\(x^2+xy+y^2+1=\left(x^2+xy+\frac{y^2}{4}\right)+\frac{3y^2}{4}+1=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1>0\)

2)\(\frac{x+y}{xy}\ge\frac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)

theo yêu cầu của bạn thì đến đâ mk làm theo cách này

ÁP Dụng cô si ta có:\(x+y\ge2\sqrt{xy}\)\(\Rightarrow\left(x+y\right)^2\ge4xy\)(luôn đúng)\(\Rightarrowđpcm\)

cách 2

\(\left(x+y\right)^2\ge4xy\Leftrightarrow x^2+2xy+y^2\ge4xy\)

\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\)(luôn đúng)

\(\Rightarrowđpcm\)

\(VD1:\) CMR: \(CMR:\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)=x^4-y^4\)

Giải:

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

\(=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4\)

\(=x^4-y^4=VP\rightarrowĐPCM.\)

VD2: Chứng minh rằng nếu: \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

thì \(\dfrac{a}{x}=\dfrac{b}{y}.\)

Giải:

Ta có: \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

\(\Rightarrow a^2x^2+b^2x^2+a^2y^2+b^2y^2=a^2x^2+2axby+b^2y^2\)

\(\Rightarrow\left(bx\right)^2-2axby+\left(ay\right)^2=0\)

\(\Rightarrow\left(ay-bx\right)^2=0\)

\(\Rightarrow ay-bx=0\)

\(\Rightarrow ay=bx\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}.\)

câu trên không có điều kiện các bạn nhé ! chỉ có thế thôi!

\(7,=\left(\sqrt{x}\right)^2+2\cdot2\sqrt{x}+2^2=\left(\sqrt{x}+2\right)^2\\ 8,=\left(\sqrt{x}\right)^2-2\cdot3\sqrt{x}+3^2=x-6\sqrt{x}+9\\ 9,=\sqrt{x^3}+\sqrt{y^3}=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)\\ 10,=\sqrt{x^3}-\sqrt{y^3}=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)\\ 11,=\sqrt{x^3}+1^3=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\\ 12,=\sqrt{x^3}-2^3=\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)\)

7: \(x+4\sqrt{x}+4=\left(\sqrt{x}+2\right)^2\)

8: \(\left(\sqrt{x}-3\right)^2=x-6\sqrt{x}+9\)

9: \(x\sqrt{x}+y\sqrt{y}=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)\)

7) \(x+4\sqrt{x}+4=\left(\sqrt{x}\right)^2+2\sqrt{x}.2+2^2=\left(\sqrt{x}+2\right)^2\)

8) \(\left(\sqrt{x}-3\right)^2=\left(\sqrt{x}\right)^2-2.\sqrt{x}.3+3^2=x-6\sqrt{x}+9\)

9) \(x\sqrt{x}+y\sqrt{y}=\sqrt{x^3}+\sqrt{y^3}=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)\)

10) \(x\sqrt{x}-y\sqrt{y}=\sqrt{x^3}-\sqrt{y^3}=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)\)

11) \(x\sqrt{x}+1=\sqrt{x^3}+1^3=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)

12) \(x\sqrt{x}-8=\sqrt{x^3}-2^3=\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)\)

7. x + \(4\sqrt{x}+4\)

= \(\left(\sqrt{x}\right)^2+2.2.\sqrt{x}+2^2\)

= \(\left(\sqrt{x}+2\right)^2\)

9: \(x-1=\left(\sqrt{x}\right)^2-1^2=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

10: \(x\cdot\sqrt{x}-1=\left(\sqrt{x}\right)^3-1^3\)

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

11: \(x-2\sqrt{x}-63\)

\(=x-9\sqrt{x}+7\sqrt{x}-63\)

\(=\sqrt{x}\left(\sqrt{x}-9\right)+7\left(\sqrt{x}-9\right)=\left(\sqrt{x}-9\right)\left(\sqrt{x}+7\right)\)

12: \(5\sqrt3-3\sqrt5=\sqrt{75}-\sqrt{45}=\sqrt{15}\cdot\sqrt5-\sqrt{15}\cdot\sqrt3=\sqrt{15}\left(\sqrt5-\sqrt3\right)\)

9: \(x-1=\left(\sqrt{x}\right)^2-1^2=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

10: \(x\cdot\sqrt{x}-1=\left(\sqrt{x}\right)^3-1^3\)

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

11: \(x-2\sqrt{x}-63\)

\(=x-9\sqrt{x}+7\sqrt{x}-63\)

\(=\sqrt{x}\left(\sqrt{x}-9\right)+7\left(\sqrt{x}-9\right)=\left(\sqrt{x}-9\right)\left(\sqrt{x}+7\right)\)

12: \(5\sqrt3-3\sqrt5=\sqrt{75}-\sqrt{45}=\sqrt{15}\cdot\sqrt5-\sqrt{15}\cdot\sqrt3=\sqrt{15}\left(\sqrt5-\sqrt3\right)\)