giá trị nhỏ nhất của biểu thức A=3x^2+y^2+2xy+4x là ...
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=3x^2+y^2+2xy+4x\)
\(\Leftrightarrow A=y^2+2xy+x^2+2x^2+4x+2-2\)
\(\Leftrightarrow A=\left(x+y\right)^2+2\left(x+1\right)^2-2\)
Vì \(\left(x+y\right)^2\ge0;2\left(x+1\right)^2\ge0\)
\(\Rightarrow\left(x+y\right)^2+2\left(x+1\right)^2-2\ge-2\)
Dấu = xảy ra khi \(\hept{\begin{cases}x+y=0\\x+1=0\end{cases}\Rightarrow}\hept{\begin{cases}y=1\\x=-1\end{cases}}\)
Vậy Min A=-2 khi \(y=1;x=-1\)
\(3x^2+y^2+2xy+4x\)
\(=x^2+2xy+y^2+2x^2+4x+2-2\)
\(=\left(x+y\right)^2+2.\left(x+1\right)^2-2\ge-2\)
Dấu bằng xảy ra khi
\(\hept{\begin{cases}x=-y\\x=-1\end{cases}\Leftrightarrow\hept{\begin{cases}y=1\\x=-1\end{cases}}}\)
Vậy Min \(3x^2+y^2+2xy+4x\)=2 khi x=-1;y=1
![](https://rs.olm.vn/images/avt/0.png?1311)
A=-x^2+2xy-y^2-x^2+4x-4-36
=-(x-y)^2-(x-2)^2-36<=-36
Dấu = xảy ra khi x=y=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x^2+2xy+y^2+2x^2+4x+2-2\)
\(A=\left(x+y\right)^2+2\left(x+1\right)^2-2\ge-2\)
\(\Rightarrow A_{min}=-2\) khi \(\left\{{}\begin{matrix}x+1=0\\x+y=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
\(A=\left(y^2+2xy+x^2\right)+\left(2x^2+4x+2\right)-2\)
\(A=\left(y+x\right)^2+2\left(x+1\right)^2-2\)
\(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\\\left(x+1\right)^2\ge0\end{matrix}\right.\)\(\Rightarrow A\ge-2\)
\(A_{min}=-2\) khi \(x=-1,y=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) = 9(x2 - 2.x/2.9 + 1/324) - 9/324 +5
GTNN A = 4,97
b) = (2x +y)2 + y2 + 2018
GTNN B = 2018 khi x=0;y=0
c) = -4(x2 - 2.3x/ 4.2 + 9/16) +9/16 +10
GTLN C = 169/16
d) = -(x-y)2 - (2x +1) +1 + 2016
GTLN D = 2017
(trg bn cho bài khó dữ z, làm hại cả não tui)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=2x^2+y^2-4x+2xy+1\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2-4x+4\right)-3\)
\(=\left(x+y\right)^2+\left(x-2\right)^2-3\ge-3\)
-Dấu bằng xảy ra khi và chỉ khi \(x=2\) và \(y=-2\).
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=4x^2-12x+100=\left(2x\right)^2-12x+3^2+91=\left(2x-3\right)^2+91\)
Ta có: \(\left(2x-3\right)^2\ge0\forall x\inℤ\)
\(\Rightarrow\left(2x-3\right)^2+91\ge91\)
hay A \(\ge91\)
Dấu "=" xảy ra <=> \(\left(2x-3\right)^2=0\)
<=> 2x-3=0
<=> 2x=3
<=> \(x=\frac{3}{2}\)
Vậy Min A=91 đạt được khi \(x=\frac{3}{2}\)
b) \(B=-x^2-x+1=-\left(x^2+x-1\right)=-\left(x^2+x+\frac{1}{4}-\frac{5}{4}\right)=-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\)
Ta có: \(-\left(x+\frac{1}{2}\right)^2\le0\forall x\)
\(\Rightarrow-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\) hay B\(\le\frac{5}{4}\)
Dấu "=" \(\Leftrightarrow-\left(x+\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x+\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy Max B=\(\frac{5}{4}\)đạt được khi \(x=\frac{-1}{2}\)
\(C=2x^2+2xy+y^2-2x+2y+2\)
\(C=x^2+2x\left(y-1\right)+\left(y-1\right)^2+x^2+1\)
\(\Leftrightarrow C=\left(x+y-1\right)^2+x^2+1\)
Ta có:
\(\hept{\begin{cases}\left(x+y-1\right)^2\ge0\forall x;y\inℤ\\x^2\ge0\forall x\inℤ\end{cases}}\)
\(\Leftrightarrow\left(x+y-1\right)^2+x^2+1\ge1\)
hay C\(\ge\)1
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x+y-1\right)^2=0\\x^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=1\\x=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=1\\x=0\end{cases}}}\)
Vậy Min C=1 đạt được khi y=1 và x=0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=3x^2+y^2+2xy+4x\)
\(=\left(2x^2+4x+2\right)+\left(x^2+y^2+2xy\right)-2\)
\(=2\left(x^2+2x+1\right)+\left(x+y\right)^2-2\)
\(=2\left(x+1\right)^2+\left(x+y\right)^2-2\)
Dễ thấy: \(2\left(x+1\right)^2+\left(x+y\right)^2\ge0\)
\(\Rightarrow2\left(x+1\right)^2+\left(x+y\right)^2-2\ge-2\)
Xảy ra khi \(\hept{\begin{cases}x+1=0\\x+y=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=-1\\x=-y\end{cases}}\Rightarrow x=-y=-1\)