tìm GTNN và GTLN của phân thức P=\(\frac{2x^2+4x+5}{x^2+1}\)
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có 3x^2+2y^2=7xy
=>3x^2+2y^2-7xy=0
=>(3x^2-6xy)+(2y^2-xy)=0
=>3x(x-2y)-y(x-2y)=0
=>(x-2y)(3x-y)=0
=>x-2y=0 hoặc 3x-y=0
=>x=2y hoặc y=3x
Xét TH x=2y vài A ta được 3x+y/7y-x+6x-9y/2x+y
=6y+y/7y-2y+12y-9y/4y+y
=7y/5y+3y/5y
=7/5+3/5
=10/5
=1/2
Xét TH y=3x có
3x+y/7y-x+6x-9y/2x+y
=3x+3x/(21x-x)+(6x-27x)/2x+3x
=6x/20x-21x/5x
=3/10-21/5
=3/10-42/10
=-39/10
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐKXĐ:\hept{\begin{cases}x\ne-1\\x\ne1\end{cases}}\)
\(\frac{x+1}{2x-2}+\frac{x-1}{2x+2}+\frac{x^2}{1-x^2}=\frac{x+1}{2\left(x-1\right)}+\frac{x-1}{2\left(x+1\right)}-\frac{x^2}{x^2-1}\)
\(=\frac{x+1}{2\left(x-1\right)}+\frac{x-1}{2\left(x+1\right)}-\frac{x^2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}-\frac{2x^2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{\left(x+1\right)^2+\left(x-1\right)^2-2x^2}{2\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1+x^2-2x+1-2x^2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{\left(x-1\right)\left(x+1\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x^2-2x-3/x^2+x=(x+x)-(3x+3)/x(x+1)=x(x+1)-3(x+1)/x(x+1)=x-3/x
x^2-4x-3/x^2-x=(x^2-x)-(3x-3)/x(x-1)=x(x-1)-3(x-1)/x(x-1)=x-3/x
=>x^2-2x-3/x^2+x=x-3/x=x^2-4x+3/x^2-x
![](https://rs.olm.vn/images/avt/0.png?1311)
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a/ x2+ xy+ y2+1
= (x2+xy+y24) +34y2 +1
= (x+y2)2 +3y24 +1 ≥1 >0 x2 + xy + y2+1
=(x2+xy+y24)+34y2+1
=(x+y2)2+3y24+1≥1>0
Với mọi x,y
CHúc bạn học tốt
\(5x^2+10y^2-6xy-4x-2y+3\)
\(=\left(x^2-6xy+9y^2\right)+\left(4x^2-4x+1\right)+\left(y^2-2y+1\right)+1\)
\(=\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐKXĐ:\hept{\begin{cases}x\ne0\\y\ne0\\x\ne y\end{cases}}\)
\(\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}=\frac{x}{y\left(x-y\right)}-\frac{2x-y}{x^2-xy}=\frac{x}{y\left(x-y\right)}-\frac{2x-y}{x\left(x-y\right)}\)
\(=\frac{x^2}{xy\left(x-y\right)}-\frac{y\left(2x-y\right)}{xy\left(x-y\right)}=\frac{x^2}{xy\left(x-y\right)}-\frac{2xy-y^2}{xy\left(x-y\right)}\)
\(=\frac{x^2-\left(2xy-y^2\right)}{xy\left(x-y\right)}=\frac{x^2-2xy+y^2}{xy\left(x-y\right)}=\frac{\left(x-y\right)^2}{xy\left(x-y\right)}=\frac{x-y}{xy}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\) (điều kiện: \(x;y\ne0\); \(x\ne\pm2y\))
\(=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2}{x+2y}+\frac{1}{x-2y}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2\left(x-2y\right)+\left(x+2y\right)+4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{3x-2y+4}{\left(x-2y\right)\left(x+2y\right)}\)
\(P=\frac{2\left(x^2+1\right)+3}{x^2+1}=2+\frac{3}{x^2+1}\)
Vì \(x^2\ge0\Rightarrow x^2+1\ge1\Rightarrow\frac{3}{x^2+1}\le3\Rightarrow2+\frac{3}{x^2+1}\le5\)\(\Leftrightarrow P\le5\)
Dấu "=' xảy ra khi \(x=0\)
Vậy: \(GTLN\)của \(P=5\)tại \(x=0\)