\(a,\text{Với }x< -2\Rightarrow3-x-x-2=4\\ \Rightarrow-2x=3\Rightarrow x=-\dfrac{3}{2}\left(ktm\right)\\ \text{Với }-2\le x< 3\Rightarrow3-x+x+2=4\\ \Rightarrow0x=-1\Rightarrow x\in\varnothing\\ \text{Với }x\ge3\Rightarrow x-3+x+2=4\\ \Rightarrow2x=5\Rightarrow x=\dfrac{5}{2}\left(ktm\right)\)

Vậy \(x\in\varnothing\)

\(b,\text{Với }x< 2\Rightarrow4-2x+18-6x=21\\ \Rightarrow22-8x=21\Rightarrow x=\dfrac{1}{8}\left(tm\right)\\ \text{Với }2\le x< 3\Rightarrow2x-4+18-6x=21\\ \Rightarrow-4x+14=21\Rightarrow x=-\dfrac{7}{4}\left(ktm\right)\\ \text{Với }x\ge3\Rightarrow2x-4+6x-18=21\\ \Rightarrow8x=43\Rightarrow x=\dfrac{43}{8}\left(tm\right)\)

Vậy \(x\in\left\{\dfrac{1}{8};\dfrac{43}{8}\right\}\)

9(x + 1)² - 16(y + 3)²

= 9(x² + 2x + 1) - 16(y² + 6y + 9)

= 9x² + 18x + 9 - 16y² - 96y - 144

= 9x² - 16y² + 18x - 96y - 135

\(C=\frac{\left(x+3\right)^2-2x^2+6+x\left(x-3\right)}{x^2-9}.\frac{2x^2-18}{6x-12}\)\(\)

\(C=\frac{x^2+6x+9-2x^2+6+x^2-3x}{x^2-9}.\frac{2\left(x^2-9\right)}{6x-12}\)\(C=\frac{3x+15}{6x-12}.2=\frac{x+5}{x-2}=1+\frac{7}{x-2}\)

Để C nguyên =>(x-2) thuộc Ư(7) \(\Rightarrow x\in\left\{3;1;9;-5\right\}\)

Bài này kêu tính j vậy cậu???

1) \(4x^3+5x^2+10x-12\)

\(=4x^3-3x^2+8x^2-6x+16x-12\)

\(=x^2\left(4x-3\right)+2x\left(4x-3\right)+4\left(4x-3\right)\)

\(=\left(4x-3\right)\left(x^2+2x+4\right)\)

Đề bài : Phân tích các đa thức sau thành nhân tử

a) X - 36 x 18 = 12

    X - 36         = 12 : 18

    X - 36         = 2/3

    X                = 2/3 + 36

    X                = \(36\frac{2}{3}\)

b) ( X - 36 ) : 18 = 12

      X - 36           = 12 x 18

      X - 38           = 216

      X                  = 216 + 38

      X                  = 254

x-36.18=12

x-36=2/3

x=24

a) X - 36 x 18 = 12

X - 36 = 12 : 18

X - 36 = 2/3

X = 2/3 + 36

X = 110/3

b) ( X - 36 ) : 18 = 12

X - 36 = 12 x 18

X - 38 = 216

X = 216 + 38

X = 254 

1.8100

2. 34 x 18 + 18 x 66

= 18 x ( 34 + 66)

= 18 x 100 = 1800

3. X × ( 8 +  12) =  160 + 20 × 12

= X x 20 = 160 + 240

= X x 20 = 400

X = 400 : 20 = 20

4. X x 12 - X x 2 = 2020

(12 - 2) x X = 2020

10 x X = 2020

X = 2020 : 10 = 202

1.8100

2. 34 x 18 + 18 x 66

= 18 x ( 34 + 66)

= 18 x 100 = 1800

3. X × ( 8 +  12) =  160 + 20 × 12

= X x 20 = 160 + 240

= X x 20 = 400

X = 400 : 20 = 20

4. X x 12 - X x 2 = 2020

(12 - 2) x X = 2020

10 x X = 2020

X = 2020 : 10 = 202

[ 25 + ( 6x - 18 ) : 3 ] . 2 = 78

     [ 25 + ( 6x - 18 ) : 3 ] = 78 : 2

     [ 25 + ( 6x - 18 ) : 3 ] = 39

                 ( 6x - 18 ) : 3 = 39 - 25

                 ( 6x - 18 ) : 3 = 14

                          6x - 18 = 14 . 3

                          6x - 18 = 42

                                 6x = 42 + 18

                                 6x = 60

                                   x = 60 : 6 

                            Vậy:x = 10 

                      Hok tốt !!!

c) \(\sqrt{\left(x-2\right)^2}=10\)

\(x-2=10\)

\(x=12\)

d) \(\sqrt{9x^2-6x+1}=15\)

\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)

\(\sqrt{\left(3x-1\right)^2}=15\)

\(3x-1=15\)

\(3x=16\)

\(x=\dfrac{16}{3}\)

a) \(đk:x\ge0\)

\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)

\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)

b) \(đk:x\ge-2\)

\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)

\(\Leftrightarrow13\sqrt{x+2}=26\)

\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)

c) \(pt\Leftrightarrow\left|x-2\right|=10\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)

\(\Leftrightarrow\left|3x-1\right|=15\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)

e) \(đk:x\ge\dfrac{8}{3}\)

\(pt\Leftrightarrow3x+4=9x^2-48x+64\)

\(\Leftrightarrow9x^2-51x+60=0\)

\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

a. \(\sqrt{18x}+2\sqrt{8x}-3\sqrt{2x}=12\)      ĐK: \(x\ge0\)

<=> \(\sqrt{9.2x}+2\sqrt{4.2x}-3\sqrt{2x}=12\)

<=> \(3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)

<=> \(\sqrt{2x}\left(3+4-3\right)=12\)

<=> \(4\sqrt{2x}=12\)

<=> \(\sqrt{2x}=12:4\)

<=> \(\sqrt{2x}=3\)

<=> 2x = 3²

<=> 2x = 9

<=> \(x=\dfrac{9}{2}\) (TM)

b. \(\sqrt{9x+18}+2\sqrt{36x+72}-\sqrt{4x+8}=26\)          ĐK: \(x\ge-2\)

<=> \(\sqrt{9\left(x+2\right)}+2\sqrt{36\left(x+2\right)}-\sqrt{4\left(x+2\right)}=26\)

<=> \(3\sqrt{x+2}+72\sqrt{x+2}-2\sqrt{x+2}=26\)

<=> \(\sqrt{x+2}\left(3+72-2\right)=26\)

<=> \(73\sqrt{x+2}=26\)

<=> \(\sqrt{x+2}=\dfrac{26}{73}\)

<=> x + 2 = \(\left(\dfrac{26}{73}\right)^2\)

<=> x + 2 = \(\dfrac{676}{5329}\)

<=> \(x=\dfrac{676}{5329}-2\)

<=> \(x=-1,873146932\) (TM)

c. \(\sqrt{\left(x-2\right)^2}=10\)

<=> \(\left|x-2\right|=10\)

<=> \(\left[{}\begin{matrix}x-2=10\left(x\ge2\right)\\x-2=-10\left(x< 2\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=12\left(TM\right)\\x=-8\left(TM\right)\end{matrix}\right.\)

d. \(\sqrt{9x^2-6x+1}=15\)

<=> \(\sqrt{\left(3x-1\right)^2}=15\)

<=> \(\left|3x-1\right|=15\)

<=> \(\left[{}\begin{matrix}3x-1=15\left(x\ge\dfrac{16}{3}\right)\\3x-1=-15\left(x< \dfrac{16}{3}\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{16}{3}\left(TM\right)\\x=\dfrac{-14}{3}\left(TM\right)\end{matrix}\right.\)

e. \(\sqrt{3x+4}=3x-8\)        ĐK: \(x\ge\dfrac{-4}{3}\)

<=> 3x + 4 = (3x - 8)²

<=> 3x + 4 = 9x² - 48x + 64

<=> 9x² - 3x - 48x + 64 - 4 = 0

<=> 9x² - 51x + 60 = 0

<=> 9x² - 36x - 15x + 60 = 0

<=> 9x(x - 4) - 15(x - 4) = 0

<=> (9x - 15)(x - 4) = 0

<=> \(\left[{}\begin{matrix}9x-15=0\\x-4=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{15}{9}\left(TM\right)\\x=4\left(TM\right)\end{matrix}\right.\)