a, 4x² - 49 = 0

⇔⇔ (2x)² - 7² = 0

⇔⇔ (2x - 7)(2x + 7) = 0

⇔{2x−7=02x+7=0⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=72x=−72⇔{2x−7=02x+7=0⇔{x=72x=−72

b, x² + 36 = 12x

⇔⇔ x² + 36 - 12x = 0

⇔⇔ x² - 2.x.6 + 6² = 0

⇔⇔ (x - 6)² = 0

⇔⇔ x = 6

e, (x - 2)² - 16 = 0

⇔⇔ (x - 2)² - 4² = 0

⇔⇔ (x - 2 - 4)(x - 2 + 4) = 0

⇔⇔ (x - 6)(x + 2) = 0

⇔{x−6=0x+2=0⇔{x=6x=−2⇔{x−6=0x+2=0⇔{x=6x=−2

f, x² - 5x -14 = 0

⇔⇔ x² + 2x - 7x -14 = 0

⇔⇔ x(x + 2) - 7(x + 2) = 0

⇔⇔ (x + 2)(x - 7) = 0

⇔{x+2=0x−7=0⇔{x=−2x=7

a) \(x^3-16x=0\)

<=> \(x\left(x^2-16\right)=0\)

<=> \(x\left(x-4\right)\left(x+4\right)=0\)

<=> \(\orbr{\begin{cases}x=0\\x=-4;4\end{cases}}\)

b) \(2x^3-50x=0\)

<=> \(2x\left(x^2-25\right)=0\)

<=> \(2x\left(x-5\right)\left(x+5\right)=0\)

<=> \(\orbr{\begin{cases}x=0\\x=5;-5\end{cases}}\)

c) \(x^3-4x^2-9x+36=0\)

<=> \(\left(x^3-4x^2\right)-\left(9x-36\right)=0\)

<=> \(x^2\left(x-4\right)-9\left(x-4\right)=0\)

<=> \(\left(x-4\right)\left(x^2-9\right)=0\)

<=> \(\left(x-4\right)\left(x-3\right)\left(x+3\right)=0\)

<=> \(\orbr{\begin{cases}x=-3;3\\x=4\end{cases}}\)

a)\(x^3-16x=0\)

   \(x\left(x^2-4^2\right)=0\)

     \(x\left(x-4\right)\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

       x + 4 =0                  x = -4

b)Giống ở câu a

c)\(x^3-4x^2-9x+36=0\)

    \(x^2\left(x-4\right)+9\left(x-4\right)=0\)

    \(\left(x^2+9\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\x^2+9=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=4\\xkoTM\end{cases}}\)

b) Ta có: \(\frac{36\left(x-2\right)^3}{32-16x}=\frac{36\left(x-2\right)^3}{16\left(2-x\right)}=\frac{-36\left(2-x\right)^3}{16\left(2-x\right)}=\frac{-9\left(2-x\right)^2}{4}\)

đề sai 

pt đã cho tương đương với (4x²-x+6)²=0

phần còn lại cậu tự giải đc

Rảnh rỗi thật sự .-.

\(a,3x+2\left(5-x\right)=0\)

\(\Rightarrow3x+10-2x=0\)

\(\Rightarrow x+10=0\)

\(\Rightarrow x=-10\)

\(b,x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Rightarrow\left(2x^2-x\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Rightarrow2x^3+9x^2-5x-2x^3-9x^2-4,5=3,5\)

\(\Rightarrow-5x-4,5=3,5\)

\(\Rightarrow-5x=8\)

\(\Rightarrow x=-\dfrac{8}{5}\)

\(c,3x^2-3x\left(x-2\right)=36\)

\(\Rightarrow3x^2-3x^2+6x=36\)

\(\Rightarrow6x=36\)

\(\Rightarrow x=6\)

\(d,\left(3x^2-x+1\right)\left(x-1\right)=x^2\left(4-3x\right)=\dfrac{5}{2}\)

\(\Rightarrow3x^3-3x^2-x^2+x+x-1+4x^2-3x^3=\dfrac{5}{2}\)

\(\Rightarrow2x-1=\dfrac{5}{2}\)

\(\Rightarrow2x=\dfrac{7}{2}\)

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

a,\(3x+2\left(5-x\right)=0\)

\(3x+10-2x=0\)

\(x+10=0\)

\(x=-10\)

giải

5x-(4-2x+x^2)(x+2)+x(x-1)(x+1)=0

5x-(4x+8-2x^2-4x+x^3+2x^2)+x(x^2-1)=0

5x-4x-8+2x^2+4x-x^3-2x^2+x^3-1x=0

(5x-4x+4x-1x)+(-8)+(2x^2-2x^2)+(-x^3+x^3)=0

4x+(-8)=0

4x=0+8

4x=8

x=8:4

x=2

D)(4x+1)(16x^2-4x+1)-16x(4x^2-5)=17

64x^3-16x^2+4x+16x^2-4x+1-64x^3+80x=17

80x+1=17

80x=17-1

80x=16

x=1/5

a: =>3x+10-2x=0

hay x=-10

c: \(\Leftrightarrow3x^2-3x^2+6x=36\)

=>6x=36

hay x=6