Cho biểu thức ban đầu là A
Đặt 3 = a ; \(\sqrt{9+\dfrac{125}{27}}\)= b
⇔A = \(\sqrt[3]{a+b} . \sqrt[3]{b-a}\) 
⇔A=  \(\sqrt[3]{(a+b)(b-a)}\)
⇔A= \(\sqrt[3]{b^2-a^2}\)
⇔A= \(\sqrt[3]{9+\dfrac{125}{27}-9}\)
⇔A= \(\sqrt[3]{\dfrac{125}{27}}\)
⇔A = \(\dfrac{5}{3}\) ( ĐPCM)
 

\(\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}\sqrt[3]{-3+\sqrt{9+\dfrac{125}{27}}}\)

=\(\sqrt[3]{-\left(3+\sqrt{9+\dfrac{125}{27}}\right)\left(3-\sqrt{9+\dfrac{125}{27}}\right)}\)

=\(\sqrt[3]{-\left[9-\left(9+\dfrac{125}{27}\right)\right]}\)

=\(\sqrt[3]{\dfrac{125}{27}}\)

=5/3

9.

\(\sqrt{20}+2\sqrt{45}+\sqrt{125}-3\sqrt{80}\)

\(=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)

\(=-\sqrt{5}\)

10.

\(\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\)

\(=5\sqrt{3}-\sqrt{5+\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2+\dfrac{2}{3}}+6\sqrt{3}\)

\(=11\sqrt{3}-\sqrt{\dfrac{16}{3}}+\dfrac{9}{2}\sqrt{\dfrac{8}{3}}\)

\(=11\sqrt{3}-\dfrac{4\sqrt{3}}{3}+3\sqrt{6}\)

\(=\dfrac{29\sqrt{3}}{3}+3\sqrt{6}\)

\(\sqrt{20}+2\sqrt{45}+\sqrt{125}-3\sqrt{80}\\ =2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}=\sqrt{5}\)

\(\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\\ =5\sqrt{3}-\dfrac{4\sqrt{3}}{3}+3\sqrt{6}+6\sqrt{3}\\ =\dfrac{15\sqrt{3}-4\sqrt{3}+6\sqrt{6}+18\sqrt{3}}{3}\\ =\dfrac{29\sqrt{3}+6\sqrt{6}}{3}\)

Bài 1 :

a, ĐKXĐ : \(\dfrac{2x+1}{x^2+1}\ge0\)

Mà \(x^2+1\ge1>0\)

\(\Rightarrow2x+1\ge0\)

\(\Rightarrow x\ge-\dfrac{1}{2}\)

Vậy ...

b, Ta có : \(\sqrt[3]{-27}+\sqrt[3]{64}-\sqrt[3]{-\dfrac{128}{2}}\)

\(=-3+4-\left(-4\right)=-3+4+4=5\)

Bài 2 :

\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)

\(=\sqrt{5}\left(2+6+5-12\right)=\sqrt{2}\)

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

\(=2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}-2\)

\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)

\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)

\(=3\)

a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)

b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)

c) \(\Leftrightarrow C=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{4}=\dfrac{3}{2}\)

d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)

\(I=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)

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

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

\(=-5\sqrt{3}.\dfrac{1}{\sqrt{3}}\)

\(=-5\)

\(K=\sqrt{125}-4\sqrt{45}+3\sqrt{20}-\sqrt{80}\)

\(=\sqrt{5^2.5}-4\sqrt{3^2.5}+3\sqrt{2^2.5}-\sqrt{4^2.5}\)

\(=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}\)

\(=\sqrt{5}.\left(5-12+6-4\right)\)

\(=-5\sqrt{5}\)

\(L=2\sqrt{9}+\sqrt{25}-5\sqrt{4}\)

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

\(=2.3+5-5.2\)

\(=1\)

\(N=2\sqrt{32}-5\sqrt{27}-4\sqrt{8}+3\sqrt{75}\)

\(=2.4\sqrt{2}-5.3\sqrt{3}-4.2\sqrt{2}+3.5\sqrt{3}\)

\(=8\sqrt{2}-8\sqrt{2}-15\sqrt{3}+15\sqrt{3}\)

\(=0\)

\(O=2\sqrt{3.5^2}-3\sqrt{3.2^2}+\sqrt{3.3^2}\)

\(=2.5\sqrt{3}-3.2\sqrt{3}+3\sqrt{3}\)

\(=10\sqrt{3}-6\sqrt{3}+3\sqrt{3}\)

\(=7\sqrt{3}\)

\(L=\dfrac{2\sqrt{3}-15\sqrt{3}+8\sqrt{3}}{\sqrt{3}}=2-15+8=-5\)

\(K=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}=-5\sqrt{5}\)

L=2*3+5-5*2=5-4=1

N=8căn 2-8căn2-15căn3+15căn 3=0

O=10căn 3-6căn3+3căn3=7căn 3

a) Ta có: \(P=\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)

\(=\dfrac{x+3+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)

\(=\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

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

=1

Thay x=1 vào P, ta được:

\(P=\dfrac{1+1}{1+3}=\dfrac{2}{4}=\dfrac{1}{2}\)

a) \(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)

=\(\sqrt[3]{16+8\sqrt{5}}+\sqrt[3]{16-8\sqrt{5}}\)

=\(\sqrt[3]{\left(1+\sqrt{5}\right)^3}+\sqrt[3]{\left(1-\sqrt{5}\right)^3}\)

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

b) \(\left(2-\sqrt{3}\right)\sqrt[3]{26+15\sqrt{3}}\)

=\(\left(2-\sqrt{3}\right)\sqrt[3]{\left(2+\sqrt{3}\right)^3}\)

=\(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3=1\)

c) xem lại đề

Bạn Thái làm sai rồi

a)do ban đầu cậu nhân 2 cho hai vế nhưng bạn chưa chia lại.mik bổ sung ý tiếp cho bạn là

2A=2=>A=1.

mik lam tiep cau b la

B=\(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)\)

=4-3

=1.

còn câu c mik pó tay :))