\(P=\frac{2\left(x^2+1\right)+3}{x^2+1}=2+\frac{3}{x^2+1}\)

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 \(x=0\)

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

5xz(x-3y+6z)

\(ĐKXĐ:\hept{\begin{cases}x\ne-1\\x\ne1\end{cases}}\)

\(\frac{x+1}{2x-2}+\frac{x-1}{2x+2}+\frac{x^2}{1-x^2}=\frac{x+1}{2\left(x-1\right)}+\frac{x-1}{2\left(x+1\right)}-\frac{x^2}{x^2-1}\)

\(=\frac{x+1}{2\left(x-1\right)}+\frac{x-1}{2\left(x+1\right)}-\frac{x^2}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}-\frac{2x^2}{2\left(x-1\right)\left(x+1\right)}\)

\(=\frac{\left(x+1\right)^2+\left(x-1\right)^2-2x^2}{2\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1+x^2-2x+1-2x^2}{2\left(x-1\right)\left(x+1\right)}\)

\(=\frac{2}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{\left(x-1\right)\left(x+1\right)}\)

Tính chất cơ bản của phân thức , rút gọn phân thức

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

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

\(5x^2+10y^2-6xy-4x-2y+3\)

\(=\left(x^2-6xy+9y^2\right)+\left(4x^2-4x+1\right)+\left(y^2-2y+1\right)+1\)

\(=\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1\)

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\y\ne0\\x\ne y\end{cases}}\)

\(\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}=\frac{x}{y\left(x-y\right)}-\frac{2x-y}{x^2-xy}=\frac{x}{y\left(x-y\right)}-\frac{2x-y}{x\left(x-y\right)}\)

\(=\frac{x^2}{xy\left(x-y\right)}-\frac{y\left(2x-y\right)}{xy\left(x-y\right)}=\frac{x^2}{xy\left(x-y\right)}-\frac{2xy-y^2}{xy\left(x-y\right)}\)

\(=\frac{x^2-\left(2xy-y^2\right)}{xy\left(x-y\right)}=\frac{x^2-2xy+y^2}{xy\left(x-y\right)}=\frac{\left(x-y\right)^2}{xy\left(x-y\right)}=\frac{x-y}{xy}\)

\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)          (điều kiện: \(x;y\ne0\); \(x\ne\pm2y\))

\(=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2}{x+2y}+\frac{1}{x-2y}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2\left(x-2y\right)+\left(x+2y\right)+4}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{3x-2y+4}{\left(x-2y\right)\left(x+2y\right)}\)