Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
câu 1) ta có : \(M=\left(x^2-x\right)^2+\left(2x-1\right)^2=x^4-2x^3+x^2+4x^2-4x+1\)
\(=\left(x^2-x+2\right)^2-3=\left(\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right)^2-3\)
\(\Rightarrow\dfrac{1}{16}\le M\le61\)
\(\Rightarrow M_{min}=\dfrac{1}{16}\)khi \(x=\dfrac{1}{2}\) ; \(M_{max}=61\) khi \(x=3\)
câu 2) điều kiện xác định : \(0\le x\le2\)
đặt \(\sqrt{2x-x^2}=t\left(t\ge0\right)\)
\(\Rightarrow M=-t^2+4t+3=-\left(t-2\right)^2+7\)
\(\Rightarrow3\le M\le7\)
\(\Rightarrow M_{min}=3\)khi \(x=0\) ; \(M_{max}=7\) khi \(x=2\)câu 3) ta có : \(M=\left(x-2\right)^2+6\left|x-2\right|-6\ge-6\)
\(\Rightarrow M_{min}=-6\) khi \(x=2\)
4) điều kiện xác định \(-6\le x\le10\)
ta có : \(M=5\sqrt{x+6}+2\sqrt{10-x}-2\)
áp dụng bunhiacopxki dạng căn ta có :
\(-\sqrt{\left(5^2+2^2\right)\left(x+6+10-x\right)}\le5\sqrt{x+6}+2\sqrt{10-x}\le\sqrt{\left(5^2+2^2\right)\left(x+6+10-x\right)}\)
\(\Leftrightarrow-4\sqrt{29}\le5\sqrt{x+6}+2\sqrt{10-x}\le4\sqrt{29}\)
\(\Rightarrow-2-4\sqrt{29}\le B\le-2+4\sqrt{29}\)
\(\Rightarrow M_{max}=-2+4\sqrt{29}\) khi \(\dfrac{\sqrt{x+6}}{5}=\dfrac{\sqrt{10-x}}{2}\Leftrightarrow x=\dfrac{226}{29}\)
\(\Rightarrow M_{min}=-2-4\sqrt{29}\) dấu của bđt này o xảy ra câu 5 lm tương tự
1.
\(f\left(x\right)=\frac{x-7}{\left(x-4\right)\left(4x-3\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{\frac{3}{4};4\right\}\)
\(f\left(x\right)=0\Rightarrow x=7\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< x< 4\\x>7\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3}{4}\\4< x< 7\end{matrix}\right.\)
2.
\(f\left(x\right)=\frac{11x+3}{-\left(x-\frac{5}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=-\frac{3}{11}\)
\(f\left(x\right)>0\Rightarrow x< -\frac{3}{11}\)
\(f\left(x\right)< 0\Rightarrow x>-\frac{3}{11}\)
3.
\(f\left(x\right)=\frac{3x-2}{\left(x-1\right)\left(x^2-2x-2\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{3}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\frac{2}{3}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< 1-\sqrt{3}\\\frac{2}{3}< x< 1\\x>1+\sqrt{3}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}1-\sqrt{3}< x< \frac{2}{3}\\1< x< 1+\sqrt{3}\end{matrix}\right.\)
4.
\(f\left(x\right)=\frac{\left(x-2\right)\left(x+6\right)}{\sqrt{6}\left(x+\frac{\sqrt{6}}{4}\right)^2+\frac{8\sqrt{2}-3\sqrt{6}}{8}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)
\(A=\left(2x+\dfrac{1}{3}\right)^4-1\ge-1\forall x\in R\)
Dấu "=" xảy ra khi\(2x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{6}\)
\(B=-2\left(x-3\right)^2-\dfrac{7}{11}\left|3y+7\right|-2011\ge-2011\forall x,y\in R\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-3=0\\3y+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-\dfrac{7}{3}\end{matrix}\right.\)
\(C=\left|2x+1\right|+\left|3-2x\right|\ge\left|2x+1+3-2x\right|=4\)
Dấu "=" xảy ra khi \(\left[{}\begin{matrix}2x+1=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(\left(x^3+y^3\right)\left(x+y\right)=xy\left(1-x\right)\left(1-y\right)\Leftrightarrow\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)=\left(1-x\right)\left(1-y\right)\left(1\right)\)
Ta có : \(\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)\ge4xy\)
và \(\left(1-x\right)\left(1-y\right)=1-\left(x+y\right)+xy\le1-2\sqrt{xy}+xy\)
\(\Rightarrow1-2\sqrt{xy}+xy\ge4xy\Leftrightarrow0\) <\(xy\le\frac{1}{9}\)
Dễ chứng minh : \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\le\frac{1}{1+xy};\left(x,y\in\left(0;1\right)\right)\)
\(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\le\sqrt{2\left(\frac{1}{1+x^2}+\frac{1}{1+y^2}\right)}\le\sqrt{2\left(\frac{2}{1+xy}\right)}=\frac{2}{\sqrt{1+xy}}\)
\(3xy-\left(x^2+y^2\right)=xy-\left(x-y\right)^2\le xy\)
\(\Rightarrow P\le\frac{2}{\sqrt{1+xy}}+xy=\frac{2}{\sqrt{1+t}}+t\), \(\left(t=xy\right)\), (0<\(t\le\frac{1}{9}\)
Xét hàm số :
\(f\left(t\right)=\frac{2}{\sqrt{t+1}}+t\) , (0<\(t\le\frac{1}{9}\)