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

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

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

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

a) <=> 4x^3 - 12x^2 - x^2 + 3x + 6x - 18 = 0

<=> 4x^2 (x - 3) - x(x - 3) + 6(x - 3) = 0

<=> (x - 3)(4x^2 - x + 6) = 0

xét 2 th

. x - 3 = 0 <=> x = 3

. 4x^2 - x + 6 = 0

<=> 4x^2 + 2.(1/2)x + 1/4 + 23/4 = 0

<=> (4x + 1/2)^2 = -23/4

.... phần sau bạn tự làm nhé 

vậy pt trên có nghiệm là ...

. mik bận nên chỉ làm như vậy thôi.. những ý sau thì tách tương tự

c) => x³ + 2x² - 6x²  - 12x + 4x + 8 = 0

=> (x³ + 2x²)  -  (6x²  + 12x)  + (4x + 8) = 0

=> x². (x +2) - 6x. (x + 2) + 4.(x + 2) =0

=> (x +2).(x²  - 6x + 4) = 0

=> x+ 2 = 0 hoặc x²  - 6x + 4 = 0

+) x+ 2 =0 => x = -2

+) x²  - 6x + 4 = 0 => x²  - 2.x.3  + 9  - 5 = 0 => (x -3)²  = 5

=> x - 3 = \(\sqrt{5}\) hoặc x - 3 = - \(\sqrt{5}\)

=> x = 3 + \(\sqrt{5}\) hoặc x = 3 - \(\sqrt{5}\)

vậy...

1/ (x-63)(x+10)(4x² -188x-2520)

2/ 9(x-1)(2x-1)(64x² + 208x+32)/8

3/ (x+2)(x-6)(x² -4x -41)

a) \(\dfrac{9x^2-6x+1}{9x^2-1}\)

\(=\dfrac{\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right)}\)

\(=\dfrac{3x-1}{3x+1}\)

\(=\dfrac{3\cdot\left(-3\right)-1}{3\cdot\left(-3\right)+1}=\dfrac{-9-1}{-9+1}=\dfrac{-10}{-8}=\dfrac{5}{4}\)

b) Ta có: \(\dfrac{x^2-6x+9}{3x^2-9x}\)

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

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

\(=\dfrac{-\dfrac{1}{3}-3}{3\cdot\dfrac{-1}{3}}=\dfrac{-\dfrac{10}{3}}{-1}=\dfrac{10}{3}\)

c) Ta có: \(\dfrac{x^2-4x+4}{2x^2-4x}\)

\(=\dfrac{\left(x-2\right)^2}{2x\left(x-2\right)}\)

\(=\dfrac{x-2}{2x}\)

\(=\dfrac{\dfrac{-1}{2}-2}{2\cdot\dfrac{-1}{2}}=\dfrac{-\dfrac{5}{2}}{-1}=\dfrac{5}{2}\)

a. \(=4x^3-12x^2-x^2+3x+6x-18=\left(x-3\right)\left(4x^2-x+6\right)\)

b.  \(=-x^3+x^2-7x^2+7x-x+1=\left(x-1\right)\left(-x^2-7x-1\right)\)

c.  \(=x^3+2x^2-6x^2-12x+4x+8=\left(x+2\right)\left(x^2-6x+4\right)\)

11: \(2x^2-12xy+18y^2\)

\(=2\left(x^2-6xy+9y^2\right)\)

\(=2\left(x-3y\right)^2\)

12: \(\left(x^2+x\right)^2+3\left(x^2+x\right)+2\)

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

18 - 4\(x\) = -20 - 6\(x\) 

-4\(x\) + 6\(x\) = - 20  - 18

      2\(x\)    = - 38

         \(x\)  = - 19

h, -15 \(\times\) 24 = -7\(x\) + 32

     7\(x\) = 360 + 32

      7\(x\) = 392

         \(x\) =  392:7

          \(x\) = 56

i, 15\(x\) -3.(4\(x\) - 6) = -12 + 36

    15\(x\) - 12\(x\) + 18 =  24

      3\(x\) = 24 - 18

       3\(x\) = 6

         \(x\) = 2

k, -10\(x\) - 27 = -7\(x\) + 33

    -27 - 33 = -7\(x\) + 10\(x\)

     3\(x\) = -60

        \(x\) = -20

m, -17\(x\) - 24 = - 9\(x\) - 40

     - 24 + 40 =  -9\(x\) + 17\(x\)

          8\(x\) = 16

             \(x\) = 2

n, -23 \(\times\) 25 = -18\(x\) + 75

     -575 = -18\(x\) + 75

       18\(x\) = 575 + 75

        18\(x\) = 650

              \(x\) = \(\dfrac{325}{9}\)

a: ĐKXĐ: x>=-3/2

\(\sqrt{x^2+4}=\sqrt{2x+3}\)

=>\(x^2+4=2x+3\)

=>\(x^2-2x+1=0\)

=>\(\left(x-1\right)^2=0\)

=>x-1=0

=>x=1(nhận)

b: \(\sqrt{x^2-6x+9}=2x-1\)(ĐKXĐ: \(x\in R\))

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

=>\(\left\{{}\begin{matrix}\left(2x-1\right)^2=\left(x-3\right)^2\\x>=\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+2\right)\left(3x-4\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)

=>x=4/3(nhận) hoặc x=-2(loại)

c:

Sửa đề: \(\sqrt{4x+12}=\sqrt{9x+27}-5\)

ĐKXĐ: \(x>=-3\)

\(\sqrt{4x+12}=\sqrt{9x+27}-5\)

=>\(2\sqrt{x+3}=3\sqrt{x+3}-5\)

=>\(-\sqrt{x+3}=-5\)

=>x+3=25

=>x=22(nhận)

d: ĐKXĐ: \(\left[{}\begin{matrix}x< =\dfrac{3-\sqrt{5}}{4}\\x>=\dfrac{3+\sqrt{5}}{4}\end{matrix}\right.\)
\(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)

=>\(\sqrt{\left(4x^2-6x+1\right)}=\sqrt{4x^2-20x+25}\)

=>\(4x^2-6x+1=4x^2-20x+25\)

=>\(-6x+20x=25-1\)

=>\(14x=24\)

=>x=12/7(nhận)

1: =>|2x-1|=5

=>2x-1=5 hoặc 2x-1=-5

=>2x=6 hoặc 2x=-4

=>x=3 hoặc x=-2

2: \(\Leftrightarrow2\sqrt{x-3}+\dfrac{1}{3}\cdot3\sqrt{x-3}-\sqrt{x-3}=4\)

\(\Leftrightarrow\sqrt{x-3}=2\)

=>x-3=4

hay x=7

5: \(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)

=>x-2=0 hoặc x+2=1

=>x=2 hoặc x=-1