Question

Tìm x:

a) x^2 - 25x=0

b) (x-3)^2 - 36x^2=0

c) 2x(3-x)+2x^2=12

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

Answer 1

a) x² - 25x = 0

=> x(x - 25) = 0

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

b) (x - 3)² - 36x² = 0

=> (x - 3)² - (6x)² = 0

=> \(\left(x+6x-3\right)\left(x-6x-3\right)=0\)

=> \(\orbr{\begin{cases}7x-3=0\\-5x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{7}\\x=-\frac{3}{5}\end{cases}}\)

c) 2x(3 - x) + 2x² = 12

=> 6x - 2x² + 2x² = 12

=> 6x = 12

=> x = 2

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

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

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

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

Answer 2

a. x²  - 25x = 0

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

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

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

Vậy ...

b.(x-3)² - 36x² = 0

\(\Leftrightarrow\left(x-3-6x\right)\left(x-3+6x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-5x-3=0\\7x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-3}{5}\\x=\frac{3}{7}\end{cases}}\)

Vậy...

c.2x(3-x)+2x² = 12 

<=> 6x - 2x² + 2x² = 12

<=> 6x = 12

<=> x = 2

d. x (x-2) - x + 2 =0

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}\)

Vậy...