tìm giá trị nhỏ nhất của biểu thức
c= x2+y2-x+6x+10
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1:
=x^2-6x+9-4=(x-3)^2-4>=-4
Dấu = xảy ra khi x=3
3: =-y^2-4y-4+13
=-(y+2)^2+13<=13
Dấu = xảy ra khi y=-2
4: D=x^2-8>=-8
Dấu = xảy ra khi x=0
Bài 1:
\(A=x^2+6x+9+x^2-10x+25\)
\(=2x^2+4x+34\)
\(=2\left(x^2+2x+17\right)\)
\(=2\left(x+1\right)^2+32>=32\forall x\)
Dấu '=' xảy ra khi x=-1
a) \(P=x^2-2x+5\)
\(=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\ge4\)
\(MinP=4\Leftrightarrow x-1=0\Rightarrow x=1\)
b) \(Q=2x^2-6x\)
\(=2\left(x^2-3x\right)\)
\(=2\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}\right)\)
\(=2\left(\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right)\)
\(=-\frac{9}{2}-2\left(x-\frac{3}{2}\right)^2\le\frac{-9}{2}\)
\(MinQ=\frac{-9}{2}\Leftrightarrow x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)
M=x^2+y^2-x+6y+10
M=(x^2-x+1/4)+(y^2+6y+9)+3/4
M=(x-1/2)^2+(y+3)^2+3/4
\(minM=\frac{3}{4}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}\)
a: ta có: \(P=x^2+10x+27\)
\(=x^2+10x+25+2\)
\(=\left(x+5\right)^2+2\ge2\forall x\)
Dấu '=' xảy ra khi x=-5
Bài 1:
a: \(M=x^2-10x+3\)
\(=x^2-10x+25-22\)
\(=\left(x^2-10x+25\right)-22\)
\(=\left(x-5\right)^2-22>=-22\forall x\)
Dấu '=' xảy ra khi x-5=0
=>x=5
b: \(N=x^2-x+2\)
\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi x-1/2=0
=>x=1/2
c: \(P=3x^2-12x\)
\(=3\left(x^2-4x\right)\)
\(=3\left(x^2-4x+4-4\right)\)
\(=3\left(x-2\right)^2-12>=-12\forall x\)
Dấu '=' xảy ra khi x-2=0
=>x=2
\(x^2+y^2+xy=3\)
Có \(x^2+y^2\ge2xy\) \(\Rightarrow3=x^2+y^2+xy\ge2xy+xy\) \(\Leftrightarrow xy\le1\)
\(x^2+y^2\ge-2xy\) \(\Rightarrow3=x^2+y^2+xy\ge-2xy+xy\) \(\Leftrightarrow-3\le xy\)
Đặt A= \(x^2+y^2-xy=\left(3-xy\right)-xy=3-2xy\)
mà \(-3\le xy\le1\) \(\Rightarrow9\ge3-2xy\ge1\)
=> minA=1 <=> \(\left\{{}\begin{matrix}xy=1\\x=y\end{matrix}\right.\) <=>x=y=1
maxA=9 <=>\(\left\{{}\begin{matrix}xy=-3\\x=-y\end{matrix}\right.\) <=>\(\left(x;y\right)=\left(\sqrt{3};-\sqrt{3}\right);\left(-\sqrt{3};\sqrt{3}\right)\)
Đặt \(P=x^2+y^2-xy\)
\(\Rightarrow\dfrac{P}{3}=\dfrac{x^2+y^2-xy}{3}=\dfrac{x^2+y^2-xy}{x^2+y^2+xy}\)
\(\dfrac{P}{3}=\dfrac{3x^2+3y^2-3xy}{3\left(x^2+y^2+xy\right)}=\dfrac{x^2+y^2+xy+2\left(x^2+y^2-2xy\right)}{3\left(x^2+y^2+xy\right)}\)
\(\dfrac{P}{3}=\dfrac{1}{3}+\dfrac{2\left(x-y\right)^2}{3\left(x^2+y^2+xy\right)}\ge\dfrac{1}{3}\Rightarrow P\ge1\)
\(P_{min}=1\) khi \(x=y=1\)
\(\dfrac{P}{3}=\dfrac{x^2+y^2-xy}{x^2+y^2+xy}=\dfrac{3\left(x^2+y^2+xy\right)-2\left(x^2+y^2+2xy\right)}{x^2+y^2+xy}=3-\dfrac{2\left(x+y\right)^2}{x^2+y^2+xy}\le3\)
\(\Rightarrow P\le9\)
\(P_{max}=9\) khi \(\left(x;y\right)=\left(\sqrt{3};-\sqrt{3}\right);\left(-\sqrt{3};\sqrt{3}\right)\)
Bài 1:
a: \(A=x^2+2x+4\)
\(=x^2+2x+1+3\)
\(=\left(x+1\right)^2+3>=3\forall x\)
Dấu '=' xảy ra khi x+1=0
=>x=-1
Vậy: \(A_{min}=3\) khi x=-1
b: \(B=x^2-20x+101\)
\(=x^2-20x+100+1\)
\(=\left(x-10\right)^2+1>=1\forall x\)
Dấu '=' xảy ra khi x-10=0
=>x=10
Vậy: \(B_{min}=1\) khi x=10
c: \(C=x^2-2x+y^2+4y+8\)
\(=x^2-2x+1+y^2+4y+4+3\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+3>=3\forall x\)
Dấu '=' xảy ra khi x-1=0 và y+2=0
=>x=1 và y=-2
Vậy: \(C_{min}=3\) khi (x,y)=(1;-2)
Bài 2:
a: \(A=5-8x-x^2\)
\(=-\left(x^2+8x\right)+5\)
\(=-\left(x^2+8x+16-16\right)+5\)
\(=-\left(x+4\right)^2+16+5=-\left(x+4\right)^2+21< =21\forall x\)
Dấu '=' xảy ra khi x+4=0
=>x=-4
b: \(B=x-x^2\)
\(=-\left(x^2-x\right)\)
\(=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}< =\dfrac{1}{4}\forall x\)
Dấu '=' xảy ra khi \(x-\dfrac{1}{2}=0\)
=>\(x=\dfrac{1}{2}\)
c: \(C=4x-x^2+3\)
\(=-x^2+4x-4+7\)
\(=-\left(x^2-4x+4\right)+7\)
\(=-\left(x-2\right)^2+7< =7\forall x\)
Dấu '=' xảy ra khi x-2=0
=>x=2
d: \(D=-x^2+6x-11\)
\(=-\left(x^2-6x+11\right)\)
\(=-\left(x^2-6x+9+2\right)\)
\(=-\left(x-3\right)^2-2< =-2\forall x\)
Dấu '=' xảy ra khi x-3=0
=>x=3
10?
\(C=x^2+y^2-x+6x+10\\ =x^2+5x+y^2+10\\ =x^2+2\cdot\dfrac{5}{2}x+\dfrac{25}{4}+y^2+\dfrac{15}{4}\\ =\left(x+\dfrac{5}{2}\right)^2+y^2+\dfrac{15}{4}\)
Mà \(\left(x+\dfrac{5}{2}\right)^2+y^2\ge0\forall x,y\)
\(\Rightarrow\left(x+\dfrac{5}{2}\right)^2+y^2+\dfrac{15}{4}\ge\dfrac{15}{4}\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{5}{2}=0\Leftrightarrow x=-\dfrac{5}{2}\\y=0\end{matrix}\right.\)
Vậy GTNN của C là \(\dfrac{15}{4}\) khi x = \(-\dfrac{5}{2}\) và y = 0