Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(3x^2-9x+5=3\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\ge-\frac{7}{4}\)
Dấu "=" xảy ra khi x = 3/2
Vậy BT đạt giá trị nhỏ nhất bằng -7/4 khi x = 3/2
b/ \(x^2+y^2+x-y-1=\left(x^2+x+\frac{1}{4}\right)+\left(y^2-y+\frac{1}{4}\right)-\frac{3}{2}=\left(x+\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2-\frac{3}{2}\ge-\frac{3}{2}\)
Dấu "=" xảy ra khi \(\begin{cases}x=-\frac{1}{2}\\y=\frac{1}{2}\end{cases}\)
Vậy BT đạt giá trị nhỏ nhất bằng -3/2 khi (x;y) = (-1/2;1/2)
c/ \(2x^2+2x+1=2\left(x+\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)
Dấu "=" xảy ra khi x = -1/2
Vậy BT đạt giá trị nhỏ nhất bằng 1/2 khi x = -1/2
\(A=x^2+4x+5=\left(x+2\right)^2+1\ge1\)
Dấu \("="\Leftrightarrow x=-2\)
\(B=x^2+10x-1=\left(x+5\right)^2-26\ge-26\)
Dấu \("="\Leftrightarrow x=-5\)
\(C=5-4x+4x^2=\left(2x-1\right)^2+4\ge4\)
Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)
\(D=x^2+y^2-2x+6y-3=\left(x-1\right)^2+\left(y+3\right)^2-13\ge-13\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
\(E=2x^2+y^2+2xy+2x+3=\left(x+y\right)^2+\left(x+1\right)^2+2\ge2\)
Dấu \("="\Leftrightarrow x=-y=-1\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
\(A=x^2+4x+5\)
\(=x^2+4x+4+1\)
\(=\left(x+2\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi x=-2
\(C=4x^2-4x+5\)
\(=4x^2-4x+1+4\)
\(=\left(2x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
a/ \(M=x^2+y^2-x+6y+10=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+10-\frac{1}{4}-9\)
\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Suy ra Min M = 3/4 <=> (x;y) = (1/2;-3)
b/
1/ \(A=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
Suy ra Min A = 7 <=> x = 2
2/ \(B=x-x^2=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Suy ra Min B = 1/4 <=> x = 1/2
3/ \(N=2x-2x^2-5=-2\left(x^2-x+\frac{1}{4}\right)-5+\frac{1}{2}=-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\)
\(\ge-\frac{9}{2}\)
Suy ra Min N = -9/2 <=> x = 1/2
a) \(A=9x^2+5x+1\)
\(A=9x^2+5x+\frac{25}{36}+\frac{11}{36}\)
\(A=\left(3x+\frac{5}{6}\right)^2+\frac{11}{36}\)
Có: \(\left(3x+\frac{5}{6}\right)^2\ge0\)
\(\Rightarrow\left(3x+\frac{5}{6}\right)^2+\frac{11}{36}\ge\frac{11}{36}\)
Dấu = xảy ra khi: \(\left(3x+\frac{5}{6}\right)^2=0\Rightarrow3x+\frac{5}{6}=0\)
\(\Rightarrow x=-\frac{5}{18}\)
Vậy: \(Min_A=\frac{11}{36}\) tại \(x=-\frac{5}{18}\)
b) \(B=4x^2+12x-8\)
\(B=4x^2+12x+9-17\)
\(B=\left(2x+3\right)^2-17\)
Có: \(\left(2x+3\right)^2\ge0\)
\(\Rightarrow\left(2x+3\right)^2-17\ge-17\)
Dấu = xảy ra khi: \(\left(2x+3\right)^2=0\Rightarrow2x+3=0\)
\(\Rightarrow x=-\frac{3}{2}\)
Vậy: \(Min_B=-17\) tại \(x=-\frac{3}{2}\)
a) \(A=9x^2-2x+15\)
\(A=9x^2-2x+\frac{1}{9}+\frac{134}{9}\)
\(A=\left(3x+\frac{1}{3}\right)^2+\frac{134}{9}\)
Có: \(\left(3x+\frac{1}{3}\right)^2\ge0\Rightarrow\left(3x+\frac{1}{3}\right)^2+\frac{134}{9}\ge\frac{134}{9}\)
Dấu '=' xảy ra khi: \(\left(3x+\frac{1}{3}\right)^2=0\Rightarrow3x+\frac{1}{3}=0\Rightarrow x=-\frac{1}{9}\)
Vậy: \(Min_A=\frac{134}{9}\) tại \(x=-\frac{1}{9}\)
b) \(B=3x^2+x+1\)
\(B=3x^2+x+\frac{1}{12}+\frac{11}{12}\)
\(B=\left(\sqrt{3}x+\sqrt{\frac{1}{12}}\right)^2+\frac{11}{12}\)
Có: \(\left(\sqrt{3}x+\sqrt{\frac{1}{12}}\right)^2\ge0\Rightarrow\left(\sqrt{3}x+\sqrt{\frac{1}{12}}\right)^2+\frac{11}{12}\ge\frac{11}{12}\)
Dấu '=' xảy ra khi: \(\left(\sqrt{3}x+\sqrt{\frac{1}{12}}\right)^2=0\Rightarrow\sqrt{3}x+\sqrt{\frac{1}{12}}=0\Rightarrow x=-\frac{1}{6}\)
Vậy: \(Min_B=\frac{11}{12}\) tại \(x=-\frac{1}{6}\)
c) \(C=x^2-6y+4x+y^2+38\)
\(C=\left(x^2+4x+4\right)+\left(y^2-6y+9\right)+25\)
\(C=\left(x+2\right)^2+\left(y-3\right)^2+25\)
Có: \(\left(x+2\right)^2+\left(y-3\right)^2\ge0\Rightarrow\left(x+2\right)^2+\left(y-3\right)^2+25\ge25\)
Dấu = xảy ra khi: \(\hept{\begin{cases}\left(x+2\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x+2=0\\y-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=3\end{cases}}\)
Vậy: \(Min_C=25\) tại \(\hept{\begin{cases}x=-2\\y=3\end{cases}}\)