\(\text{a) 3.(x-2)+x.(x-2)=0}\)
\(\Leftrightarrow\)\(\text{(x-2)(3+x)=0}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\3+x=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)
\(\text{Vậy x=2 hoặc x=-3}\)
\(b,4x.\left(x-2\right)-x+2\)=0
\(\Leftrightarrow4x.\left(x-2\right)-\left(x-2\right)\)=0
\(\Leftrightarrow\left(x-2\right)\left(4x-1\right)\)=0
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\4x-1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{1}{4}\end{array}\right.\)
Vậy x=2 hoặc \(x=\frac{1}{4}\)
 

a) 3.(x-2) + x. ( x-2) = 0

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

TH1:

x - 2 = 0

x = 2

TH2:

3 + x = 0

x = -3

Vậy x = 2 hoặc x = -3

b) 4x.(x-2) -x +2 = 0

4x(x - 2) - x + 2 = 0

(x - 2)(4x - 1) = 0

TH1:

x - 2 = 0

x = 2

4x - 1 = 0

4x = 1

x = 1/4

Vậy x = 2 hoặc x = 1/4

a) 3 . ( x - 2 ) + x . ( x - 2 ) = 0

=> 3x - 6 + 2x - 2x = 0

=> 3x + 2x - 2x = 0 + 6

=> 3x = 6

=> x = 6 : 3 = 2

b) 4x . ( x - 2 ) - x + 2 = 0

=> 5x - 6x - x + 2 = 0

=> 5x - 6x - x = 0 - 2 = - 2

=> - 2x = - 2

=> x = - 2 : ( - 2 )

=> x = 1

\(\left(x-1\right)^2+\left(y+3\right)^2=0\left(1\right)\)

Ta thấy \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0,\forall x\\\left(y+3\right)^2\ge0,\forall y\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left(x+1\right)^2=0^2\\\left(y+3\right)^2=0^2\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x+1=0\\y+3=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)

a) pt

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

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

<=> x - 5 = 0 hoặc x + 4 = 0

<=> x = 5 hoặc x = -4

b) pt

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

<=> (2x - 1).(-2)=0

<=> 2x - 1 = 0

<=> x = 1/2

c) pt

<=> (x - 1)(x + 1)(x^2 + 4) = 0

<=> x - 1 = 0 hoặc x + 1 = 0 hoặc x^2 + 4 = 0

<=> x = 1 hoặc x = -1

a,x2−52−(x−5)=0<=>(x−5)(x+5)−(x−5)=0<=>(x−5)(x+4)=0=>x=5;x=−4.b,x2−x−6=0<=>x2−3x+2x−6=0<=>x(x−3)+2(x−3)=0<=>(x+2)(x−3)=0=>x=3;x=−2

a. x² - 25 - (x - 5) = 0

<=> x² - 5² - (x - 5) = 0

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

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

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

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

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

b. (2x - 1)² - (4x² - 1) = 0

<=> (2x - 1)² - (2x - 1)(2x + 1) = 0

<=> (2x - 1)(1 - 2x + 1) = 0

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

<=> \(\left[{}\begin{matrix}2x-1=0\\2-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

c. x²(x² + 4) - x² - 4 = 0

<=> x²(x² + 4) - (x² + 4) = 0

<=> (x² - 1)(x² + 4) = 0

<=> (x - 1)(x + 1)(x² + 4) = 0

<=> \(\left[{}\begin{matrix}x-1=0\\x+1=0\\x^2+4=0\left(VLí\right)\end{matrix}\right.\)

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

\(\Leftrightarrow x^2-2x+1-9x^2+36x-36=0\\ \Leftrightarrow-8x^2+34x-35=0\\ \Leftrightarrow8x^2-34x+35=0\\ \Leftrightarrow8x^2-20x-14x+35=0\\ \Leftrightarrow\left(2x-5\right)\left(4x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{7}{4}\end{matrix}\right.\)

a) \(x\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Vậy x=0 hoặc x=-2

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

\(\Rightarrow x-5=0\Rightarrow x=5\)

Vậy x=5

a) x(x-2)=0

=> x=0 hoặc x-2=0

=> x=0 hoặc x=0+2

=> x=0 hoặc x = 2

Bài 2 : Tìm x \(\in\) Z , biết :

a) x ( x-2 ) = 0

=> x = 0 hoặc x -2 = 0

* x - 2 = 0 => x = 2 .

Vậy x = 0 hoặc x = 2

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

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

=> x - 5 = 0 hoặc x\(^2\) + 2 = 0

*Nếu x - 5 = 0 => x = 5

* Nếu x\(^2\) + 2 = 0 => x\(^2\) = -2 => Không có giá trị nào để x\(^2\) = -2 => x ở trường hợp này ko xảy ra .

Vậy : x = 5

\(-2x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x-4=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

\(-2x\left(x-4\right)=0\)

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