4x^2-4x+1-x

=>-5x+1=-4

=>-5x=-5

=>x=1

d: A=B

=>x^3+3x^2+3x+1-x^3+6x^2-12x+8=9x^2+1

=>9x^2-9x+9=9x^2+1

=>-9x=-8

=>x=8/9

x + 5