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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(P=x^2-3x+\frac{1}{2x}+2=\left(x-2\right)^2+\left(\frac{x}{8}+\frac{1}{2x}\right)+\frac{7x}{8}-2\ge\frac{1}{4}\)
Đẳng thức xảy ra khi x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
Dự đoán dấu "=" khi x = 2 ; y= 1
Áp dụng bđt Cô-si cho 3 số và bđt \(\frac{a^2}{m}+\frac{b^2}{n}\ge\frac{\left(a+b\right)^2}{m+n}\) ta được
\(P=2x^2+y^2+\frac{28}{x}+\frac{1}{y}\)
\(=\left(\frac{7x^2}{4}+\frac{14}{x}+\frac{14}{x}\right)+\left(\frac{y^2}{2}+\frac{1}{2y}+\frac{1}{2y}\right)+\left(\frac{x^2}{4}+\frac{y^2}{2}\right)\)
\(\ge3\sqrt[3]{\frac{7x^2.14.14}{4.x^2}}+3\sqrt[3]{\frac{y^2.1.1}{2.2y.2y}}+\frac{\left(x+y\right)^2}{4+2}\)
\(=3.\sqrt[3]{\frac{7.14.14}{4}}+\frac{3}{\sqrt[3]{2^3}}+\frac{3^2}{6}=24\)
Dấu "=" khi x = 2 ; y = 1
Bài toán easy!
\(P=\left(2x^2+8\right)+\left(y^2+1\right)+\frac{28}{x}+\frac{1}{y}-9\)
Áp dụng BĐT AM-GM,ta có:
\(P\ge8x+2y+\frac{28}{x}+\frac{1}{y}-9\)
\(=\left(7x+\frac{28}{x}\right)+\left(y+\frac{1}{y}\right)+\left(x+y\right)-9\)
\(\ge2\sqrt{7x.\frac{28}{x}}+2\sqrt{y.\frac{1}{y}}+\left(x+y\right)-9\)
\(\ge28+2+3-9=24\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}2x^2=8\\y^2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)
Vậy \(P_{min}=24\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b1:
ĐKXĐ: \(x\ne0;x\ne\pm2\)
Ta có : \(A=\left(\frac{4x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{8x^2}{x^2-4}\right)\left(\frac{x-1}{x\left(x-2\right)}-\frac{2\left(x-2\right)}{x\left(x-2\right)}\right)\)
\(=\left(\frac{4x^2-8x-8x^2}{\left(x-2\right)\left(x+2\right)}\right)\left(\frac{x-1-2x+4}{x\left(x-2\right)}\right)\)
\(=\left(\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right)\left(\frac{3-3x}{x\left(x-2\right)}\right)\)
\(=\frac{12\left(x-1\right)}{x-2}\)
Vậy ....
Ta có : \(A< 0\Rightarrow\frac{12\left(x-1\right)}{x-2}< 0\)
Đến đây xét 2 TH 12(x-1)<0 & (x-2)>0 hoặc 12(x-1)>0 & (x-2)<0
![](https://rs.olm.vn/images/avt/0.png?1311)
2x2 + 3y2 = 5xy
=> 2x2 + 3y2 - 5xy = 0
=> 2 ( x2 - 2xy + y2 ) - xy + y2 = 0
=> 2 ( x - y ) 2 - y ( x - y ) = 0
=> ( x - y )[ 2( x - y ) - y ] = 0
=> ( x- y ) ( 2x - 2y - y ) = 0
=> ( x - y ) ( 2x - 3y ) = 0
TH1 : x - y = 0
=> x = y
Thay x = y vào \(\frac{x+2y}{3x-y}\)
=> \(\frac{x+2y}{3x-y}=\frac{y+2y}{3y-y}\)\(=\frac{3y}{2y}=\frac{3}{2}\)
TH2 : 2x - 3y = 0
=> 2x = 3y
=> \(\frac{x}{y}=\frac{3}{2}\)
=> x = \(\frac{3}{2}.y\)
Thay x = \(\frac{3}{2}.y\)vào \(\frac{x+2y}{3x-y}\)
=> \(\frac{x+2y}{3x-y}=\frac{\frac{3}{2}.y+2y}{3.\frac{3}{2}y-y}\)\(=\frac{\frac{7}{2}.y}{\frac{7}{2}.y}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
_xin hỏi bài này có cần dùng bất đẳng thức Bunhiacopski không?
\(P=x^2-3x+\dfrac{1}{2x}+\dfrac{7}{4}+\dfrac{1}{4}\)
\(P=\dfrac{4x^3-12x^2+7x+2}{4x}+\dfrac{1}{4}=\dfrac{\left(x-2\right)\left(4x^2-4x-1\right)}{4x}+\dfrac{1}{4}\)
\(P=\dfrac{\left(x-2\right)\left[4x\left(x-2\right)+\dfrac{1}{2}\left(x-2\right)+\dfrac{7x}{2}\right]}{4x}+\dfrac{1}{4}\ge\dfrac{1}{4}\)
\(P_{min}=\dfrac{1}{4}\) khi \(x=2\)
\(P=x^2-3x+\dfrac{1}{2x}+2\)
\(P=x^2-4x+4+x+\dfrac{4}{x}-\dfrac{7}{2x}-2\)
\(P=\left(x-2\right)^2+x+\dfrac{4}{x}-\dfrac{7}{2x}-2\)
Áp dụng bđt cosi và bđt x \(\ge\)2
Ta có: P \(\ge0+2\sqrt{x\cdot\dfrac{4}{x}}-\dfrac{7}{2.2}-2=\dfrac{1}{4}\)
Dấu "=" xảy ra <=> x = 2
Vậy MinP = 1/4 <=> x = 2