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Bài 3:
a) Ta có: \(A=25x^2-20x+7\)
\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)
\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)
d) Ta có: \(D=x^2-2x+2\)
\(=x^2-2x+1+1\)
\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)
Bài 1:
a) Ta có: \(A=x^2-2x+5\)
\(=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi x=1
b) Ta có: \(B=x^2-x+1\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
\(a,=3\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)
Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)
\(b,=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(c,=\left(x^2-2xy+y^2\right)+x^2+1=\left(x-y\right)^2+x^2+1\ge1\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=0\end{matrix}\right.\Leftrightarrow x=y=0\)
\(A=x^2-x+3=x^2-x+\dfrac{1}{4}-\dfrac{1}{4}+3=\left(x-2\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\left(\left(x-2\right)^2\ge0\right)\)
\(\Rightarrow Min\left(A\right)=\dfrac{11}{4}\)
\(B=x^2-4x+1=x^2-4x+4-4+1=\left(x-2\right)^2-3\ge-3\left(\left(x-2\right)^2\ge0\right)\)
\(\Rightarrow Min\left(B\right)=-3\)
Câu C bạn xem lại đề
\(D=3-4x-x^2=3+4-4-4x-x^2=7-\left(x^2+4x+4\right)=7-\left(x+2\right)^2\le7\left(-\left(x+2\right)^2\le0\right)\)
\(\Rightarrow Max\left(D\right)=7\)
\(A=x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2+\dfrac{11}{4}\\ =\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\in R\)
Vậy GTNN của A là 11/4 khi x=1/2
a) \(M=x^2-3x+10\)
\(M=x^2-2\cdot\dfrac{3}{2}\cdot x+\dfrac{9}{4}+\dfrac{31}{4}\)
\(M=\left(x^2-2\cdot\dfrac{3}{2}\cdot x+\dfrac{9}{4}\right)+\dfrac{31}{4}\)
\(M=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\)
Mà: \(\left(x-\dfrac{3}{2}\right)^2\ge0\) nên: \(M=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\ge\dfrac{31}{4}\)
Dấu "=" xảy ra
\(\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}=\dfrac{31}{4}\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)
Vậy: \(M_{min}=\dfrac{31}{4}\) với \(x=\dfrac{3}{2}\)
b) \(N=2x^2+5y^2+4xy+8x-4y-100\)
\(N=x^2+x^2+4y^2+y^2+4xy+8x-4y-120+16+4\)
\(N=\left(x^2+4xy+4y^2\right)+\left(x^2+8x+16\right)+\left(y^2-4y+4\right)-120\)
\(N=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\)
Mà:
\(\left\{{}\begin{matrix}\left(x+2y\right)^2\ge0\\\left(x+4\right)^2\ge0\\\left(y-2\right)^2\ge0\end{matrix}\right.\) nên \(N=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\ge120\)
Dấu "=" xảy ra:
\(\left\{{}\begin{matrix}\left(x+2y\right)^2=0\\\left(x+4\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-4+2y=0\\x=-4\\y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-4\\y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
Vậy: \(N_{min}=120\) khi \(\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(a,P=x^2-2x+5=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)
Dấu \("="\Leftrightarrow x=1\)
\(b,Q=2x^2-6x=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}-\dfrac{9}{4}\right)=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
Dấu \("="\Leftrightarrow x=\dfrac{3}{2}\)
\(c,M=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)
a: Ta có: \(P=x^2-2x+5\)
\(=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi x=1
Bài 1:
Ta có: \(5x^3-3x^2+2x+a⋮x+1\)
\(\Leftrightarrow5x^3+5x^2-8x^2-8x+10x+10+a-10⋮x+1\)
\(\Leftrightarrow a-10=0\)
hay a=10
Bài 1: \(A=2x^2-8x=2\left(x^2-4x\right)\)
\(=2\left(x^2-4x+4\right)-8=2\left(x-2\right)^2-8\ge-8\)
Vậy MinA= -8 \(\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)
\(B=3x^2-3x=3\left(x^2-x\right)=3\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{3}{4}\)
\(=3\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\ge-\dfrac{3}{4}\)
Vậy \(Min_B=-\dfrac{3}{4}\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)
\(C=x^2+y^2-2x+4y+7=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+2\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+2\ge2\)
Vậy \(Min_C=2\Leftrightarrow x=1;y=-2\)
\(D=x^2+4y^2+x+4y+2=\left(x^2+x+\dfrac{1}{4}\right)+\left(4y^2+4y+1\right)+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\left(2y+1\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy \(Min_D=\dfrac{3}{4}\Leftrightarrow x=y=-\dfrac{1}{2}\)
Bài 2: \(A=x-x^2=-\left(x^2-x\right)=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
Vậy \(Max_A=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)
\(B=3x-2x^2=-2\left(x^2-\dfrac{3}{2}x\right)\)
\(=-2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)+\dfrac{9}{8}\)
\(=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{9}{8}\le\dfrac{9}{8}\)
Vậy \(Max_B=\dfrac{9}{8}\Leftrightarrow x=\dfrac{3}{4}\)
\(C=2x-2x^2-3=-2\left(x^2-x+\dfrac{3}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{5}{4}\right)=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{2}\le-\dfrac{5}{2}\)
Vậy \(Max_C=-\dfrac{5}{2}\Leftrightarrow x=\dfrac{1}{2}\)