a.
\(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2x+10-x^2-5x=0\)
\(\Leftrightarrow-x^2-3x+10=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow\left(x^2+5x\right)-\left(2x+10\right)=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
b.
\(2x^2+3x-5=0\)
\(\Leftrightarrow2x^2-2x+5x-5=0\)
\(\Leftrightarrow\left(2x^2-2x\right)+\left(5x-5\right)=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-5}{2}\end{matrix}\right.\)
a) 2(x + 5) - x2 - 5x = 0
\(\Leftrightarrow\) 2x + 10 - x2 - 5x = 0
\(\Leftrightarrow\) -x2 - 3x + 10 = 0
\(\Leftrightarrow\) -(x2 + 3x - 10) = 0
\(\Leftrightarrow\) x2 + 3x - 10 = 0
\(\Leftrightarrow\) x2 - 2x + 5x - 10 = 0
\(\Leftrightarrow\) x(x - 2) + 5(x - 2) = 0
\(\Leftrightarrow\) (x + 5)(x - 2) = 0
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy.........
b) 2x2 + 3x - 5 = 0
\(\Leftrightarrow\) 2x2 - 2x + 5x - 5 = 0
\(\Leftrightarrow\) 2x(x - 1) + 5(x - 1) = 0
\(\Leftrightarrow\) (2x + 5)(x - 1) = 0
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)
Vậy.........
c) (x- 1)2 + 4(x + 2) - (x2 - 3) = 0
\(\Leftrightarrow\) x2 - 2x + 1 + 4x + 8 - x2 + 3 = 0
\(\Leftrightarrow\) 2x + 4 = 0
\(\Leftrightarrow\) 2(x + 2) = 0
\(\Leftrightarrow\) x = -2
Vậy.............
Mấy bn đọc bài mk xong nhớ tik nha
c.
\(\left(x-1\right)^2+4\left(x+2\right)-\left(x^2-3\right)=0\)
\(\Leftrightarrow x^2-2x+1+4x+8-x^2+3=0\)
\(\Leftrightarrow2x+12=0\)
\(\Leftrightarrow x=-6\)
