Tìm GTNN của biểu thức A=x2+y2+z2-yz-4x-3y+2027
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PK
1
22 tháng 8 2021
\(A=x^2+y^2+z^2-yz-4x-3y+2027\)
\(\Rightarrow4A=4x^2+4y^2+4z^2-4yz-16x-12y+8108=4x^2-16x+16+3y^2+12y+12+y^2-4yz+4z^2+8080=4\left(x-2\right)^2+3\left(y+2\right)^2+\left(y-2z\right)^2+8080\)
Vì \(4\left(x-2\right)^2\ge0\)
\(3\left(y+2\right)^2\ge0\)
\(\left(y-2z\right)^2\ge0\)
\(\Rightarrow4A\ge8080\Rightarrow A\ge2020\)
\(ĐTXR\Leftrightarrow x=2,y=-2,z=-1\)
NN
0
20 tháng 5 2021
\(\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)
\(\Rightarrow2\ge3x^2+2y^2+2z^2+y^2+z^2\)
\(\Leftrightarrow2\ge3\left(x^2+y^2+z^2\right)\)
Có: \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\le2\)
\(\Rightarrow\)\(A^2\le2\) \(\Leftrightarrow A\in\left[-\sqrt{2};\sqrt{2}\right]\)
minA=-1\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y+z=-\sqrt{2}\\x=y=z\end{matrix}\right.\) \(\Rightarrow x=y=z=-\dfrac{\sqrt{2}}{3}\)
maxA=1\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=\sqrt{2}\\x=y=z\end{matrix}\right.\) \(\Rightarrow x=y=z=\dfrac{\sqrt{2}}{3}\)
HQ
Hà Quang Minh
Giáo viên
9 tháng 8 2023
\(a,M=x^2-4x+5=\left(x-2\right)^2+5\\ \Rightarrow M\ge5\)
Dấu "=" xảy ra \(\Leftrightarrow x=2\)
\(b,N=y^2-y-3=\left(y-\dfrac{1}{2}\right)^2-\dfrac{13}{4}\\ \Rightarrow N\ge-\dfrac{13}{4} \)
Dấu "=" xảy ra \(\Leftrightarrow y=\dfrac{1}{2}\)
\(P=x^2+y^2-4x+y+7=\left(x-2\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\\ \Rightarrow P\ge\dfrac{11}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{2}\end{matrix}\right.\)
9 tháng 8 2023
a: M=x^2-4x+4+1
=(x-2)^2+1>=1
Dấu = xảy ra khi x=2
b: N=y^2-y+1/4-13/4
=(y-1/2)^2-13/4>=-13/4
Dấu = xảy ra khi y=1/2
c: P=x^2-4x+4+y^2+y+1/4+11/4
=(x-2)^2+(y+1/2)^2+11/4>=11/4
Dấu = xảy ra khi x=2 và y=-1/2
HA
0
TH
2
T
16 tháng 10 2023
\(B=y^2-y+1\)
\(=y^2-2\cdot y\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}+1\)
\(=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta thấy: \(\left(y-\dfrac{1}{2}\right)^2\ge0\forall y\)
\(\Rightarrow\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall y\)
Dấu \("="\) xảy ra \(\Leftrightarrow y-\dfrac{1}{2}=0\Leftrightarrow y=\dfrac{1}{2}\)
Vậy \(B_{min}=\dfrac{3}{4}\) khi \(y=\dfrac{1}{2}\).
\(---\)
\(C=x^2-4x+y^2-y+5\)
\(=\left(x^2-4x+4\right)+\left(y^2-y+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=\left(x^2-2\cdot x\cdot2+2^2\right)+\left[y^2-2\cdot y\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{3}{4}\)
\(=\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta thấy: \(\left(x-2\right)^2\ge0\forall x\)
\(\left(y-\dfrac{1}{2}\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2\ge0\forall x;y\)
\(\Rightarrow\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x;y\)
Dấu \("="\) xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y-\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(C_{min}=\dfrac{3}{4}\) khi \(x=2;y=\dfrac{1}{2}\).
\(Toru\)
MH
16 tháng 10 2023
\(B=y^2-y+1\)
\(=y^2-2.y.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Vì \(\left(y-\dfrac{1}{2}\right)^2\ge0\forall y\Rightarrow B\ge\dfrac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow y=\dfrac{1}{2}\)
\(C=x^2-4x+y^2-y+5\)
\(=x^2-4x+4+y^2-y+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2\)
Vì \(\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2\ge0\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)
5 tháng 3 2022
a, xem lại đề
\(b,x^2-4x+y^2-6y+1\\ =\left(x^2-4x+4\right)+\left(y^2-6y+9\right)-12\\ =\left(x-2\right)^2+\left(y-3\right)^2-12\ge-12\)
Dấu "=" xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy ...
\(c,x^2-4xy+5y^2-2y+5\\ =\left(x^2-4xy+4y^2\right)+\left(y^2-2y+1\right)+4\\ =\left(x-2y\right)^2+\left(y-1\right)^2+4\ge4\)
Dấu "=" xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy ...
5 tháng 3 2022
a,
b,x2−4x+y2−6y+1=(x2−4x+4)+(y2−6y+9)−12=(x−2)2+(y−3)2−12≥−12b,x2−4x+y2−6y+1=(x2−4x+4)+(y2−6y+9)−12=(x−2)2+(y−3)2−12≥−12
Dấu "=" xảy ra⇔{x=2y=3⇔{x=2y=3
Vậy ...
c,x2−4xy+5y2−2y+5=(x2−4xy+4y2)+(y2−2y+1)+4=(x−2y)2+(y−1)2+4≥4c,x2−4xy+5y2−2y+5=(x2−4xy+4y2)+(y2−2y+1)+4=(x−2y)2+(y−1)2+4≥4
Dấu "=" xảy ra⇔{x=2y=1⇔{x=2y=1
Vậy ...
t hong bíc nè
\(A=x^2+y^2+z^2-yz-4x-3y+2027\)
\(\Rightarrow4A=4x^2+4y^2+4z^2-4yz-16x-12y+8108\)
\(=\left(4x^2-16x+16\right)+\left(3y^2-12y+12\right)+\left(y^2-4yz+4z^2\right)+8080\)
\(=4.\left(x^2-4x+4\right)+3.\left(y^2-4y+4\right)+\left(y-2z\right)^2+8080\)
\(=4.\left(x-2\right)^2+3.\left(y-2\right)^2+\left(y-2z\right)^2+8080\)
Mà: \(\hept{\begin{cases}4.\left(x-2\right)^2\ge0\\3.\left(y-2\right)^2\ge0\\\left(y-2z\right)^2\ge0\end{cases}}\)
\(\Rightarrow4.\left(x-2\right)^2+3.\left(y-2\right)^2+\left(y-2z\right)^2\ge0\)
\(\Rightarrow4.\left(x-2\right)^2+3.\left(y-2\right)^2+\left(y-2z\right)^2+8080\ge8080\)
\(\Rightarrow A\ge8080\)
Dấu '' = '' xảy ra khi:
\(\hept{\begin{cases}4.\left(x-2\right)^2=0\\3.\left(y-2\right)^2=0\\\left(y-2z\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y=2\\z=1\end{cases}}\)
Vậy giá trị nhỏ nhất của \(A=2020\) khi \(\hept{\begin{cases}x=y=2\\z=1\end{cases}}\)