1 ) a ) Sai đề
b ) \(x^3+3x^2y+3xy^2+y^3-x-y\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-1\right)\)
2 ) a ) \(6x^2-10x=10-6x\)
\(\Leftrightarrow6x^2-10x-10+6x=0\)
\(\Leftrightarrow6x^2-4x-10=0\)
\(\Leftrightarrow6x^2+6x-10x-10=0\)
\(\Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\)
\(\Leftrightarrow\left(6x-10\right)\left(x+1\right)=0\)
\(\Leftrightarrow2\left(3x-5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=5\\x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
Vậy ...
b ) \(x^2+3x+2=0\)
\(\Leftrightarrow x^2+2x+x+2=0\)
\(\Leftrightarrow x\left(x+2\right)+x+2=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-1\end{matrix}\right.\)
Vậy ...
3 ) \(A=x^2+x+3=x^2+x+\dfrac{1}{4}+\dfrac{11}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy Min A là : \(\dfrac{11}{4}\Leftrightarrow x=-\dfrac{1}{2}\)
3:
Ta có:
\(A=x^2+x+3\)
\(=x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)
Ta lại có:
\(\left(x+\dfrac{1}{2}\right)^2\ge0\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)
\(\Rightarrow A\ge\dfrac{11}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x+\dfrac{1}{2}=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy \(Min_A=\dfrac{11}{4}\Leftrightarrow x=-\dfrac{1}{2}\)
