\(2\sqrt{20}+\sqrt{50}+3\sqrt{80}-\sqrt{320}\)

\(=2\sqrt{4.5}+\sqrt{2.25}+3\sqrt{16.5}-\sqrt{64.5}\)

\(=2.2\sqrt{5}+5\sqrt{2}+3.4\sqrt{5}-8\sqrt{5}\)

\(=\left(4+12-8\right)\sqrt{5}+5\sqrt{2}\)

\(=8\sqrt{5}+5\sqrt{2}\)

P/s: Em mới lớp 5 nên làm đại, sai thì thông cảm ạ.

\(2\sqrt{20}+\sqrt{50}+3\sqrt{80}-\sqrt{320}\)

\(=2\sqrt{4.5}+\sqrt{25.2}+3\sqrt{16.5}-\sqrt{64.5}\)

\(=2.2\sqrt{5}+3.4\sqrt{5}-8\sqrt{5}+5\sqrt{2}\)

\(=4\sqrt{5}+12\sqrt{5}-8\sqrt{5}+5\sqrt{2}\)

\(=8\sqrt{5}+5\sqrt{2}\)

Ta có : \(a^3+1=\left(a+1\right)\left(a^2-a+1\right)=\left(a+1\right)\left(a^2-a+\frac{1}{4}\right)+\frac{3}{4}\left(a+1\right)\)

do đó : \(a^3-\frac{3}{4}a+\frac{1}{4}=\left(a+1\right)\left(a-\frac{1}{2}\right)^2\ge0\)với \(a\ge-1\)

Tương tự : \(b^3-\frac{3}{4}b+\frac{1}{4}\ge0,c^3-\frac{3}{4}c+\frac{1}{4}\ge0\)với \(b,c\ge-1\)

\(a^3+b^3+c^3-\frac{3}{4}\left(a+b+c\right)+\frac{3}{4}\ge0\Rightarrow a^3+b^3+c^3\ge-\frac{3}{4}\)

Dấu " = " xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}a=-1\text{ hoặc }a=\frac{1}{2}\\.....\\a+b+c=0\end{cases}}\)

Vậy GTNN của A là \(\frac{-3}{4}\)\(\Leftrightarrow\) a,b,c có 2 số bằng \(\frac{1}{2}\)và 1 số bằng -1

Ta có : \(x+3-4\sqrt{x-1}=\left(\sqrt{x-1}-2\right)^2\)và \(x+15-8\sqrt{x-1}=\left(\sqrt{x-1}-4\right)^2\)
Suy ra: B=\(\sqrt{x-1}-2+\sqrt{x-1}-4=2\sqrt{x-1}-6\)
Ta lại có : \(x-1\ge0\)=>\(B\ge-6\)dấu ''='' xảy ra khi: x-1=0 <=>x=1
Vậy minB=-6 khi x=1

Ta có \(2^x+\left(x^2+1\right)\left(y-2\right)\left(y-4\right)=0\)

Mà \(2^x>0,x^2+1>0\)

=> \(\left(y-2\right)\left(y-4\right)< 0\)

=> \(2< y< 4\)

=> \(y=3\)

Thay y=3 vào đề bài ta có:

\(2^x-\left(x^2+1\right)=0\)

=> \(2^x=x^2+1\)

Mà \(2^x\)chẵn với \(x>0\)

=> \(x\)lẻ

Đặt \(x=2k+1\)(k không âm)

Khi đó \(2^{2k+1}=\left(2k+1\right)^2+1\)

=> \(2.2^{2k}=4k^2+4k+2\)

=> \(2^{2k}=2k^2+2k+1\)

+ k=0 => \(2^0=1\)thỏa mãn 

=> \(x=1\)

+ \(k>0\)=> \(2^k\)chẵn 

Mà \(2k^2+2k+1\)lẻ với mọi k

=> không giá trị nào của k thỏa mãn

Vậy x=1,y=3

a)\(\sqrt{-8x}\)có nghĩa khi \(-8x\ge0\Leftrightarrow x\le0\)

b)\(\sqrt{\left(\sqrt{3}-x\right)^2}\)có nghĩa khi \(\left(\sqrt{3}-x\right)^2\ge0\Leftrightarrow\sqrt{3}-x\ge0\Leftrightarrow x\le\sqrt{3}\)

c)\(\frac{16x-1}{\sqrt{x-7}}\)có nghĩa khi \(\hept{\begin{cases}\sqrt{x-7}\ne0\\x-7\ge0\end{cases}\Leftrightarrow x-7}>0\Leftrightarrow x>7\)

 \(a,-8x>0\Rightarrow x< 0\)

\(b,x\in R\)

\(c,\hept{\begin{cases}\sqrt{x-7}\ne0\\x-7>0\Rightarrow x>7\end{cases}}\)

\(\sqrt{2+\sqrt{2+\sqrt{2}}}< \sqrt{2+\sqrt{2+\sqrt{4}}}=\sqrt{2+\sqrt{2+2}}=\sqrt{2+\sqrt{4}}=\sqrt{2+2}=2\)

\(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)

\(=\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}.\left(\sqrt{5}-1\right).\sqrt{2}.\sqrt{3+\sqrt{5}}\)

\(=\sqrt{9-5}\left(\sqrt{5}-1\right)\sqrt{6+2\sqrt{5}}\)

\(=2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\)

\(=2\left(5-1\right)\)

\(=8\)

\(\left(4+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\left(4+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(=\left(4+\sqrt{5}\right)\left(8-2\sqrt{15}\right)\)

Em thử ạ. Bài dài quá em chẳng biết có tính sai chỗ nào hay không nữa ;(

Từ giả thiết ta có: 

\(\hept{\begin{cases}x+y=-\frac{2}{3}\left(z+1\right)\\xy=-\frac{1}{3}\end{cases}}\Rightarrow x^2+y^2=\left(x+y\right)^2-2xy=\frac{4}{9}\left(z+1\right)^2+\frac{2}{3}\)

Và \(\left(x-y\right)^2=\left(x+y\right)^2-4xy=\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}\)

Ta có: \(A=\frac{\left(x-y\right)\left(x^2+xy+y^2\right)+\left(z+1\right)\left(x-y\right)\left(x+y\right)-\left(x-y\right)}{\left(x-y\right)^3}\)

\(=\frac{\left(x-y\right)\left(x^2+y^2-\frac{1}{3}\right)+\left(z+1\right)\left(x-y\right)\left(x+y\right)-\left(x-y\right)}{\left(x-y\right)^3}\)

\(=\frac{\left(x-y\right)\left(x^2+y^2-\frac{1}{3}+\left(z+1\right)\left(x+y\right)-1\right)}{\left(x-y\right)^3}\)

\(=\frac{\left(x^2+y^2-\frac{1}{3}+\left(z+1\right)\left(x+y\right)-1\right)}{\left(x-y\right)^2}\)

\(=\frac{\left(\frac{4}{9}\left(z+1\right)^2+\frac{1}{3}-\frac{2}{3}\left(z+1\right)^2\right)}{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}=\frac{-\frac{2}{9}\left(z+1\right)^2+\frac{1}{3}}{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}\)

\(=\frac{\left(\frac{4}{9}\left(z+1\right)^2+\frac{1}{3}-\frac{2}{3}\left(z+1\right)^2\right)}{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}=\frac{-\frac{2}{9}\left(z+1\right)^2+\frac{1}{3}}{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}\)

Ơ....hình như em tính sai chỗ nào rồi:(

Nguyễn Khang 

\(A=\frac{\left(x^2+y^2-\frac{1}{3}+\left(z+1\right)\left(x+y\right)-1\right)}{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}\)

\(=\frac{\left(\frac{4}{9}\left(z+1\right)^2+\frac{1}{3}-\frac{2}{3}\left(z+1\right)^2-1\right)}{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}\) ( như này mới đúng, e thiếu -1 ở tử ) 

\(=\frac{\frac{-2}{9}\left(z+1\right)^2-\frac{2}{3}}{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}=-\frac{1}{2}.\frac{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}{\frac{4}{9}\left(z+1\right)^2+\frac{4}{3}}=\frac{-1}{2}\)