Đáp án:
$6)\\ a)min_A= 1 \Leftrightarrow x=10\\ b)min_B=\ge 1 \Leftrightarrow x=-\dfrac{1}{2}\\ c)min_C=2 \Leftrightarrow \left\{\begin{array}{l} x=-3\\y=1\end{array} \right.\\ d)min_D= -\dfrac{9}{2} \Leftrightarrow x=\dfrac{3}{2}\\ 7)\\ a)max_M=7 \Leftrightarrow x=2\\ b)max_N=\dfrac{1}{4} \Leftrightarrow x=\dfrac{1}{2}\\ c)max_P=-\dfrac{9}{2} \Leftrightarrow x=\dfrac{1}{2}$
Giải thích các bước giải:
$5)\\ a)VT=(a-b)^3=[-(b-a)]^3=-(b-a)^3=VP\\ b)(-a-b)^2=[-(a+b)]^2=(a+b)^2=VP\\ c)VP=x(x-3y)^2+y(y-3x)^2\\ =x(x^2-6xy+9y^2)+y(y^2-6xy+9x^2)\\ =x^3-6x^2y+9xy^2+y^3-6xy^2+9x^2y\\ =x^3-6x^2y+9x^2y+9xy^2-6xy^2+y^3\\ =x^3+3x^2y+3xy^2+y^3\\ =(x+y)^3\\ =VT\\ d)VT=(x+y)^3-(x-y)^3\\ =x^3+3x^2y+3xy^2+y^3-(x^3-3x^2y+3xy^2-y^3)\\ =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\\ =3x^2y+y^3+3x^2y+y^3\\ =2y^3+6x^2y\\ =2y(y^2+3x^2)\\ =VP\\ 6)\\ a)A=x^2-20x+101\\ =x^2-2.x.10+100+1\\ =(x-10)^2+1 \ge 1 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x-10=0\Leftrightarrow x=10$
$b)B=4x^2+4x+2\\ =4x^2+4x+1+1\\ =(2x+1)^2+1 \ge 1 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow 2x+1=0\Leftrightarrow x=-\dfrac{1}{2}$
$c)C=x^2-4xy+5y^2+10x-22y+28\\ =x^2-4xy+4y^2+y^2+10x-22y+28\\ =(x-2y)^2+10(x-2y)+y^2-2y+28\\ =(x-2y)^2+2.(x-2y).5+25+y^2-2y+1+2\\ =(x-2y+5)^2+(y-1)^2+2 \ge 2 \ \forall \ x,y$
Dấu "=" xảy ra $\Leftrightarrow \left\{\begin{array}{l} x-2y+5=0\\y-1=0\end{array} \right.\Leftrightarrow \left\{\begin{array}{l} x=-3\\y=1\end{array} \right.$
$d)D=2x^2-6x\\ =2(x^2-3x)\\ =2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ =2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2} \ge -\dfrac{9}{2} \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}$
$7)\\ a)M=4x-x^2+3\\ =-x^2+4x-4+7\\ =-(x-2)^2+7 \le 7 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x-2=0\Leftrightarrow x=2$
$b)N=x-x^2\\ =-x^2+x-\dfrac{1}{4}+\dfrac{1}{4}\\ =-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4} \le \dfrac{1}{4} \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}$
$c)P=2x-2x^2-5\\ =-2(x^2-x)-5\\ =-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}\\ =-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2} \le -\dfrac{9}{2} \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}$
