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\(\dfrac{1}{\sqrt{11-2\sqrt{30}}}-\dfrac{3}{\sqrt{7-2\sqrt{10}}}+\dfrac{4}{\sqrt{8+4\sqrt{3}}}\)
\(=\dfrac{1}{\sqrt{11-2.\sqrt{6}.\sqrt{5}}}-\dfrac{3}{\sqrt{7-2.\sqrt{5}.\sqrt{2}}}+\dfrac{4}{\sqrt{2\left(4+2\sqrt{3}\right)}}\)
\(=\dfrac{1}{\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}}-\dfrac{3}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)2}}+\dfrac{4}{\sqrt{2\left(\sqrt{3}+1\right)^2}}\)
\(=\dfrac{1}{\sqrt{6}+\sqrt{5}}-\dfrac{3}{\sqrt{5}+\sqrt{2}}+\dfrac{2\sqrt{2}}{\sqrt{3}+1}\)
\(=\dfrac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}-\dfrac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}+\dfrac{2\sqrt{2}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\sqrt{6}-\sqrt{5}+\sqrt{5}-\sqrt{2}+\sqrt{6}-\sqrt{2}=2\sqrt{6}-2\sqrt{2}\)
Câu 1:
\(\sqrt{33-8\sqrt{7}}=\sqrt{33-2\cdot\sqrt{112}}\)
Câu 2:
\(\Leftrightarrow2\sqrt{x}-3\sqrt{x}+8\sqrt{x}=18\)
\(\Leftrightarrow7\sqrt{x}=18\)
=>căn x=18/7
hay x=324/49
bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không
a) \(\dfrac{30\left(\sqrt{6}-1\right)}{6-1}+\dfrac{2\left(\sqrt{6}+2\right)}{6-4}-\dfrac{6\left(3+\sqrt{6}\right)}{9-6}\)
= \(6\sqrt{6}-6+\sqrt{6}+2-6-2\sqrt{6}=5\sqrt{6}+2\)
b) \(\dfrac{\sqrt{6+2\sqrt{5}}}{2}-\dfrac{\sqrt{5}-1}{2}=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{2}=1\)
c: \(=-\sqrt{3}+\dfrac{4+2\sqrt{3}}{2}=\dfrac{-2\sqrt{3}+4+2\sqrt{3}}{2}=\dfrac{4}{2}=2\)
a: \(=6\left(\sqrt{6}-1\right)+\sqrt{6}+2-6-2\sqrt{6}\)
\(=6\sqrt{6}-6-\sqrt{6}-4=5\sqrt{6}-10\)
b:\(=\dfrac{\sqrt{6+2\sqrt{5}}}{2}-\dfrac{\sqrt{5}-1}{2}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{2}=\dfrac{2}{2}=1\)
3.
Ta có: \(VT=\)\(8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}\)
\(=8+8+\left(2\sqrt{10+2\sqrt{5}}-2\sqrt{10+2\sqrt{5}}\right)\)
\(=16\ne VP\)
⇒ Đề sai
1. Ta có: \(\sqrt{4x}\)- 3\(\sqrt{x}\)+2\(\sqrt{15x}\)=18
⇌2\(\sqrt{x}\)-3\(\sqrt{x}\) +2\(\sqrt{15x}\)=18
⇌\(-\sqrt{x}\) +2\(\sqrt{15x}\)-15 = 3
⇌-(\(\sqrt{x}\) -2\(\sqrt{15x}\)+15 )=3
⇌(\(\sqrt{x}\)-\(\sqrt{15}\))=-3 (vô lí)
Vậy không tìm được giá trị x thỏa mãn bài toán
2.Ta có: B=\(\dfrac{1}{\sqrt{11-2\sqrt{30}}}-\dfrac{3}{7-2\sqrt{10}}\)
= \(\dfrac{1}{\sqrt{6-2\sqrt{6.5}+5}}-\dfrac{3}{2-2\sqrt{2.5}+5}\)
=\(\dfrac{1}{\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}}-\dfrac{3}{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
=\(\dfrac{1}{\sqrt{6}-\sqrt{5}}-\dfrac{3}{\sqrt{3}-\sqrt{2}}\)
hình như đề sai
\(A=\dfrac{\sqrt{3}-3}{\sqrt{2-\sqrt{3}}+2\sqrt{2}}+\dfrac{\sqrt{3}+3}{\sqrt{2+\sqrt{3}}-2\sqrt{2}}\)
\(A=\dfrac{\sqrt{2}\left(\sqrt{3}-3\right)}{\sqrt{2}.\left(\sqrt{2-\sqrt{3}}+2\sqrt{2}\right)}+\dfrac{\sqrt{2}.\left(\sqrt{3}+3\right)}{\sqrt{2}.\left(\sqrt{2+\sqrt{3}}-2\sqrt{2}\right)}\)
\(A=\dfrac{\sqrt{6}-3\sqrt{2}}{\sqrt{4-2\sqrt{3}}+4}+\dfrac{\sqrt{6}+3\sqrt{2}}{\sqrt{4+2\sqrt{3}}-4}\)
\(A=\dfrac{\sqrt{6}-3\sqrt{2}}{\sqrt{\left(\sqrt{3}-1\right)^2}+4}+\dfrac{\sqrt{6}+3\sqrt{2}}{\sqrt{\left(\sqrt{3}+1\right)^2}-4}\)
\(A=\dfrac{\sqrt{6}-3\sqrt{2}}{\sqrt{3}-1+4}+\dfrac{\sqrt{6}+3\sqrt{2}}{\sqrt{3}+1-4}\)
\(A=\dfrac{\sqrt{3}\left(\sqrt{2}-\sqrt{6}\right)}{\sqrt{3}\left(1+\sqrt{3}\right)}+\dfrac{\sqrt{3}\left(\sqrt{2}+\sqrt{6}\right)}{\sqrt{3}\left(1-\sqrt{3}\right)}\)
\(A=\dfrac{\sqrt{2}-\sqrt{6}}{1+\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{6}}{1-\sqrt{3}}=\dfrac{\left(\sqrt{2}-\sqrt{6}\right)\left(1-\sqrt{3}\right)+\left(\sqrt{2}+\sqrt{6}\right)\left(1+\sqrt{3}\right)}{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}\)
\(A=\dfrac{\sqrt{2}-\sqrt{6}-\sqrt{6}+3\sqrt{2}+\sqrt{2}+\sqrt{6}+\sqrt{6}+3\sqrt{2}}{1-3}=\dfrac{8\sqrt{2}}{-2}=-4\sqrt{2}\)
* \(B=\dfrac{\sqrt{11+2\sqrt{30}}-\sqrt{11-2\sqrt{30}}}{\sqrt{5}}\) \(=\dfrac{\sqrt{6+2.\sqrt{6}.\sqrt{5}+5}-\sqrt{6-2.\sqrt{6}.\sqrt{5}+5}}{\sqrt{5}}\)\(=\dfrac{\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}}{\sqrt{5}}\)
\(=\dfrac{\sqrt{6}+\sqrt{5}-\sqrt{6}+\sqrt{5}}{\sqrt{5}}=\dfrac{2\sqrt{5}}{\sqrt{5}}=2\)
* \(C=2\sqrt{3+\sqrt{5}}-\left(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\right)\)
Đặt:\(y=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\Rightarrow y^2=4+\sqrt{15}+4-\sqrt{15}+2\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}=8+2=10\Rightarrow y=\sqrt{10}\)
Suy ra: \(C=\sqrt{12+4\sqrt{5}}-y=\sqrt{\left(\sqrt{10}+\sqrt{2}\right)^2}-\sqrt{10}=\sqrt{10}+\sqrt{2}-\sqrt{10}=\sqrt{2}\)* \(D=\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\dfrac{\left(\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}\right)+\left(\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2-\sqrt{3}}\right)}{\left(\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}\right)}=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{1}=4\)
a: \(=\dfrac{2+\sqrt{3}}{2-\sqrt{3}}-\dfrac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\dfrac{7+4\sqrt{3}-7+4\sqrt{3}}{1}=8\sqrt{3}\)
b: \(=\sqrt{2}-1-\sqrt{2}=-1\)
\(=\dfrac{1}{\sqrt{11-2\sqrt{5}.\sqrt{6}}}-\dfrac{3\left(7+2\sqrt{10}\right)}{\left(7-2\sqrt{10}\right)\left(7+2\sqrt{10}\right)}\\ =\dfrac{1}{\sqrt{\left(\sqrt{5}-\sqrt{6}\right)^2}}-\dfrac{3\left(7+2\sqrt{10}\right)}{49-40}\\ =\dfrac{1}{\left|\sqrt{5}-\sqrt{6}\right|}-\dfrac{7+2\sqrt{10}}{3}\\ =\dfrac{1}{\sqrt{6}-\sqrt{5}}-\dfrac{7+2\sqrt{10}}{3}\\ =\dfrac{\sqrt{6}+\sqrt{5}}{6-5}-\dfrac{7+2\sqrt{10}}{3}\\ =\sqrt{6}+\sqrt{5}+\dfrac{7+2\sqrt{10}}{3}\\ =\dfrac{3\sqrt{6}+3\sqrt{5}+7+2\sqrt{10}}{3}\)
\(=\dfrac{1}{\sqrt{6}-\sqrt{5}}+\dfrac{7+2\sqrt{10}}{3}\)
\(=\sqrt{6}+\sqrt{5}+\dfrac{7}{3}+\dfrac{2}{3}\sqrt{10}\)