Tìm gtnn của C=2x2-2x+4xy-6y+10y2+2022
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5 tháng 10 2021
\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)
\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)
5 tháng 10 2021
\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)
Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)
AD
1
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 8 2021
\(A=\left(x^2+4y^2+1-4xy+2x-4y\right)+\left(x^2-4x+4\right)-3\)
\(A=\left(x-2y+1\right)^2+\left(x-2\right)^2-3\ge-3\)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(2;\dfrac{3}{2}\right)\)
NT
0
22 tháng 10 2021
A=−2x2−10y2+4xy+4x+4y+2016A=−2x2−10y2+4xy+4x+4y+2016
=−2.(x2+5y2−4xy−4x−4y)+2016=−2.(x2+5y2−4xy−4x−4y)+2016
=−2.(x2+4y2+4−4xy−4x+8y+y2−12y+36)+2.36+2016=−2.(x2+4y2+4−4xy−4x+8y+y2−12y+36)+2.36+2016
=−2.[(x−2y−2)2+(y−6)2]+2088=−2.[(x−2y−2)2+(y−6)2]+2088
Ta có: (x−2y−2)2+(y−6)2≥0(x−2y−2)2+(y−6)2≥0
⇒−2.[(x−2y−2)2+(y−6)2]≤0⇒−2.[(x−2y−2)2+(y−6)2]≤0
⇒−2.[(x−2y−2)2+(y−6)2]+2088≤2088⇒−2.[(x−2y−2)2+(y−6)2]+2088≤2088
⇒A≤2088⇒A≤2088
Vậy giá trị lớn nhất của A=2088A=2088 khi: \hept{x−2y−2=0y=6⇒\hept{x=2y+2y=6⇒\hept{x=14y=6\hept{x−2y−2=0y=6⇒\hept{x=2y+2y=6⇒\hept{x=14y=6
Thu gọn
22 tháng 10 2021
\(A=-2\left(x^2+2xy+y^2\right)+4\left(x+y\right)-2-8y^2+2018\\ A=-2\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]-8y^2+2018\\ A=-2\left(x+y-1\right)^2-8y^2+2018\le2018\\ A_{max}=2018\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)
LP
2
4 tháng 10 2021
a) \(4x^2+12x+1=\left(4x^2+12x+9\right)-8=\left(2x+3\right)^2-8\ge-8\)
\(ĐTXR\Leftrightarrow x=-\dfrac{3}{2}\)
b) \(4x^2-3x+10=\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{151}{16}=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\)
\(ĐTXR\Leftrightarrow x=\dfrac{3}{8}\)
c) \(2x^2+5x+10=\left(2x^2+5x+\dfrac{25}{8}\right)+\dfrac{55}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\)
\(ĐTXR\Leftrightarrow x=-\dfrac{5}{4}\)
d) \(x-x^2+2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{9}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)
\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)
e) \(2x-2x^2=-2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{2}=-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\le\dfrac{1}{2}\)
\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)
f) \(4x^2+2y^2+4xy+4y+5=\left(4x^2+4xy+y^2\right)+\left(y^2+4y+4\right)+1=\left(2x+y\right)^2+\left(y+2\right)^2+1\ge1\)
\(ĐTXR\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
4 tháng 10 2021
a: Ta có: \(4x^2+12x+1\)
\(=4x^2+12x+9-8\)
\(=\left(2x+3\right)^2-8\ge-8\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)
b: Ta có: \(4x^2-3x+10\)
\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{5}{2}\right)\)
\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{151}{64}\right)\)
\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{8}\)
c: Ta có: \(2x^2+5x+10\)
\(=2\left(x^2+\dfrac{5}{2}x+5\right)\)
\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{55}{16}\right)\)
\(=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{5}{4}\)
LT
0
NT
1
M
16 tháng 8 2018
\(2x^2-4xy+8y^2+7x+6y-15.\)
= \(x^2+x^2-4xy+4y^2+4y^2+7x+6y-15\)
= \(\left(x^2-4xy+4y^2\right)+\left[x^2+7x+\left(\frac{7}{2}\right)^2\right]+\left[4y^2+6y+\left(\frac{3}{2}\right)^2\right]-\left(\frac{7}{2}\right)^2-\left(\frac{3}{2}\right)^2-15\)
= \(\left(x-2y\right)^2+\left(x+\frac{7}{2}\right)^2+\left(2y+\frac{3}{2}\right)^2-\frac{59}{2}\)
Vì \(\left(x-2y\right)^2+\left(x+\frac{7}{2}\right)^2+\left(2y+\frac{3}{2}\right)^2\ge0\forall x;y\)
=> \(\left(x-2y\right)^2+\left(x+\frac{7}{2}\right)^2+\left(2y+\frac{3}{2}\right)^2-\frac{59}{2}\ge0-\frac{59}{2}\forall x;y\)
=> \(\left(x-2y\right)^2+\left(x+\frac{7}{2}\right)^2+\left(2y+\frac{3}{2}\right)^2-\frac{59}{2}\ge-\frac{59}{2}\)
Vậy GTNN của bt là \(\frac{-59}{2}\Leftrightarrow\hept{\begin{cases}x-2y=0\\x+\frac{7}{2}=0\\2y+\frac{3}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2y\Rightarrow\orbr{\begin{cases}x=-\frac{7}{4}\\y=-\frac{3}{2}\end{cases}}\\x=-\frac{7}{2}\\y=-\frac{3}{4}\end{cases}}\)
19 tháng 12 2020
A= -x2+2x+3
=>A= -(x2-2x+3)
=>A= -(x2-2.x.1+1+3-1)
=>A=-[(x-1)2+2]
=>A= -(x+1)2-2
Vì -(x+1)2 ≤0=> A≤-2
Dấu "=" xảy ra khi
-(x+1)2=0 => x=-1
Vây A lớn nhất= -2 khi x= -1
19 tháng 12 2020
B=x2-2x+4y2-4y+8
=> B= (x2-2x+1)+(4y2-4y+1)+6
=> B=(x-1)2+(2y+1)2+6
=> B lớn nhất=6 khi x=1 và y=-1/2
29 tháng 6 2021
b) Ta có: \(B=x^2+2x+y^2-4y+6\)
\(=x^2+2x+1+y^2-4y+4+1\)
\(=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Vậy: \(B_{min}=1\) khi (x,y)=(-1;2)
c) Ta có: \(C=4x^2+4x+9y^2-6y-5\)
\(=4x^2+4x+1+9y^2-6y+1-7\)
\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(C_{min}=-7\) khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
29 tháng 6 2021
\(A=2x^2+x=2\left(x^2+\dfrac{1}{2}x\right)=2\left(x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}-\dfrac{1}{16}\right)\)
\(=2\left[\left(x+\dfrac{1}{4}\right)^2-\dfrac{1}{16}\right]\ge-\dfrac{1}{8}\) dấu"=' xảy ra<=>x=\(-\dfrac{1}{4}\)
\(B=x^2+2x+y^2-4y+6\)
\(=x^2+2x+1+y^2-4y+4+1=\left(x+1\right)^2+\left(y-2\right)^2+1\)
\(\ge1\) dấu"=" xảy ra<=>x=-1;y=2
\(C=4x^2+4x+9y^2-6y-5\)
\(=4x^2+4x+1+9y^2-6y+1-7\)
\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)
dấu"=" xảy ra<=>x=\(-\dfrac{1}{2},y=\dfrac{1}{3}\)
\(D=\left(2+x\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)^2\)
=\(x^2+6x+8-\left(x-1\right)\left(x+3\right)^2\)
\(=\left(x+3\right)^2-1-\left(x-1\right)\left(x+3\right)^2\)
\(=\left(x+3\right)^2\left(2-x\right)-1\ge-1\)
dấu"=" xảy ra\(< =>\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
\(C=2x^2-2x+4xy-6y+10y^2+2022\)
\(=\frac{1}{2}\left(4x^2+4y^2+1+8xy-4x-4y\right)+\frac{1}{2}\left(16y^2-8y+1\right)+2021\)
\(=\frac{1}{2}\left[\left(2x+2y-1\right)^2+\left(4y-1\right)^2\right]+2021\ge2021\)
Dấu \(=\)khi \(\hept{\begin{cases}2x+2y-1=0\\4y-1=0\end{cases}}\Leftrightarrow x=y=\frac{1}{4}\).