x+y=1

=>x=1-y

M=5x^2+y^2

=5(1-y)^2+y^2

\(=5y^2-10y+5+y^2\)

\(=6y^2-10y+5\)

\(=6\left(y^2-\dfrac{5}{3}y+\dfrac{5}{6}\right)\)

\(=6\left(y^2-2\cdot y\cdot\dfrac{5}{6}+\dfrac{25}{36}+\dfrac{5}{36}\right)\)

\(=6\left(y-\dfrac{5}{6}\right)^2+\dfrac{5}{6}>=\dfrac{5}{6}\)

Dấu = xảy ra khi y=5/6

=>\(M_{min}=\dfrac{5}{6}\) khi y=5/6 và x=1/6

Ta có 

x+y=1 => x=1-y

thay vào phương trình 

\(\Rightarrow M=5.\left(1-y\right)^2+y^2\)

\(\Rightarrow M=5.\left(1-2y+y^2\right)+y^2\)

\(\Rightarrow M=5-10y+5y^2+y^2\)

\(\Rightarrow M=6y^2-10y+5\)

\(\Rightarrow M=6\left(y^2-\frac{5}{3}y+\frac{5}{6}\right)\)

\(\Rightarrow M=6\left(y^2-2.\frac{5}{6}y+\frac{25}{36}-\frac{25}{36}+\frac{5}{6}\right)\)

\(\Rightarrow M=6\left[\left(y-\frac{5}{6}\right)^2+\frac{5}{36}\right]\)

\(\Rightarrow M=6\left(y-\frac{5}{6}\right)^2+\frac{5}{6}\ge\frac{5}{6}\)

Vậy \(M_{min}=\frac{5}{6}\Leftrightarrow\hept{\begin{cases}x+y=1\\y-\frac{5}{6}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1-y\\y=\frac{5}{6}\end{cases}}}\Leftrightarrow\hept{\begin{cases}x=1-\frac{5}{6}=\frac{1}{6}\\y=\frac{5}{6}\end{cases}}\)

T I C K chọn mình nha bạn cảm ơn chúc bạn học tốt

\(\)

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^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

Áp dụng BĐT Cauchy-Schwarz dạng Engel:

\(P=\frac{x^2}{\frac{1}{5}}+\frac{y^2}{1}\ge\frac{\left(x+y\right)^2}{\frac{1}{5}+1}=\frac{5}{6}\)

Dấu "=" xảy ra khi \(\frac{x}{\frac{1}{5}}=\frac{y}{1}\Leftrightarrow5x=y\Rightarrow x=\frac{1}{6}\Rightarrow y=\frac{5}{6}\)

Vậy ...

\(x+y=1\Rightarrow y=1-x\)

\(P=5x^2+\left(1-x\right)^2=6x^2-2x+1=6\left(x-\frac{1}{6}\right)^2+\frac{5}{6}\ge\frac{5}{6}\)

\(P_{min}=\frac{5}{6}\) khi \(\left\{{}\begin{matrix}x=\frac{1}{6}\\y=\frac{5}{6}\end{matrix}\right.\)

a: Thay m=2 vào hệ, ta được:

\(\begin{cases}x-2y=3-2=1\\ 2x+y=3\cdot\left(2+2\right)=3\cdot4=12\end{cases}\Rightarrow\begin{cases}2x-4y=2\\ 2x+y=12\end{cases}\)

=>\(\begin{cases}2x-4y-2x-y=12-2\\ 2x+y=12\end{cases}\Rightarrow\begin{cases}-5y=10\\ 2x=12-y\end{cases}\)

=>\(\begin{cases}y=-2\\ 2x=12-\left(-2\right)=12+2=14\end{cases}\Rightarrow\begin{cases}y=-2\\ x=7\end{cases}\)

b: Để hệ có nghiệm duy nhất thì 1/2<>-2/1(luôn đúng)

=>Hệ luôn có nghiệm duy nhất

c: \(\begin{cases}x-2y=3-m\\ 2x+y=3\left(m+2\right)\end{cases}\Rightarrow\begin{cases}2x-4y=6-2m\\ 2x+y=3m+6\end{cases}\)

=>\(\begin{cases}2x-4y-2x-y=6-2m-3m-6\\ x-2y=3-m\end{cases}\Rightarrow\begin{cases}-5y=-5m\\ x=2y+3-m\end{cases}\)

=>\(\begin{cases}y=m\\ x=2m+3-m=m+3\end{cases}\)

\(A=x^2+y^2\)

\(=\left(m+3\right)^2+m^2=2m^2+6m+9\)

\(=2\left(m^2+3m+\frac92\right)=2\left(m^2+3m+\frac94+\frac94\right)=2\left(m+\frac32\right)^2+\frac92\ge\frac92\forall m\)

Dấu '=' xảy ra khi m+3/2=0

=>m=-3/2

d: 5x-y=3

=>5(m+3)-m=3

=>5m+15-m=3

=>4m=3-15=-12

=>m=-3

Câu 2:

\(A-4=2x+3y\Rightarrow\left(A-4\right)^2=\left(2x+3y\right)^2\)

\(\left(A-4\right)^2\le\left(2^2+3^2\right)\left(x^2+y^2\right)=676\)

\(\Rightarrow-26\le A-4\le26\)

\(\Rightarrow-22\le A\le30\)

\(A_{max}=30\) khi \(\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)

\(A_{min}=-22\) khi \(\left\{{}\begin{matrix}x=-4\\y=-6\end{matrix}\right.\)

\(2x+3y=1\Rightarrow y=\frac{1-2x}{3}\)

Do \(x;y\ge0\Rightarrow0\le x\le\frac{1}{2}\)

\(A=x^2+3\left(\frac{1-2x}{3}\right)^2=x^2+\frac{1}{3}\left(4x^2-4x+1\right)=\frac{7}{3}x^2-\frac{4}{3}x+\frac{1}{3}\)

\(A=\frac{7}{3}\left(x-\frac{2}{7}\right)^2+\frac{1}{7}\ge\frac{1}{7}\)

\(\Rightarrow A_{min}=\frac{1}{7}\) khi \(x=\frac{2}{7};y=\frac{1}{7}\)

Mặt khác \(A=\frac{1}{3}x\left(7x-4\right)+\frac{1}{3}\)

Do \(x\le\frac{1}{2}\Rightarrow7x-4< 0\Rightarrow x\left(7x-4\right)\le0\)

\(\Rightarrow A\le\frac{1}{3}\Rightarrow A_{max}=\frac{1}{3}\) khi \(x=0;y=\frac{1}{3}\)