2) \(x^2\left(x+3\right)+y^2\left(y+5\right)-\left(x+y\right)\left(x^2-xy+y^2\right)=0\Leftrightarrow x^3+3x^2+y^3+5y^2-x^3-y^3=0\Leftrightarrow3x^2+5y^2=0\)Do \(3x^2\ge0\) và \(5y^2\ge0\) => 3x²+5y²\(\ge\)0.Dấu "=" xảy ra khi x=y=0

1b) \(5x\left(x+3\right)^2-5\left(x+1\right)^3+15\left(x+2\right)\left(x-2\right)=5\Leftrightarrow5x\left(x^2+6x+9\right)-5\left(x^3+3x^2+3x+1\right)+15\left(x^2-4\right)=5\Leftrightarrow30x-65=5\Leftrightarrow30x=70\Leftrightarrow x=\dfrac{7}{3}\)

Ta có : B= 3xy(x+3y) - 2xy(x+4y) - x²(y-1)+y²(1-x)+36 = 3x²ý+9xy²-2x²y-8xy²-x²y+x²+y²-xy²+36 =x²+y²+36 \(\ge36\) Dấu "=" xảy ra khi x=y=0 Vậy Min B=36 <=> x=y=0

đề có j sai ko bạn

Ta có: a+b+c=9 <=> (a+b+c)²=a²+b²+c²+2ab+2bc+2ca=81 mà a²+b²+c²=53 => 2(ab+bc+ca) = 81 -53 =28

Ta có :\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}=\dfrac{a^3+b^3+c^3}{abc}\) Lại có: a³+b³+c³= [a³+3ab(a+b)+b³] +c³ - 3ab(a+b) =(a+b)³+c³-3ab(a+b) = (a+b+c)[a²+2ab+b²-ac-bc+c²] -3ab(a+b) = -3ab(a+b) = -3ab(-c) = 3abc => A=\(\dfrac{3abc}{abc}=3\)

\(K=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x\sqrt{x}-x+\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}}{x+1}\right)=\left(\dfrac{x-1}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right).\left(\dfrac{x+1}{x+1-\sqrt{x}}\right)=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+1}{x-\sqrt{x}+1}\)để rk là ok rồi

\(D=\left(\dfrac{1}{\sqrt{5}-2}-\dfrac{1}{\sqrt{5}+2}+1\right).\dfrac{1}{\left(\sqrt{2}+1\right)^2}=\left(\dfrac{\sqrt{5}+2-\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+1\right).\dfrac{1}{\left(\sqrt{2}+1\right)^2}=\dfrac{4+1}{5-4}.\dfrac{1}{3+2\sqrt{2}}=\dfrac{5}{3+2\sqrt{2}}=\dfrac{5\left(3-2\sqrt{2}\right)}{9-8}=15-10\sqrt{2}\)

\(C=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+3\sqrt{\dfrac{1}{3}}\right)=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(2+\sqrt{\dfrac{9}{3}}\right)=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}=\sqrt{2}\)

\(C=\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2.2+\sqrt{5+2\sqrt{5}+1}}+\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2.2-\sqrt{5-2\sqrt{5}+1}}=\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{4+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{4-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{2.}\left(\dfrac{3+\sqrt{5}}{5+\sqrt{5}}+\dfrac{3-\sqrt{5}}{5-\sqrt{5}}\right)=\sqrt{2}\left(\dfrac{\left(3+\sqrt{5}\right)\left(5-\sqrt{5}\right)+\left(3-\sqrt{5}\right)\left(5+\sqrt{5}\right)}{\left(5+\sqrt{5}\right)\left(5-\sqrt{5}\right)}\right)=\sqrt{2}.\dfrac{15+5\sqrt{5}-3\sqrt{5}-5+15-5\sqrt{5}+3\sqrt{5}-5}{25-5}=\sqrt{2}.\dfrac{20}{20}=\sqrt{2}\)