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

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

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

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

a: x>2

y>2

=>x+y>2+2=4

x>y>2

=>xy>2^2=4

b: x^2-xy=x(x-y)

x-y>0; x>0

=>x(x-y)>0

=>x^2-xy>0

y>2

=>y-2>0

=>y(y-2)>0

=>y^2-2y>0

x>y và y>2

=>y>0 và x-y>0

=>y(x-y)>0

=>xy-y^2>0

a) \(A=\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}\)

\(A=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)

\(A=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

b) \(B=\dfrac{x\sqrt{y}-y\sqrt{x}}{x-2\sqrt{xy}+y}\)

\(B=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)

\(B=\dfrac{\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\)

c) \(C=\dfrac{3\sqrt{a}-2a-1}{4a-4\sqrt{a}+1}\)

\(C=\dfrac{-\left(2a-3\sqrt{a}+1\right)}{\left(2\sqrt{a}\right)^2-2\sqrt{a}\cdot2\cdot1+1^2}\)

\(C=\dfrac{-\left(\sqrt{a}-1\right)\left(2\sqrt{a}-1\right)}{\left(2\sqrt{a}-1\right)^2}\)

\(C=\dfrac{-\sqrt{a}+1}{2\sqrt{a}-1}\)

d) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)

\(D=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}\)

\(D=\sqrt{a}+2-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)

\(D=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)\)

\(D=0\)

1 , sai đề

2/ xy-x-y+1=0

x(y-1)-(y-1)=0

(y-1)(x-1)=0

->y-1=o hoặc x-1=0

y-1=0            y=1

x-1=0           x=1

vậy x=y=1

3, 

c) \(^{x^2}\)+xy-x-2=0

bn vào trang wed này mik chỉ cho, cứ nhắn tin cho mik đi rồi mik sẽ ns.

\(VT\ge\dfrac{4}{x^2+y^2+2xy}=\dfrac{4}{\left(x+y\right)^2}\ge4\) (vì \(x+y\le1\) )

Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)

Ta có đpcm