\(A^2=\left(\sqrt{13+4\sqrt{3}}+\sqrt{13-4\sqrt{3}}\right)^2\)

\(=13+4\sqrt{3}+13-4\sqrt{3}+2\sqrt{\sqrt{13+4\sqrt{3}}\cdot\sqrt{13-4\sqrt{3}}}\)

\(=26+2\sqrt{13^2-\left(4\sqrt{3}\right)^2}\)

\(=26+2\sqrt{121}=26+22=48\)

\(\Rightarrow A^2=48\Rightarrow A=\sqrt{48}\)

a, \(\frac{\sqrt{3-\sqrt{5}}\times''3+\sqrt{5}''}{\sqrt{10}+\sqrt{2}}\)

\(=\frac{-9.976153125}{4.576491223}\)

b,\(\frac{''\sqrt{5}+2''^2-8\sqrt{5}}{2\sqrt{5}-4}\)

\(=\frac{0.05572809}{0.472135955}\)

P/s; Em không chắc đâu ạ. Mới lớp 5 lên 6 thôi

Ta có:

\(VT=\sqrt{3x^2-6x+19}+\sqrt{x^2-2x+26}\)

\(=\sqrt{3\left(x-1\right)^2+16}+\sqrt{\left(x-1\right)^2+25}\ge4+5=9\)

\(VP=8-x^2+2x=9-\left(x-1\right)^2\le9\)

Dấu = xảy ra khi \(x=1\)

Xét \(y=-1\)\(\Rightarrow x^3=2\left(l\right)\)

Xét \(y=-2\)\(\Rightarrow x=1\)

Xét \(y\ne-1,-2\)thì ta có:

\(\left(y-1\right)^3< y^3+2y^2+1=x^3\le\left(y+1\right)^3\)

\(\Rightarrow\orbr{\begin{cases}y^3+2y^2+1=y^3\\y^3+2y^2+1=\left(y+1\right)^3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=-3\\y=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)

Đặt \(\left(\sqrt{a};\sqrt{b}\right)\rightarrow\left(x;y\right)\);\(GT\Leftrightarrow x+y\ge2\)& cm \(x^3+y^3\ge x^2+y^2\)

we had: \(x^3+x^3+1\ge3x^2\)

.... \(2VT+2\ge3\left(x^2+y^2\right)=2VF+x^2+y^2\)

\(x^2+y^2\ge x+y\ge2\)

\(\frac{\left(\sqrt{a}\right)^3-\left(\sqrt{b}\right)^3}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}-\sqrt{b}}=a+b+\sqrt{ab}\)

we had abc+(4-a)(4-b)(4-c)\(\ge0\). khai triển ta có \(ab+bc+ca\ge8\)( maybe)

\(P=\left(a+b+c\right)^2-\left(ab+bc+ca\right)\le6^2-8=28\)

Dấu = xảy ra (a,b,c)~(0;2;4) và các hoán vị