Question

giải phương trình:

a) (2x - 5)² - (x + 2)² = 0

b) (x + 5)(4x - 1) + x² - 25 = 0

Answer 1

a, (2x-5)²-(x+2)²=0

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

\(\Leftrightarrow\left(x-7\right)\left(3x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)\(\)

b, \(\left(x+5\right)\left(4x-1\right)+x^2-25=0\)

\(\Leftrightarrow\left(x+5\right)\left(4x-1\right)+\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(4x-1+x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(5x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\5x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{6}{5}\end{matrix}\right.\)

Answer 2

dễ thế này cũng phải đưa lên

Answer 3

a, (2x-5)²- (x+2)²=0

<=> (2x-5-x-2)(2x-5+x+2)=0

<=> (x-7)(3x-3)=0

<=> (x-7)(x-1)3=0

<=>\(\left[{}\begin{matrix}x-7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

Vậy S=\(\left\{7;1\right\}\)

b, (x+5)(4x-1)+x²-25=0

<=> (x+5)(4x-1)+(x-5)(x+5)=0

<=> (x+5)(4x-1+x-5)=0

<=> (x+5)(5x-6)=0

<=>\(\left[{}\begin{matrix}x+5=0\\5x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{6}{5}\end{matrix}\right.\)

Vậy S=\(\left\{-5;\dfrac{6}{5}\right\}\)