Question

[dương thị thanh vân]

bài 1 rút gọn biểu thức sau:

a)\(\sqrt{16+6\sqrt{7}}\)- \(\sqrt{8-2\sqrt{7}}\)           b)K=\(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)

c)\(\sqrt{60-24\sqrt{6}}\)+\(\sqrt{40-16\sqrt{6}}\)        d)B=(3+\(\sqrt{3}\))\(\sqrt{12-6\sqrt{13}}\)

e)\(\sqrt{6-4\sqrt{2}}\)-\(\sqrt{\left(\sqrt{2}-\sqrt{6}\right)^2}\)

bài 2 cho biểu thức A=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{3}{\sqrt{x}-3}\right).\dfrac{\sqrt{x}+3}{x+9}\)( với x≥0 và x≠ 9)

a) rút gọn biểu thức A

b) tính giá trị biểu thức\(x=4+2\sqrt{3}\)

Answer 1

[Nguyễn Hoàng Minh]

\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)

\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)

Answer 2

[dương thị thanh vân]

cảm ơn bạn