D

D. 52

D

a) \(x^2-2x=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)

b) \(\left(3x-1\right)^2-16=0\)

\(\left(3x-1\right)^2-4^2=0\)

\(\left(3x-1-4\right)\left(3x-1+4\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}3x-5=0\\3x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}\)

c) \(x^2-25x=0\)

\(x\left(x-25\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}}\)

d) \(\left(4x-1\right)^2-9=0\)

\(\left(4x-1\right)^2-3^2=0\)

\(\left(4x-1-3\right)\left(4x-1+3\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}4x-4=0\\4x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-1}{2}\end{cases}}}\)

a) \(x^2-2x=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)

vậy..

b) \(\left(3x-1\right)^2-16=0\)

\(\left(3x-1\right)^2=16\)

\(\left(3x-1\right)^2=4^2=\left(-4\right)^2\)

\(\Rightarrow\orbr{\begin{cases}3x-1=4\\3x-1=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}\)

vậy ...

c) \(x^2-25x=0\)

\(x.\left(x-25\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}}\)

vậy ....

d) \(\left(4x-1\right)^2-9=0\)

\(\left(4x-1\right)^2=3^2=\left(-3\right)^2\)

\(\Rightarrow\orbr{\begin{cases}4x-1=3\\4x-1=-3\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)

vậy ...

a)\(x^2-2x=0\)

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

\(\Rightarrow\hept{\begin{cases}x=0\\x-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=2\end{cases}}\)

b)\(\left(3x-1\right)^2-16=0\)

\(\left(3x-1-4\right)\left(3x-1+4\right)=0\)

\(\Rightarrow\hept{\begin{cases}3x-1-4=0\\3x-1+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)

c)\(x^2-25x=0\)

\(x\left(x-25\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=25\end{cases}}\)

d)\(\left(4x-1\right)^2-9=0\)

\(\left(4x-1-3\right)\left(4x-1+3\right)=0\)

\(\Rightarrow\hept{\begin{cases}4x-1-3=0\\4x-1+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\x=-\frac{1}{2}\end{cases}}}\)

\(a,\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\\ b,\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(1,5+\dfrac{-3}{x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{16}\\-\dfrac{3}{x}=-1,5=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=2\end{matrix}\right.\)

a: \(\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

b: \(\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(\dfrac{1}{5}+\left(-3\right):x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{16}\\\left(-3\right):x=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{16}:\dfrac{3}{4}=\dfrac{9}{16}\cdot\dfrac{4}{3}=\dfrac{3}{4}\\x=\left(-3\right):\dfrac{-1}{5}=15\end{matrix}\right.\)

a)   \(x^2+8x+16=0\)

<=>  \(\left(x+4\right)^2=0\)

<=> \(x=-4\)

Vậy...

b) mk chỉnh đề

 \(4x^2-12x=-9\)

<=>  \(4x^2-12x+9=0\)

<=>  \(\left(2x-3\right)^2=0\)

<=>  \(x=\frac{3}{2}\)

Vậy....

a﴿ x + 8x + 16 = 0

<=> x + 4 = 0

<=> x = −4

Vậy x=-4

1C

2A

2: \(3x\left(x-4\right)+2x-8=0\)

=>\(3x\left(x-4\right)+2\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(3x+2\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)

3: 4x(x-3)+x²-9=0

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

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

=>\(\left(x-3\right)\left(5x+3\right)=0\)

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

4: \(x\left(x-1\right)-x^2+3x=0\)

=>\(x^2-x-x^2+3x=0\)

=>2x=0

=>x=0

5: \(x\left(2x-1\right)-2x^2+5x=16\)

=>\(2x^2-x-2x^2+5x=16\)

=>4x=16

=>x=4

a) 4x² - 9=0

(2x)² - 3² = 0

=》(2x - 3)(2x+3) =0 

=》 2x - 3 = 0 hoặc 2x +3 = 0 

=》x = 1,5 hoặc x = - 1,5 

b) (x + 1)² - 16 = 0

=》( x + 1)² - 4² = 0

=》( x - 3 )( x + 5 ) =0 

=》 x - 3 = 0 hoặc x + 5 = 0

=》 x = 3 hoặc x = -5 

c) ( x + 1)² - (2x + 3)² = 0

=》 ( x + 1 - 2x - 3)(x+1 +2x +3 ) =0

=》 ( -x - 2 )( 3x + 4 ) = 0 

=》 -x -2 =0 hoặc 3x + 4 = 0

=》 x = -2 hoặc x = -4/3

d) 4(3x +2)² - 9( x + 1 )² =0

=》 [ 2(3x +2) ]² - [3 (x + 1)] ² = 0

=> ( 6x +4 )² - ( 3x + 3)² = 0

=》 ( 6x +4 -3x -3 )( 6x + 4 + 3x + 3 )=0

=》 (3x +1)(9x + 7 ) =0 

=》 3x + 1 =0 hoặc 9x + 7 =0 

=》 x = -1/3 hoặc x = -7/9

a) Thực hiện rút gọn VT = -2x – 64

Giải phương trình -2x – 64 = 0 thu được x = -32.

b) Thực hiện rút gọn VT = -62 x +12

Giải phương trình -62x + 12 = -50 thu được x = 1.

x^2-2.2.x+2^2+12=0

=>(x-2)^2=-12

vì (x-2)^2>=0

=>x thuộc tập rỗng