b) \(=\left(5+2\sqrt{3}\right).\sqrt{25-2.5.2\sqrt{3}+12}\\ =\left(5+2\sqrt{3}\right).\sqrt{\left(5-2\sqrt{3}\right)^2}=\left(5+2\sqrt{3}\right).\left(5-2\sqrt{3}\right)=25-12=13\)

c: \(\left(2-\dfrac{5-\sqrt{5}}{1-\sqrt{5}}\right)\left(\dfrac{\sqrt{10}+\sqrt{5}}{\sqrt{2}+1}-2\right)\)

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

=1

\(b,=\left(5+2\sqrt{3}\right)\sqrt{\left(5-2\sqrt{3}\right)^2}=\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)=25-12=13\\ c,=\left(2-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{1-\sqrt{5}}\right)\left(\dfrac{\sqrt{5}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-2\right)\\ =\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)=5-4=1\)

Bài 4: 

a) Ta có: \(x^2-3x=0\)

\(\Leftrightarrow x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

c) Ta có: \(4x^3-x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Bài 3: 

c) Ta có: \(C=xy\left(x+y\right)-2x-2y\)

\(=\left(x+y\right)\left(xy-2\right)\)

\(=7\cdot\left(8-2\right)=7\cdot6=42\)

Chọn A

Số phần trứng còn lại là

1-2/5-1/5=2/5(phần)

a: =4-10

=-6

a, \(\dfrac{\sqrt{80}}{\sqrt{5}}\)-\(\sqrt{5}\).\(\sqrt{20}\)= \(\sqrt{16}\)-10=-6

b, (\(\sqrt{28}\)-\(\sqrt{12}\)-\(\sqrt{7}\))\(\sqrt{7}\)+2\(\sqrt{21}\)=\(\sqrt{196}\)-\(\sqrt{84}\)-7+2 \(\sqrt{21}\)=14-7=7

c, \(\sqrt[3]{2}\).\(\sqrt[3]{32}\)+\(\sqrt{2}\).\(\sqrt{32}\)=\(\sqrt[3]{64}\)+\(\sqrt{64}\)=4+8=12

d, \(2\sqrt{8\sqrt{3}}\)-\(\sqrt{2\sqrt{3}}\)-\(\sqrt{9\sqrt{12}}\)=\(4\sqrt{12}\)-\(\sqrt{12}\)-\(3\sqrt{12}\)=0

1: Thay x=16 vào A, ta được:

\(A=\dfrac{16-4}{4-2}=\dfrac{12}{2}=6\)

2: \(B=\dfrac{x-4+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\dfrac{x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)