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\(a) (x+y)+(y-x)\)

\(= x+y+y-x\)

\(=(x-x)+(y+y)\)

\(= 2y\)

\(b) (x+y)-(y-x)\)

\(= x+y-y+x\)

\(= (x+x)+(y-y)\)

\(= 2x\)

\(c) (x-y)+(y-x)\)

\(= x-y+y-x\)

\(= (x-x)+(y-y) \)

\(=0\)

\(e) (x^2+xy-1)+(3x^2+xy+1)\)

\(= x^2+xy-1+3x^2+xy+1\)

\(= (x^2+3x^2)+(xy+xy)+(1-1)\)

\(= 4x^2 + 2xy\)

\(g) (x^2+xy-1)-(3x^2+xy+1)\)

\(= x^2+xy-1-3x^2-xy-1\)

\(= (x^2-3x^2)+(xy-xy)-(1+1)\)

\(= -2x^2 - 2\)

x²y + xy² - x - y

= (x²y + xy²) - (x + y)

= xy(x + y) - (x + y)

= (x + y)(xy - 1)

\(\begin{array}{l}T + H = 3{x^2}y - 2x{y^2} + xy + \left( { - 2{x^2}y + 3x{y^2} + 1} \right)\\ = 3{x^2}y - 2x{y^2} + xy - 2{x^2}y + 3x{y^2} + 1\\ = \left( {3{x^2}y - 2{x^2}y} \right) + \left( { - 2x{y^2} + 3x{y^2}} \right) + xy + 1\\ = {x^2}y + x{y^2} + xy + 1\\T - H = 3{x^2}y - 2x{y^2} + xy - \left( { - 2{x^2}y + 3x{y^2} + 1} \right)\\ = 3{x^2}y - 2x{y^2} + xy + 2{x^2}y - 3x{y^2} - 1\\ = \left( {3{x^2}y + 2{x^2}y} \right) + \left( { - 2x{y^2} - 3x{y^2}} \right) + xy - 1\\ = 5{x^2}y - 5x{y^2} + xy - 1\end{array}\)

Chọn B.

B

Dễ ợt ak

Hướng dẫn thôi nhé:

Lời giải:

a)\(xy+x+y+1=0\)

\(\Rightarrow x\left(y+1\right)+1\left(y+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(y+1\right)=0\)

b)\(xy-x-y=0\)

\(\Rightarrow xy-x-y+1=1\)

\(\Rightarrow x\left(y-1\right)-1\left(y-1\right)=1\)

\(\Rightarrow\left(x-1\right)\left(y-1\right)=1\)

c)\(xy-x-y-1=0\)

\(\Rightarrow xy-x-y+1=2\)

\(\Rightarrow x\left(y-1\right)-1\left(y-1\right)=2\)

\(\Rightarrow\left(x-1\right)\left(y-1\right)=2\)

d) \(xy-x-y+1=0\)

\(\Rightarrow x\left(y-1\right)-1\left(y-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(y-1\right)=0\)

e)\(xy+2x+y+11=0\)

\(\Rightarrow xy+2x+y+2=-9\)

\(\Rightarrow x\left(y+2\right)+1\left(y+2\right)=-9\)

\(\Rightarrow\left(x+1\right)\left(y+2\right)=-9\)

\(B=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2+\left(xy+\frac{1}{xy}\right)^2\)

\(-\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)\left(xy+\frac{1}{xy}\right)\)

\(\Rightarrow B=x^2+2+\frac{1}{x^2}+y^2+2+\frac{1}{y^2}+x^2y^2+2+\frac{1}{x^2y^2}-x^2y^2\) 

\(-2-x^2-y^2-\frac{1}{y^2}-\frac{1}{x^2}-\frac{1}{x^2y^2}\)

\(\Rightarrow B=x^2y^2-x^2y^2+x^2-x^2+1.\frac{1}{x^2}+1.\frac{1}{x^2y^2}-1.\frac{1}{x^2}-1\)

\(.\frac{1}{x^2y^2}+1.\frac{1}{y^2}-1.\frac{1}{y^2}+y^2-y^2+2+2+2-2\)

\(\Rightarrow B=4\)

\(6xy=x+y\ge2\sqrt[]{xy}\Rightarrow\sqrt{xy}\ge\dfrac{1}{3}\Rightarrow xy\ge\dfrac{1}{9}\Rightarrow\dfrac{1}{xy}\le9\)

\(M=\dfrac{\dfrac{x+1}{xy+1}+\dfrac{xy+x}{1-xy}+1}{1+\dfrac{xy+x}{1-xy}-\dfrac{x+1}{xy+1}}=\dfrac{\dfrac{x+1}{xy+1}+\dfrac{x+1}{1-xy}}{\dfrac{x+1}{1-xy}-\dfrac{x+1}{xy+1}}=\dfrac{\dfrac{1}{1-xy}+\dfrac{1}{1+xy}}{\dfrac{1}{1-xy}-\dfrac{1}{1+xy}}\)

\(M=\dfrac{1+xy+1-xy}{1+xy-1+xy}=\dfrac{2}{2xy}=\dfrac{1}{xy}\le9\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{3}\)