PugMath

có ai giải 5b chưa ạ cho em tham khảo với ??? 

$-2(2-x)\sqrt{x+y}+3\sqrt{x+y}=6\sqrt{2-x}-(x+y)\sqrt{2-x}<=>\sqrt{(2-x)(x+y)}(2\sqrt{2-x}-\sqrt{x+y})+3(2\sqrt{2-x}-\sqrt{x+y})=0=>2\sqrt{2-x}=\sqrt{x+y}=>8-4x=x+y<=>y=8-5x$

=> $x^2+11x+8-5x=4+\sqrt{6x+8-5x}<=>x^2+6x+4=\sqrt{x+8}<=>(x^2+5x+1)(x^2+7x+8)=0=>.........$

$\frac{3}{2}a+\frac{a}{2}+\frac{b}{4a}+b^2\geqslant \frac{3}{2}a+\frac{3}{2}b\geqslant \frac{3}{2}$

câu 1b) $xt=\frac{x^2(x^2+x+1)}{x^4+x^2+1}=A(x^2+x+1)=>\frac{t}{A}=\frac{x^2+x+1}{x}=\frac{x^2-x+1+2x}{x}=\frac{1}{t}+2=>A=\frac{t^2}{1+2t}$

câu 2b) $2x^2+xy-y^2-5x+y+2=0<=>2x(x+y-2)-y(x+y-2)-(x+y-2)=0<=>(x+y-2)(2x-y-1)=0=>.$

câu 3b) $a=\sqrt[3]{3x^2-x+1};b=\sqrt[3]{3x^2-7x+2} =>a-b-\sqrt[3]{2}=\sqrt[3]{a^3-b^3-2}<=>a^3-b^3-2-3ab(a-b)-3(a-b)\sqrt[3]{2}(a-b-\sqrt[3]{2})=a^3-b^3-2<=>ab(a-b)+(a-b)\sqrt[3]{2}(a-b-\sqrt[3]{2})=0<=>(a-\sqrt[3]{2})(a-b)(b+\sqrt[3]{2})=0$

$\sum{\frac{a}{\frac{a+b}{2}+\frac{a+c}{2}+a+\frac{a+b}{2}+\frac{a+c}{2}}}\leqslant \frac{1}{25} \left ( \frac{4a}{a+b}+\frac{4b}{a+b}+\frac{4c}{a+c}+\frac{4a}{a+c}+\frac{4b}{c+b}+\frac{4c}{c+b}+3 \right )=\frac{1}{25}.15=\frac{3}{5}$