a) \(x^2-5x=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}}\)

b) \(x^4=125x\)

\(x^4-125x=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=0\\x^3-125=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}}\)

c) \(3^x+3^{x+2}=90\)

\(3^x\left(1+3^2\right)=90\)

\(3^x\cdot10=90\)

\(3^x=9=3^2\)

\(\Rightarrow x=2\)

a) x² - 5x  = 0

x.(x-5) = 0

=> x = 0

x-5 = 0 => x = 5

KL:...

b) x⁴ = 125x

=> x³ = 125 = 5³

=> x = 5

c) 3^x + 3^x+2 = 90

3^x + 3^x.9 = 90

3^x.(1+9) = 90

3^x.10 = 90

3^x = 9 = 3²

=> x = 2

     \(x^2-5x=0\)

\(\Rightarrow x\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)

     \(x^4=125x\)

\(\Rightarrow x^4-125x=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=0\\x^3=125\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^3=5^3\end{cases}\Rightarrow}}\orbr{\begin{cases}x=0\\x=5\end{cases}}\)

     \(3^x+3^{x+2}=90\)

\(\Rightarrow3^x+3^x.3^2=90\)

\(\Rightarrow3^x\left(3^2+1\right)=90\)

\(\Rightarrow3^x.10=90\)

\(\Rightarrow3^x=9\Rightarrow3^x=3^2\Rightarrow x=2\)

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

\(B=x^3-8+\left(x-2\right)\left(5x+4\right)\)

\(=\left(x-2\right)\left(x^2+2x+4+5x+4\right)\)

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

a, \(\left(x+2\right)^3-x\left(x^2+6x-3\right)=0\Leftrightarrow x^3+4x^2+4x+2x^2+8x+8-x^3-6x^2+3x=0\)

\(\Leftrightarrow15x+8=0\Leftrightarrow x=-\frac{8}{15}\)

b, \(\left(x+4\right)^3-x\left(x+6\right)^2=7\Leftrightarrow12x+64=0\Leftrightarrow x=-\frac{19}{4}\)làm tắt:P 

Tự làm nốt nhé 

Có:

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

\(\Rightarrow x.\left(125x^2-5\right)=0\)

\(\Rightarrow\) x=0 hoặc 125x²-5 =0

\(\Rightarrow\) x=0 hoặc 125x²=5

\(\Rightarrow\) x=0 hoặc x²=\(\frac{5}{125}=\frac{1}{25}\)

\(\Rightarrow\) x=0 , \(x=-\frac{1}{5}\)hoặc  \(x=\frac{1}{5}\)

Vậy \(x\in\left\{-\frac{1}{5};0;\frac{1}{5}\right\}\)

ta có:P(x) = 125x^3 - 5x=0

f(x)g(x)<=>f(x)=0 hoặc g(x)=0

125x^3 - 5x=5x(5x-1)(5x+1) (*)

thay (*) vào VT ta đc

5x(5x-1)(5x+1)=0

Th1:5x=0

=>x=0

Th2:5x-1=0

=>5x=1

=>x=\(\frac{1}{5}\)

Th3:5x+1=0

=>5x=-1

=>x=\(-\frac{1}{5}\)

vậy các nhiệm các đa thức trên là x=±\(\frac{1}{5}\);0

(125x³ - 1) : (25x² + 5x + 1)

= (5x - 1)(25x² + 5x + 1) : (25x² + 5x + 1)

= 5x - 1

a, \(\left(x^2+2xy+y^2\right):\left(x+y\right)=\left(x+y\right)^2:\left(x+y\right)=x+y\)

b, \(\left(125x^3+1\right):\left(5x+1\right)=\left(5x+1\right)\left(25x^2-5x+1\right):\left(5x+1\right)=25x^2-5x+1\)

c, \(\left(x^2-2xy+y^2\right):\left(y-x\right)=\left(x-y\right)^2:\left(y-x\right)=\left(y-x\right)^2:\left(y-x\right)=y-x\)

kick mình nha

làm giúp mik theo cột nha mn

\(=\sqrt{\dfrac{125x^2}{25x}}=\sqrt{5x}\)

\(=\dfrac{5\sqrt{5}\left|x\right|}{5\sqrt{x}}\)

\(=\dfrac{\sqrt{5}x}{\sqrt{x}}\)(vì x>0)

\(=\sqrt{5x}\)

** Sửa: $125x^3-5x-3y+27y^3$
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Lời giải:

$125x^3-5x-3y+27y^3=(125x^3+27y^3)-(5x+3y)$

$=[(5x)^3+(3y)^3]-(5x+3y)$

$=(5x+3y)(25x^2-15xy+9y^2)-(5x+3y)$

$=(5x+3y)(25x^2-15xy+9y^2-1)$