Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có \(\left(x-y\right)^2\ge0\forall x,y\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\Leftrightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}..\)
Theo giả thiết \(x^2+y^2=\left(x+y\right)\left(\sqrt{x}+\sqrt{y}-1\right)\)
\(\Rightarrow\left(x+y\right)\left(\sqrt{x}+\sqrt{y}-1\right)\ge\frac{\left(x+y\right)^2}{2}\)
Mà x,y>1/4\(\Rightarrow\sqrt{x}+\sqrt{y}-1\ge\frac{x+y}{2}\)
\(\Leftrightarrow x+y\le2\sqrt{x}+2\sqrt{y}-2\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-2\sqrt{y}+1\right)\le0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2\le0\)
Mà \(\hept{\begin{cases}\left(\sqrt{x}-1\right)^2\ge0\\\left(\sqrt{y}-1\right)^2\ge0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\left(\sqrt{x}-1\right)^2=0\\\left(\sqrt{y}-1\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}\sqrt{x}=1\\\sqrt{y}=1\end{cases}\Leftrightarrow}x=y=1\left(TMĐK\right).\)
7. \(S=9y^2-12\left(x+4\right)y+\left(5x^2+24x+2016\right)\)
\(=9y^2-12\left(x+4\right)y+4\left(x+4\right)^2+\left(x^2+8x+16\right)+1936\)
\(=\left[3y-2\left(x+4\right)\right]^2+\left(x-4\right)^2+1936\ge1936\)
Vậy \(S_{min}=1936\) \(\Leftrightarrow\) \(\hept{\begin{cases}3y-2\left(x+4\right)=0\\x-4=0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x=4\\y=\frac{16}{3}\end{cases}}\)
8. \(x^2-5x+14-4\sqrt{x+1}=0\) (ĐK: x > = -1).
\(\Leftrightarrow\) \(\left(x+1\right)-4\sqrt{x+1}+4+\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow\) \(\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2=0\)
Với mọi x thực ta luôn có: \(\left(\sqrt{x+1}-2\right)^2\ge0\) và \(\left(x-3\right)^2\ge0\)
Suy ra \(\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2\ge0\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}\left(\sqrt{x+1}-2\right)^2=0\\\left(x-3\right)^2=0\end{cases}}\) \(\Leftrightarrow\) x = 3 (Nhận)
7. \(S=9y^2-12\left(x+4\right)y+\left(5x^2+24x+2016\right)\)
\(=9y^2-12\left(x+4\right)y+4\left(x+4\right)^2+\left(x^2+8x+16\right)+1936\)
\(=\left[3y-2\left(x+4\right)\right]^2+\left(x-4\right)^2+1936\ge1936\)
Vậy \(S_{min}=1936\) \(\Leftrightarrow\) \(\hept{\begin{cases}3y-2\left(x+4\right)=0\\x-4=0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x=4\\y=\frac{16}{3}\end{cases}}\)
ĐKXĐ: \(x\ge1;y\ge1\)
Ta có: \(\frac{x^2-4}{x}+\frac{y^2-4}{y}+8=4\left(\sqrt{x-1}+\sqrt{y-1}\right)\)
\(\Leftrightarrow\frac{x^2-4}{x}+\frac{y^2-4}{y}=4\left[\left(\sqrt{x-1}-1\right)+\left(\sqrt{y-1}+1\right)\right]\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(x+2\right)}{x}+\frac{\left(y-2\right)\left(y+2\right)}{y}=4\left(\frac{x-1-1}{\sqrt{x-1}+1}+\frac{y-1-1}{\sqrt{y-1}+1}\right)\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{x+2}{x}-\frac{4}{\sqrt{x-1}+1}\right)+\left(y-2\right)\left(\frac{y+2}{y}-\frac{4}{\sqrt{y-1}+1}\right)=0\)
\(\Leftrightarrow\left(x-2\right)\frac{x\sqrt{x-1}+2\sqrt{x-1}+2+x-4x}{x\left(\sqrt{x-1}+1\right)}+\left(y-2\right)\frac{y\sqrt{y-1}+2\sqrt{y-1}+y-4y}{y\left(\sqrt{y-1}+1\right)}=0\)
\(\Leftrightarrow\left(x-2\right)\frac{\left( x-1\right)\sqrt{x-1}+3\sqrt{x-1}-3\left(x-1\right)-1}{x\left(\sqrt{x-1}+1\right)}\)
\(+\left(y-2\right)\frac{\left(y-1\right)\sqrt{y-1}+3\sqrt{y-1}-3\left(y-1\right)-1}{y\left(\sqrt{y-1}+1\right)}=0\)
\(\Leftrightarrow\left(x-2\right)\frac{\left(\sqrt{x-1}-1\right)^3}{x\left(\sqrt{x-1}+1\right)}+\left(y-2\right)\frac{\left(\sqrt{y-1}-1\right)^3}{y\left(\sqrt{y-1}+1\right)}=0\)
\(\Leftrightarrow\left(x-2\right)\frac{\left(\sqrt{x-1}-1\right)^3\left(\sqrt{x-1}+1\right)^3}{x\left(\sqrt{x-1}+1\right)^4}+\left(y-2\right)\frac{\left(\sqrt{y-1}-1\right)^3\left(\sqrt{y-1}+1\right)^3}{y\left(\sqrt{y-1}+1\right)^4}=0\)
\(\Leftrightarrow\frac{\left(x-2\right)^4}{x\left(\sqrt{x-1}+1\right)^4}+\frac{\left(y-2\right)^4}{y\left(\sqrt{y-1}+1\right)^4}=0\)
Vì \(x\ge1;y\ge1\Rightarrow\frac{\left(x-2\right)^4}{x\left(\sqrt{x-1}+1\right)^4}\ge0;\frac{\left(y-2\right)^4}{y\left(\sqrt{y-1}+1\right)^4}\ge0\)\(\Rightarrow\frac{\left(x-2\right)^4}{x\left(\sqrt{x-1}+1\right)^4}+\frac{\left(y-2\right)^4}{y\left(\sqrt{y-1}+1\right)^4}\ge0\)
Do đó dấu ''='' xảy ra khi \(\frac{\left(x-2\right)^4}{x\left(\sqrt{x-1}+1\right)^4}=\frac{\left(y-2\right)^4}{y\left(\sqrt{y-1}+1\right)^4}=0\Leftrightarrow x-2=y-2=0\Leftrightarrow x=y=2\)
Vậy \(x=y=2\).
Bài 2 :
Tìm min : Bình phương
Tìm max : Dùng B.C.S ( bunhiacopxki )
Bài 3 : Dùng B.C.S
KP9
nói thế thì đừng làm cho nhanh bạn ạ
Người ta cũng có chút tôn trọng lẫn nhau nhé đừng có vì dăm ba cái tích
ĐKXĐ: \(x\ge1\)
Ta có: \(\frac{x^2-4}{x}+4+\frac{y^2-4}{y}+4=4\left(\sqrt{x-1}+\sqrt{y-1}\right)\)
Lại có: \(\frac{x^2-4}{x}+4=x+\frac{4x-4}{x}\ge4\sqrt{x-1}\)
Tương tự: \(\frac{y^2-4}{y}+4\ge4\sqrt{y-1}\)
Cộng từng vế: \(\frac{x^2-4}{x}+\frac{y^2-4}{y}+8\ge4\left(\sqrt{x-1}+\sqrt{y-1}\right)\)
Dấu "=" xảy ra khi: x=y=2
Vậy (x;y)=(2'2)