căn (40-x)=a , căn (45-x)=b,căn(72-x)=c (a,b,c >=0 )

đưa về hệ: ab+bc+ca=40-a^2 -> ab+bc+ca+a^2=40 

                 ab+bc+ca=45-b^2......

                 ab+bc+ca=72-c^2.....

đến đó ok rồi 

HELP ME♥

\(\sqrt{x+72}=72-x^2\Rightarrow x+72=72^2-144x^2+x^4\Rightarrow0=5184-72+x^4-144x^2-x=5112+x^4-144x^2-x\) =>\(0=x^4-9x^3+9x^3-81x^2-63x^2+567x-568x+5112\)

=>\(0=\left(x-9\right)\left(x^3+9x^2-63x-568\right)\)

=>\(0=\left(x-9\right)\left(x+8\right)\left(x^2+x-71\right)\)=>x=9 hoac x=-8

\(=3\sqrt{2}-3\sqrt{2}+2\sqrt{2}+6\sqrt{2}=8\sqrt{2}\)

\(\sqrt{36x-72}-15\sqrt{\dfrac{x-2}{25}}=20+4\sqrt{x-2}\)

\(\Leftrightarrow6\sqrt{x-2}-3\sqrt{x-2}-4\sqrt{x-2}=20\)

\(\Leftrightarrow-\sqrt{x-2}=20\)(vô lý)

Ta có :

\(a^2=72+\sqrt{72+\sqrt{72+\sqrt{72+.......}}}\)

\(\Leftrightarrow a^2=72+a\Leftrightarrow a^2-a-72=0\Leftrightarrow\left(a-9\right)\left(a+8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a=9\\a=-8\end{cases}}\)

Mà a > 0 nên a = 9 \(\Rightarrow\left[a\right]=9\)

\(\left(\sqrt[3]{\dfrac{1}{9}}+4\cdot\sqrt[3]{\dfrac{1}{72}}-\sqrt[3]{4}\right)\left(\sqrt[3]{72}+\sqrt[3]{96}+\sqrt[3]{128}\right)\)

\(=\left(\dfrac{1}{3}\cdot\sqrt[3]{3}+4\cdot\dfrac{1}{6}\cdot\sqrt[3]{3}-2\sqrt[3]{\dfrac{1}{2}}\right)\left(2\sqrt[3]{9}+2\sqrt[3]{12}+4\sqrt[3]{2}\right)\)

\(=\left(\sqrt[3]{3}-2\sqrt[3]{\dfrac{1}{2}}\right)\left(6\sqrt[3]{3}+2\sqrt[3]{12}+4\sqrt[3]{2}\right)\)

\(=6\cdot3+2\sqrt[3]{36}+4\sqrt[3]{6}-12\sqrt[3]{\dfrac{3}{2}}-4\sqrt[3]{6}-8\)

\(=10+12\sqrt[3]{\dfrac{1}{6}}-6\sqrt[3]{12}\)