a) \(\Leftrightarrow4x=20\Leftrightarrow x=5\)

b) \(\Leftrightarrow18x-288=27.4+8.9-144\)

\(\Leftrightarrow18x=108+72-144+288=324\)

\(\Leftrightarrow x=18\)

c)Ta có:

\(x\in B\left(6\right)\) và \(22< x\le36\)

mà \(B\left(6\right)=\left\{0;6;12;18;24;30;36;42;48;...\right\}\)

\(\Rightarrow x\in\left\{0;6;12;18;24;30;36;42;48;...\right\}\) và \(22< x\le36\)

\(\Rightarrow x\in\left\{24;30;36\right\}\)

Vậy \(x\in\left\{24;30;36\right\}\)

`4)x^2-5x+6`

`=x^2-2x-3x+6`

`=x(x-2)-3(x-2)=(x-2)(x-3)`

`5)x^2+7x+10`

`=x^2+5x+2x+10`

`=x(x+5)+2(x+5)=(x+5)(x+2)`

`6)x+7\sqrt{x}+10`    `ĐK: x >= 0`

`=(\sqrt{x})^2+5\sqrt{x}+2\sqrt{x}+10`

`=\sqrt{x}(\sqrt{x}+5)+2(\sqrt{x}+5)=(\sqrt{x}+5)(\sqrt{x}+2)`

`7)3x^4+7x^2+4`

`=3x^4+3x^2+4x^2+4`

`=3x^2(x^2+1)+4(x^2+1)=(x^2+1)(3x^2+4)`

`8)x^2-x-2`

`=x^2-2x+x-2`

`=x(x-2)+(x-2)=(x-2)(x+1)`

`9)x^6-x^3-2`

`=x^6+x^3-2x^3-2`

`=x^3(x^3+1)-2(x^3+1)`

`=(x^3+1)(x^3-2)`.

Phân tích đa thức thành nhân tử:

a) \(\Rightarrow72-20x-36x+84=30x-240-6x-84\)

\(\Rightarrow80x=480\Rightarrow x=6\)

b) \(\Rightarrow15x+25-8x+12=5x+6x+36+1\)

\(\Rightarrow4x=0\Rightarrow x=0\)

c) \(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)

dái wa      

Giúp mình đi mình đang cần gấp

đề bài là j vậy hả bn

a,  4.(3\(x\) - 2) - 3.(5\(x\) + 6) = 35

           12\(x\) - 8 - 15\(x\) - 18 = 35

      \(x\)(12 - 15) - ( 8 + 18) = 35

       - 3\(x\)         -      26      = 35

         3\(x\)                         = - 26 - 35

        3\(x\)                        = - 61

          \(x\)                        = - \(\dfrac{61}{3}\) 

c, \(x^2\) - 5\(x\) = 0

  \(x\).(\(x-5\)) =0

  \(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

vậy \(x\) \(\in\) {0; 5}

a) -152 - (3x + 1) = (-2).(-3)³

-152 - 3x - 1 = (-2).(-27)

-3x - 153 = 54

-3x = 54 + 153

-3x = 207

x = -69

b) x - 43 = (35 - x) - 48

x - 43 = 35 - x - 48

x - 43= -x - 13

x = -x - 13 + 43

x = 30 + x

x + x = 30

2x = 30

x = 15

a) Ta có: \(\left(2x+1\right)^2-\left(3x-4\right)^2=0\)

\(\Leftrightarrow\left(2x+1-3x+4\right)\left(2x+1+3x-4\right)=0\)

\(\Leftrightarrow\left(5-x\right)\left(5x-3\right)=0\)

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

b) Ta có: \(5x^3-3x^2+10x-6=0\)

\(\Leftrightarrow x^2\left(5x-3\right)+2\left(5x-3\right)=0\)

\(\Leftrightarrow5x-3=0\)

hay \(x=\dfrac{3}{5}\)

a) A = 3x³ - x² + 5 - 2x³ - 3x + 16 - x³ + x² - 2x

    A = 21

b) B = 24 - 4x + 2x² + 3x³ - 5x² + 4x + 3x² - 3x³

    B = 24

Bạn thấy hay thì tim giúp mik nha. Thx bạn

a) Ta có: \(A=x\left(3x^2-x+5\right)-\left(2x^3+3x-16\right)-x\left(x^2-x+2\right)\)

\(=3x^3-x^2+5x-2x^3-3x+16-x^3+x^2-2x\)

\(=16\)

b) Ta có: \(B=4\left(6-x\right)+x^2\left(2+3x\right)-x\left(5x-4\right)+3x^2\left(1-x\right)\)

\(=24-4x+2x^2+3x^3-5x^2+4x+3x^2-3x^3\)

\(=24\)

5x-x=29-36:4

4x   =20

x     =20:4

x     =5

Vậy x=5

      5x-x=29-36:4
<=>4x=29-36:4

<=>4x=29-9

<=>4x=20

<=>x=5

Vậy...bla bla

\(5.x-x=29-36:4\)

\(5.x-x=29-9\)

\(5.x-x=20\)

\(5.x-x.1=20\)

\(\Rightarrow x.\left(5-1\right)=20\)

\(x.4=20\)

\(\Rightarrow x=5\)