( \(\dfrac{1}{125}\) - \(x^3\) ) ( \(x^2\) + 2².6⁶) = 0

\(\left[{}\begin{matrix}\dfrac{1}{125}-x^3=0\\x^2-2^2.6^6=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x^3=\dfrac{1}{125}\\x^2=2^2(6^3)^2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x^2=(2.6^3)^2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=432^2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=432\\x=-432\end{matrix}\right.\)

\(x\) ϵ { -432; \(\dfrac{1}{5}\); 432; }

\(x^3\) + 125 + (\(x\) + 5)(\(x\) - 25) = 0

(\(x^3\) + 5³) + (\(x\) + 5)(\(x\) - 25) = 0

(\(x\) + 5)(\(x^2\) - 5\(x\) + 25) + (\(x\) + 5)(\(x\) - 25) =0

(\(x\) + 5)(\(x^2\) - 5\(x\) + 25 + \(x\) - 25) = 0

(\(x\) + 5)(\(x^2\) - 4\(x\)) = 0

\(x\)(\(x\) + 5)(\(x\) - 4) = 0

\(\left[{}\begin{matrix}x=0\\x+5=0\\x-4=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=-5\\x=4\end{matrix}\right.\)

`x^3 + 125 + (x + 5)(x - 25) = 0`

`<=>(x + 5)(x^2 + 5x + 25) + (x + 5)(x - 25) = 0`

`<=>(x + 5)(x^2 + 6x) = 0`

`<=>x(x + 5)(x + 6) = 0`

`<=>x = 0` hoặc `x = -5` hoặc `x=-6`

Sửa:

`x^3 + 125 + (x + 5)(x - 25) = 0`

`<=>(x + 5)(x^2 - 5x + 25) + (x + 5)(x - 25) = 0`

`<=>(x+5)(x^2-4x)=0`

`<=>x(x+5)(x-4)=0`

`<=>x=0` hoặc `x=-5` hoặc `x=4`

\(a,=\left(3+x\right)\left(9-3x+x^2\right)\\ b,=\left(4x+0,1\right)\left(16x^2-0,4x+0,01\right)\\ c,=\left(2-3x\right)\left(4+6x+9x^2\right)\\ d,=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)

a) \(27+x^3=3^3+x^3=\left(3+x\right)\left(9-3x+x^2\right)\)

b) \(64x^3+0,001=\left(4x\right)^3+\left(\dfrac{1}{10}\right)^3=\left(4x+\dfrac{1}{10}\right)\left(16x^2-\dfrac{4x}{10}+\dfrac{1}{100}\right)\)

a/\(27+x^3=\left(3+x\right)\left(9-3x+x^2\right)\)

b/ \(64x^3+0,001=\left(4x+0,1\right)\left(16x^2-0,4x+0,01\right)\)

c/ \(8-27x^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

d/ \(\dfrac{x^3}{125}-\dfrac{y^3}{27}=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)

\(\left(3.4^x-3\right).\left(x^3-125\right)=0\\ \rightarrow\left[{}\begin{matrix}3.4^x-3=0\\x^3-125=0\end{matrix}\right.\\ \rightarrow\left[{}\begin{matrix}3.4^x=3\\x^3=125\end{matrix}\right.\)

\(\rightarrow\left[{}\begin{matrix}4^x=1=4^0\\x^3=5^3\end{matrix}\right.\\ \rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

\(\left(3\cdot4^x-3\right)\left(x^3-125\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3\cdot4^x-3=0\\x^3-125=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3\left(4^x-1\right)=0\\x^3-5^3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4^x=1\\x=5\end{matrix}\right.\)

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

x=5 và 0

c

\(\sqrt{x}+9=4\)

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

Vì \(\sqrt{x}\) ≥ 0

⇒ Không tồn tại x

C

b) x3 – x2 – 5x + 125

= (x3 + 125) - (x2 + 5x)

= (x + 5)(x2 - 5x + 25) - x(x + 5)

= (x + 5)(x2 - 5x + 25 - x)

= (x + 5)(x2 - 6x + 25)

a) Kết quả - x 2  + 2.               b) Kết quả − 1 2 ( 4 x 2 + 10 x + 25 ) .  

c) Kết quả - ( x 3   +   1 ) 2 .

\(9,=\left(5+2x\right)^3\\ 10,=\left(y+4\right)^3\\ 11,=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\\ 12,=\left(x+5y\right)\left(x^2-5xy+25y^2\right)\)

\(A=x^3+15x^2+75x+125=\left(x+5\right)^3=-125\)

\(B=4x^2+12xy+9y^2=\left(2x+3y\right)^2=\left(3+6\right)^2=81\)

c: =(x-3-2y)(x-3+2y)