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coi như giải hệ pt
\(\hept{\begin{cases}y=x+1\left(1\right)\\y^2-3y\sqrt{x}+2x=0\left(2\right)\end{cases}}\)
\(\left(2\right)\Leftrightarrow\left(y^2-3\sqrt{x}.y+\frac{9x}{4}\right)=\frac{9x}{4}-2x=\frac{x}{2}\\ \)
\(\left(y-\frac{3\sqrt{x}}{2}\right)^2=\left(\frac{\sqrt{x}}{2}\right)^2\Rightarrow\orbr{\begin{cases}y=\frac{3\sqrt{x}}{2}-\frac{\sqrt{x}}{2}=\sqrt{x}\\y=\frac{3\sqrt{x}}{2}+\frac{\sqrt{x}}{2}=2\sqrt{x}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=x+1\left(3\right)\\2\sqrt{x}=x+1\left(4\right)\end{cases}}\)
\(\left(3\right)\Leftrightarrow\orbr{\begin{cases}\left(\sqrt{x}-\frac{1}{2}\right)^2=\frac{1}{4}-1\left(vonghiem\right)\\\left(\sqrt{x}-1\right)^2=0\Rightarrow\sqrt{x}=1\Rightarrow x=1\end{cases}}\)
Vậy chỉ có điểm x=1; y=2 thỏa mãn
a, \(P=\left(x^4-8x^3+16x^2\right)+12x^2-48x+35\)
\(=\left(x^2-4x\right)^2+12\left(x^2-4x\right)+36-1\)
\(=\left(x^2-4x+6\right)^2-1\)
\(=\left[\left(x-2\right)^2+2\right]^2-1\)
\(\ge2^2-1=3\)
Cách khác \(P=\left(x-2\right)^2\left[\left(x-2\right)^2+4\right]+3\ge3\)
Đẳng thức xảy ra khi \(x=2.\)
b, \(xy\le\frac{\left(x+y\right)^2}{4}=9\)
Áp dụng bđt Co6si: \(\frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2}.\frac{1}{y^2}}=\frac{2}{xy}\)
\(Q\ge\frac{102}{xy}+xy=xy+\frac{81}{xy}+\frac{21}{xy}\ge2\sqrt{xy.\frac{81}{xy}}+\frac{21}{9}=\frac{61}{3}.\)
Dấu bằng xảy ra khi \(x=y=3.\)
\(A=\dfrac{1-\sqrt{x}+x}{\sqrt{x}}\\ =\dfrac{\left(1-\sqrt{x}+x\right)\sqrt{x}}{x}\\ =\dfrac{\sqrt{x}-x+x\sqrt{x}}{x}\)
lm tiếp...
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=\(\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1+\left(4x^2-4x+1\right)+2014\)
=\(\left(\left(x+y\right)^2+2\left(x+y\right)+1\right)+\left(2x-1\right)^2+2014\)
=\(\left(x+y+1\right)^2+\left(2x-1\right)^2+2014\ge2014\)
\(\Rightarrow M\ge2014\Leftrightarrow minM=2014\)
\(\Leftrightarrow\hept{\begin{cases}x+y+1=0\\2x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0,5\\y=1,5\end{cases}}\)
Luôn có \(\left(x-y\right)^2+\left(x-z\right)^2+\left(y-x\right)^2\ge0\Leftrightarrow2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\ge0\)
\(\Leftrightarrow\left(x^2+y^2+z^2\right)\ge xy+yz+xz\ge-1\)
\(P_{min}=-1\)dấu "=" sảy ra khi (x,y,z) là hoán vị của 3 phần tử (0,0,-1)
Ta có:
\(xy+yz+zx=-1\)
\(\Leftrightarrow2\left(xy+yz+zx\right)=-2\)
\(\Leftrightarrow2\left(xy+yz+zx\right)+x^2+y^2+z^2=-2+x^2+y^2+z^2\)
\(\Leftrightarrow P=x^2+y^2+z^2=\left(x+y+z\right)^2+2\ge2\)
Dấu = xảy ra khi \(\hept{\begin{cases}x+y+z=0\\xy+yz+zx=-1\end{cases}}\)
Chỉ ra 1 bộ số thỏa mãn cái đấy nhé là: \(\hept{\begin{cases}x=0\\y=1\\z=-1\end{cases}}\)
\(B=\dfrac{xy}{xy}+\dfrac{\left(x-y\right)x}{x\left(x-y\right)}-\dfrac{y\left(x-y\right)}{y\left(x-y\right)}=1\)
\(A=\dfrac{x^2+2x+17}{2\left(x+1\right)}=\dfrac{x^2-6x+9+8x+8}{2\left(x+1\right)}\\ =\dfrac{\left(x-3\right)^2+8\left(x+1\right)}{2\left(x+1\right)}=\dfrac{\left(x-3\right)^2}{2\left(x+1\right)}+4\)
Vì \(\left(x-3\right)^2\ge0;2\left(x+1\right)>0\) (do \(x>-1\))
\(\Rightarrow\dfrac{\left(x-3\right)^2}{2\left(x+1\right)}\ge0\Leftrightarrow A=\dfrac{\left(x-3\right)^2}{2\left(x+1\right)}+4\ge4\)
Dấu "=" xảy ra khi \(x=3\)
Vậy....
a) \(A=x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=3\). \(min_A=1\)
b) \(B=3x^2+x-2=3\left(x^2+\dfrac{1}{3}x-\dfrac{2}{3}\right)=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}-\dfrac{25}{36}\right)=3\left(x+\dfrac{1}{6}\right)^2-\dfrac{25}{12}\ge\dfrac{-25}{12}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{1}{6}\). \(min_B=\dfrac{-25}{12}\)
c) \(C=\dfrac{4}{x^2}-\dfrac{3}{x}-1=\left(\dfrac{4}{x^2}-\dfrac{3}{x}+\dfrac{9}{16}\right)-\dfrac{25}{16}=\left(\dfrac{2}{x}+\dfrac{2}{3}\right)^2-\dfrac{25}{16}\ge\dfrac{-25}{16}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-3\). \(min_C=\dfrac{-25}{16}\)
d) \(D=x^2+y^2-x+3y+7=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+3y+\dfrac{9}{4}\right)+\dfrac{9}{2}=\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{3}{2}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-3}{2}\end{matrix}\right.\). \(min_D=\dfrac{9}{2}\)
Áp dụng BĐT Cauchy ta có:
P = \(x^2+y^2+\dfrac{33}{xy}\) \(\ge\) \(\dfrac{\left(x+y\right)^2}{2}+\dfrac{33}{x+y}\) = \(\dfrac{4^2}{2}+\dfrac{33}{4}=\dfrac{65}{4}\)
=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}x+y=4\\x=y\end{matrix}\right.\) <=> \(x=y=2\)
Vậy ...............................