tìm GTNN của M=\(5x^2+y^2+z^2-4x-2xy-z-1\)
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\(M=\left(x^2-2xy+y^2\right)+\left(4x^2-4x+1\right)+\left(z^2-z+\frac{1}{4}\right)-\frac{5}{4}\)
\(M=\left(x-y\right)^2+\left(2x-1\right)+\left(z-\frac{1}{2}\right)^2-\frac{5}{4}>=-\frac{5}{4}\)
=>M min\(=-\frac{5}{4}\)
<=>x=y=z=1/2
Ta có:A = 5x2 + y2 + z2 - 4x - 2xy - z - 1
A = (x2 - 2xy + y2) + (4x2 - 4x + 1) + (z2 - z + 1/4) - 9/4
A = (x - y)2 + (2x - 1)2 + (z - 1/2)2 - 9/4 \(\ge\)- 9/4 \(\forall\)x;y
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y=0\\2x-1=0\\z-\frac{1}{2}=0\end{cases}}\) <=> \(\hept{\begin{cases}x=y\\x=\frac{1}{2}\\z=\frac{1}{2}\end{cases}}\) <=> x = y = z = 1/2
Vậy MinA = -9/4 khi x = y = z = 1/2
=)) mình cũng làm ntn mà rút gọn ngu -9/4=-3/2 kq sai :v
Ta có \(C=5x^2+y^2+z^2-4x-2xy-z-1\)
\(=x^2-2xy+y^2+4x^2-4x+1+z^2-z+\dfrac{1}{4}-1-\dfrac{1}{4}-1\)
\(=\left(x-y\right)^2+\left(2x-1\right)^2+\left(z-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\)
Ta có \(\left(x-y\right)^2\ge0;\left(2x-1\right)^2\ge0;\left(z-\dfrac{1}{2}\right)^2\ge0\)
=> \(C\ge-\dfrac{9}{4}\)
=> C đạt giá trị nhỏ nhất là \(-\dfrac{9}{4}\) khi
\(\left\{{}\begin{matrix}x-y=0\\2x-1=0\\z-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=y=\dfrac{1}{2}\\x=\dfrac{1}{2}\\z=\dfrac{1}{2}\end{matrix}\right.\)
=> \(x=y=z=\dfrac{1}{2}\)
Vậy MinC = \(-\dfrac{9}{4}\)khi x=y=z = \(\dfrac{1}{2}\)
câu 1
x^2 -5x +y^2+xy -4y +2014
=(y^2+xy +1/4x^2) -4(y+1/2x)+4 +3/4x^2-3x+2010
=(y+1/2x-2)^2 +3/4(x^2-4x+4)+2007
=(y+1/2x-2)^2 +3/4(x-2)^2 +2007
GTNN là 2007<=> x=2 và y=1
ta có
\(x^2+y^2+z^2\)\(=200\)
\(2xy-yz-zx=M\)
\(\Leftrightarrow M+200=x^2+y^2+z^2+2xy-yz-zx\)
\(\Leftrightarrow M+200=\left(x+y\right)^2-z\left(x+y\right)+z^2\)
\(\Leftrightarrow\left(x+y-\frac{z}{2}\right)^2+\frac{3}{4}z^2\ge0\)
\(\Leftrightarrow M\ge-200\)
Điều kiện có 2 nghiệm phân biệt tự làm nha
Theo vi-et ta có:
\(\hept{\begin{cases}x_1+x_2=5\\x_1.x_2=m-2\end{cases}}\)
\(2\left(\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}\right)=3\)
\(\Leftrightarrow4\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{2}{\sqrt{x_1.x_2}}\right)=9\)
\(\Leftrightarrow4\left(\frac{5}{m-2}+\frac{2}{\sqrt{m-2}}\right)=9\)
Làm nốt nhé
Câu 1:
M=\(\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1+\left(4x^2-4x+1\right)+2014\)
=\(\left(\left(x+y\right)^2+2\left(x+y\right)+1\right)+\left(2x-1\right)^2+2014\)
=\(\left(x+y+1\right)^2+\left(2x-1\right)^2+2014\ge2014\)
\(\Rightarrow M\ge2014\Leftrightarrow minM=2014\)
\(\Leftrightarrow\hept{\begin{cases}x+y+1=0\\2x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0,5\\y=1,5\end{cases}}\)
Câu 1:
\(x\left(x-2\right)\left(x+2\right)-\left(x+2\right)\left(x^2-2x+4\right)=4\)
\(\Leftrightarrow x\left(x^2-4\right)-\left(x^3+8\right)=4\)
\(\Leftrightarrow x^3-4x-x^3-8=4\)
\(\Leftrightarrow-4x-8=4\)
\(\Leftrightarrow-4x=12\)
\(\Leftrightarrow x=-3\)
Vậy \(x=-3\)
a) x2+y2-4x+4y+8=0
⇔ (x-2)2+(y+2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)
b)5x2-4xy+y2=0
⇔ x2+(2x-y)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
c)x2+2y2+z2-2xy-2y-4z+5=0
⇔ (x-y)2+(y-1)2+(z-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)
b: Ta có: \(5x^2-4xy+y^2=0\)
\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)
\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
\(M=5x^2+y^2+z^2-4x-2xy-z-1\)
\(=\left(4x^2-4x+1\right)+\left(x^2-2xy+y^2\right)+\left(z^2-z+\dfrac{1}{4}\right)-\dfrac{9}{4}\)
\(=\left(2x-1\right)^2+\left(x-y\right)^2+\left(z-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\)
Vậy \(M_{min}=-\dfrac{9}{4}\) khi \(x=\dfrac{1}{2}\) ; \(y=\dfrac{1}{2}\)