Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(x^2-4x+4\right)+4=\left(x-2\right)^2+4\ge4\)
\(minA=4\Leftrightarrow x=2\)
\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)
\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)
\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)
\(minC=-8\Leftrightarrow x=-1\)
\(D=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)
\(maxD=-4\Leftrightarrow x=1\)
\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)
\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)
\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)
\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)
\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)
\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)
\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(A=\left(x^2+2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{5}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\\ A_{min}=-\dfrac{5}{4}\Leftrightarrow x=-\dfrac{3}{2}\\ B=\left(x^2+2xy+y^2\right)+\left(x^2+6x+9\right)+3\\ B=\left(x+y\right)^2+\left(x+3\right)^2+3\ge3\\ B_{min}=3\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\\ C=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1\le1\\ C_{max}=1\Leftrightarrow x=1\)
1. \(A=2x^2-6x-2xy+y^2+10\)
\(\Leftrightarrow A=\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)+1\)
\(\Leftrightarrow A=\left(x-y\right)^2+\left(x-3\right)^2+1\)
Vì \(\left(x-y\right)^2\ge0\) ; \(\left(x-3\right)^2\ge0\)\(\forall x;y\)
\(\Rightarrow A=\left(x-y\right)^2+\left(x-3\right)^2+1\ge1\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=3\)
Vậy minA = 1 \(\Leftrightarrow x=y=3\)
2. \(A=5+2xy+14y-x^2-5y^2-2x\)
\(\Leftrightarrow A=-\left(x^2-2xy+y^2+2x-2y+1\right)-\left(4y^2-12y+9\right)+15\)
\(\Leftrightarrow A=-\left(x-y+1\right)^2-\left(2y-3\right)^2+15\)
Vì \(\left\{{}\begin{matrix}\left(x-y+1\right)^2\ge0\\\left(2y-3\right)^2\ge0\end{matrix}\right.\)\(\forall x;y\)
\(\Rightarrow A=-\left(x-y+1\right)^2-\left(2y-3\right)^2+15\le15\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y+1\right)^2=0\\\left(2y-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\y=\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{3}{2}\end{matrix}\right.\)
Vậy maxA = 15 \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{3}{2}\end{matrix}\right.\)
1. A=2x2−6x−2xy+y2+10A=2x2−6x−2xy+y2+10
⇔A=(x2−2xy+y2)+(x2−6x+9)+1⇔A=(x2−2xy+y2)+(x2−6x+9)+1
⇔A=(x−y)2+(x−3)2+1⇔A=(x−y)2+(x−3)2+1
Vì (x−y)2≥0(x−y)2≥0 ; (x−3)2≥0(x−3)2≥0∀x;y∀x;y
⇒A=(x−y)2+(x−3)2+1≥1⇒A=(x−y)2+(x−3)2+1≥1
Dấu "=" xảy ra ⇔{(x−y)2=0(x−3)2=0⇔x=y=3⇔{(x−y)2=0(x−3)2=0⇔x=y=3
Vậy minA = 1 ⇔x=y=3⇔x=y=3
2. A=5+2xy+14y−x2−5y2−2xA=5+2xy+14y−x2−5y2−2x
⇔A=−(x2−2xy+y2+2x−2y+1)−(4y2−12y+9)+15⇔A=−(x2−2xy+y2+2x−2y+1)−(4y2−12y+9)+15
⇔A=−(x−y+1)2−(2y−3)2+15⇔A=−(x−y+1)2−(2y−3)2+15
Vì {(x−y+1)2≥0(2y−3)2≥0{(x−y+1)2≥0(2y−3)2≥0∀x;y∀x;y
⇒A=−(x−y+1)2−(2y−3)2+15≤15⇒A=−(x−y+1)2−(2y−3)2+15≤15
Dấu "=" xảy ra ⇔{(x−y+1)2=0(2y−3)2=0⇔{x−y=−1y=32⇔{x=12y=32⇔{(x−y+1)2=0(2y−3)2=0⇔{x−y=−1y=32⇔{x=12y=32
Vậy maxA = 15 ⇔{x=12y=32
E = 2x^2 - 5x -2 = 2( x^2 -5/2x -1) = 2(x^2 - 2.x.5/4 +25/16 - 41/16) = 2(x - 5/4 )^2 + 41/8
Vậy GTNN của biểu thức là 41/8 tại x = 5/4
F = x^2 + 5y^2 + 2xy -y +3 = (x^2 + 2xy +y^2) + (4y^2 - 2.2y.1/4 + 1/16) +47/16
(x + y)^2 + (2y - 1/4)^2 + 47/16
Vậy GTNN của BT là 47/16 tại x = y = 1/8
a) \(A=\left(x+2\right)^2-4\)
Vì \(\left(x+2\right)^2\ge0\)
Nên \(A=\left(x+2\right)^2-4\ge-4\)
Dấu "=" xảy ra khi \(x+2=0\Rightarrow x=-2\)
Vậy \(A_{min}=-4\) khi \(x=-2\)
\(H=x^2+2xy+y^2+2x+2y+x^2+4x+2019=\left(x+y\right)^2+2\left(x+y\right)+\left(x+2\right)^2+2015\)
\(=\left(x+y+1\right)^2+\left(x+2\right)^2+2014\ge2014\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=-2;y=1\)
\(I=\left(1-x\right)^2+\left(-2-y\right)^2+\left(x+y\right)^2\ge\frac{\left(1-x-2-y+x+y\right)^2}{3}=\frac{1}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(1-x=-2-y=x+y\)\(\Leftrightarrow\)\(x=\frac{4}{3};y=\frac{-5}{3}\)
a) \(A=x^2-6x+15\)
\(A=x^2+6x+9+6\)
\(A=\left(x+3\right)^2+6\ge6\)
vậy Min A=6\(\Leftrightarrow\)x=-3
b) Min B=4x
c) \(C=2x^2-6x+4\)
d) \(D=x^2+x+1\)
\(=x^2+2\cdot\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
vậy Min D\(=\frac{3}{4}\Leftrightarrow x=-\frac{1}{2}\)
Ta có : A = x2 - 6x + 15
=> A = x2 - 2.x.3 + 9 + 6
=> A = x2 - 2.x.3 + 32 + 6
=> A = (x - 3)2 + 6
Mà : (x - 3)2 \(\ge0\forall x\in R\)
Nên : (x - 3)2 + 6 \(\ge6\forall x\in R\)
Vậy GTNN của A là 6 khi x = 3