a) \(\left(x+5\right)^2\left(2+3x\right)=\left(x+5\right)^2x^2\)
\(\Leftrightarrow\left(x+5\right)^2\left(2+3x\right)-\left(x+5\right)^2x^2=0\)\(\Leftrightarrow\left(x+5\right)^2\left(2+3x-x^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2+3x-x^2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3+\sqrt{17}}{2}\\x=\dfrac{3-\sqrt{17}}{2}\end{matrix}\right.\)
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b) \(\left(x+3\right)\left(4-x\right)=x^2+6x+9\)
\(\Leftrightarrow\left(x+3\right)\left(4-x\right)=\left(x+3\right)^2\) \(\Leftrightarrow\left(x+3\right)\left(4-x\right)-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)\left(4-x-x-3\right)=0\Leftrightarrow\left(x+3\right)\left(1-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\1-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
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b, ( chép lại đề )
<=> (x+3) (4-x) - ( x + 3 ) 2 = 0
<=> (x+3) ( (4-x)-(x+3) ) = 0
<=> ( x +3 ) ( 4-x-x-3)=0
<=> ( x+3) ( -2x +1 )= 0
<=> x+3 = 0
hoặc -2x+1 = 0
<=> x=-3
hoặc x=1/2
Vậy...
