a) x³-9x²+27x-27=0

<=>(x-3)³=0

<=>x-3=0

<=>x=3

b) x³-25x=0

<=>x.(x²-25)=0

<=>x.(x-5)(x+5)=0

<=>x=0 hoặc x-5=0 hoặc x+5=0

<=>x=0 hoặc x=5 hoặc x=-5

c)9x²-1=0

<=>(3x-1)(3x+1)=0

<=>3x-1=0 hoặc 3x+1=0

<=>x=1/3 hoặc x=-1/3

a, x^3 - 9x^2 + 27x - 27 = 0 

=> ( x - 3)^3 = 0 

=> x - 3 = 0 

=> x = 3 

b, x^3 - 25x = 0 

=> x(x^2 - 25) = 0 

=> x(x-5)(x + 5) = 0 

=> x =0 hoặc x - 5 = 0 hoặc x + 5 = 0 

=> x= 0 hoặc x =5 hoặc x = -5 

c, 9x^2 -  1 = 0 

 => (3x)^2 - 1^2 = 0 

=> ( 3x- 1)(3x+ 1) = 0 

=> 3x - 1 = 0 hoặc 3x + 1 = 0 

=> x = 1/3 hoặc x = -1/3  

a) (x+5)(x-4)=0

<=> x+5=0 hoặc x-4=0

<=> x=-5 hoặc x=4

b) (x-1)(x-3)=0

<=> x-1=0 hoặc x-3=0

<=> x=1 hoặc x=3

a) (x+5).(9x-4)=0

=> x+5=0 hoặc 9x-4=0

Nếu x+5=0: x=0-5=-5

Nếu 9x-4=0: 9x=0+4=4

                     x=4/9

b) (x-1).(x-3)=0

=> x-1=0 hoặc x-3=0

Nếu x-1=0: x=0+1=1

Nếu x-3=0: x=0+3=3

c) (3-x).(x-3)=0

=> 3-x=0 hoặc x-3=0

Nếu 3-x=0: x=3-0=0

Nếu x-3=0: x=0+3=3

d) x.(x+1)=0

=> x=0 hoặc x+1=0

Nếu x+1=0: x=0-1=-1

\(a,\left(x+5\right)\left(9x-4\right)=0\)

<=>\(\orbr{\begin{cases}x+5=0\\9x-4=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=-5\\x=\frac{4}{9}\end{cases}}\)

\(b,\left(x-1\right)\left(x-3\right)=0\)

<=>\(\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

\(c,\left(3-x\right)\left(x-3\right)=0\)

<=>\(\orbr{\begin{cases}3-x=0\\x-3=0\end{cases}}\)

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

\(d,x\left(x+1\right)=0\)

<=>\(\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

a) \(x\left(2x-1\right)-6x+3=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\2x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}\)

b) \(x^2\left(x+1\right)-9x-9=0\)

\(\Leftrightarrow x^2\left(x+1\right)-9\left(x+1\right)=0\)

\(\Leftrightarrow\left(x^2-9\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\pm\sqrt{9}=\pm3\end{cases}}\)

a) x(2x - 1) - 6x + 3 = 0

=> x(2x - 1) - 3(2x - 1) = 0

=> (x - 3)(2x - 1) = 0

=> \(\orbr{\begin{cases}x-3=0\\2x-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}\)

b) x²(x + 1)  - 9(x + 1) = 0

=> (x² - 9)(x + 1) = 0

=> \(\orbr{\begin{cases}x^2-9=0\\x+1=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=\pm3\\x=-1\end{cases}}\)

a, \(5x\left(x-1\right)+\left(x+17\right)=0\)

\(\Leftrightarrow5x^2-5x+x+17=0\Leftrightarrow5x^2-4x+17=0\)

\(\Leftrightarrow5\left(x^2-\frac{4}{5}x\right)+17=0\Leftrightarrow5\left(x^2-2.\frac{2}{5}x+\frac{4}{25}-\frac{4}{25}\right)+17=0\)

\(\Leftrightarrow5\left(x-\frac{2}{5}\right)^2-\frac{4}{5}+17=0\Leftrightarrow5\left(x-\frac{2}{5}\right)^2+81\ge81>0\)

Vậy pt vô nghiệm 

b, \(3x\left(x-3\right)^2-3x\left(x+3\right)^2=0\)

\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)

\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\Leftrightarrow x.2x=0\Leftrightarrow x=0\)

c, \(2x^2-9x+7=0\Leftrightarrow2x^2-7x-2x+7=0\)

\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\Leftrightarrow\left(x-1\right)\left(2x-7\right)=0\Leftrightarrow x=1;x=\frac{7}{2}\)

Trả lời:

a, \(5x\left(x-1\right)+\left(x+17\right)=0\)

\(\Leftrightarrow5x^2-5x+x+17=0\)

\(\Leftrightarrow5x^2-4x+17=0\)

\(\Leftrightarrow5\left(x^2-\frac{4}{5}x+\frac{17}{5}\right)=0\)

\(\Leftrightarrow x^2-\frac{4}{5}x+\frac{17}{5}=0\)

\(\Leftrightarrow x^2-2.x.\frac{2}{5}+\frac{4}{25}+\frac{81}{25}=0\)

\(\Leftrightarrow\left(x-\frac{2}{5}\right)^2+\frac{81}{25}=0\)

Vì \(\left(x-\frac{2}{5}\right)^2+\frac{81}{25}\ge\frac{81}{25}>0\forall x\)

nên pt vô nghiệm 

b, \(3x\left(x-3\right)^2-3x\left(x+3\right)^2=0\)

\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)

\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\)

\(\Leftrightarrow3x.\left(-9\right).2x=0\)

\(\Leftrightarrow-54x^2=0\)

\(\Leftrightarrow x^2=0\)

\(\Leftrightarrow x=0\)

Vậy x = 0 là nghiệm của pt.

c, \(7-9x+2x^2=0\)

\(\Leftrightarrow2x^2-7x-2x+7=0\)

\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=1\end{cases}}}\)

Vậy x = 7/2; x = 1 là nghiệm của pt.

d, trùng ý c

\(x^3-9x+7x^2-63=0\)

\(\Rightarrow\left(x^3+7x^2\right)-9x-63=0\)

\(\Rightarrow x^2\left(x+7\right)-9\left(x+7\right)=0\)

\(\Rightarrow\left(x^2-9\right)\left(x+7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2-9=0\\x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=9\\x=-7\end{cases}\Rightarrow}\orbr{\begin{cases}x=\pm3\\x=-7\end{cases}}}\)

Vậy ...

x3−9x+7x2−63=0x3−9x+7x2−63=0

⇒(x3+7x2)−9x−63=0⇒(x3+7x2)−9x−63=0

⇒x2(x+7)−9(x+7)=0⇒x2(x+7)−9(x+7)=0

⇒(x2−9)(x+7)=0⇒(x2−9)(x+7)=0

⇒{x2−9=0x+7=0⇒{x2=9x=−7⇒{x=±3x=−7⇒{x2−9=0x+7=0⇒{x2=9x=−7⇒{x=±3x=−7

Vậy ...

a)(9x-21):3=2

​9x-21=6

​9x=27

​x=3

​b)(x-1)(x-3)=0

​=>x-1=0 hoặc x-3=0

​x=1 hoặc x=3

a,  (9x - 21) : 3 = 2

=>  9x - 21 = 2 x 3

=> 9x - 21 = 6

=> 9x = 6 + 21

=> 9x = 27

=> x = 27 : 9 = 3

b,  (x - 1).(x - 3) = 0

=> x - 1 = 0      và x - 3 = 0

   x - 1 = 0   => x = 0 + 1 = 1

   x - 3 = 0   => x = 0 + 3 = 3

a)\(\left(x-2\right)^2-\left(2x+3\right)^2=0\Rightarrow\left(x-2+2x+3\right)\left(x-2-2x-3\right)=0\)

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

b)\(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\Rightarrow\left[3\left(2x+1\right)+2\left(x+1\right)\right]\left[3\left(2x+1\right)-2\left(x+1\right)\right]=0\)

\(\Rightarrow\left[8x+5\right]\left[4x+1\right]=0\Rightarrow\left[{}\begin{matrix}8x+5=0\\4x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)

c)\(x^3-6x^2+9x=0\Rightarrow x\left(x^2-6x+9\right)=0\Rightarrow x\left(x-3\right)^2=0\)

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

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

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

\(\Rightarrow x\left(x+1\right)\left(x-1\right)\left(x+1\right)+x\left(x-1\right)=0\)

\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)\left(x+1\right)+1\right]=0\)

\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)^2+1\right]=0\)

Do \(\left(x+1\right)^2+1>0\)

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

Lời giải:
a. $x^2-4x-5=0$
$\Leftrightarrow (x+1)(x-5)=0$

$\Leftrightarrow x+1=0$ hoặc $x-5=0$

$\Leftrightarrow x=-1$ hoặc $x=5$

b. 

$5x^2-9x-2=0$
$\Leftrightarrow (x-2)(5x+1)=0$

$\Leftrightarrow x-2=0$ hoặc $5x+1=0$

$\Leftrightarrow x=2$ hoặc $x=\frac{-1}{5}$

c.

$(x^2+1)-5(x^2+1)+6=0$

$\Leftrightarrow a^2-5a+6=0$ (đặt $x^2+1=a$)

$\Leftrightarrow (a-2)(a-3)=0$

$\Leftrightarrow a-2=0$ hoặc $a-3=0$

$\Leftrightarrow x^2-1=0$ hoặc $x^2-2=0$

$\Leftrightarrow (x-1)(x+1)=0$ hoặc $(x-\sqrt{2})(x+\sqrt{2})=0$

$\Leftrightarrow x\in\left\{\pm 1; \pm \sqrt{2}\right\}$

d.

$(x^2+6x)-2(x+3)^2-17=0$

$\Leftrightarrow (x^2+6x+9)-2(x+3)^2-26=0$

$\Leftrightarrow (x+3)^2-2(x+3)^2-26=0$
$\Leftrightarrow -(x+3)^2-26=0$

$\Leftrightarrow (x+3)^2=-26<0$ (vô lý)

Do đó không tồn tại $x$ thỏa mãn.