Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(\Rightarrow\)\(\sqrt{2}A=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
\(\Rightarrow\)\(A=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
b) bn lm tương tự
a, \(A=\left(\sqrt{12}-2\sqrt{5}\right)\sqrt{3}+\sqrt{60}\)
\(=\left(2\sqrt{3}-2\sqrt{5}\right)\sqrt{3}+2\sqrt{15}\)
\(=2\sqrt{9}-2\sqrt{15}+2\sqrt{15}=2\sqrt{9}\)
b, \(B=\frac{\sqrt{4x}}{x-3}\sqrt{\frac{x^2-6x+9}{x}}=\frac{2\sqrt{x}}{x-3}.\sqrt{\frac{\left(x-3\right)^2}{x}}\)
\(=\frac{2\sqrt{x}}{x-3}.\frac{x-3}{\sqrt{x}}=2\)
Ta có: \(B=21\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)^2-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)^2-15\sqrt{15}\)
\(=21\cdot\left[2+\sqrt{3}+3-\sqrt{5}+2\sqrt{\left(2+\sqrt{3}\right)\left(3-\sqrt{5}\right)}\right]-6\cdot\left[2-\sqrt{3}+3+\sqrt{5}+2\cdot\sqrt{\left(2-\sqrt{3}\right)\left(3+\sqrt{5}\right)}\right]-15\sqrt{15}\)
\(=21\cdot\left(5+\sqrt{3}-\sqrt{5}+\sqrt{\left(4+2\sqrt{3}\right)\left(6-2\sqrt{5}\right)}\right)-6\cdot\left[5-\sqrt{3}+\sqrt{5}+\sqrt{\left(4-2\sqrt{3}\right)\left(6+2\sqrt{5}\right)}\right]-15\sqrt{15}\)
\(=21\cdot\left[5+\sqrt{3}-\sqrt{5}+\left(\sqrt{3}+1\right)\left(\sqrt{5}-1\right)\right]-6\cdot\left[5-\sqrt{3}+\sqrt{5}+\left(\sqrt{3}-1\right)\left(\sqrt{5}+1\right)\right]-15\sqrt{15}\)
\(=21\cdot\left(5+\sqrt{3}-\sqrt{5}+\sqrt{15}-\sqrt{3}+\sqrt{5}-1\right)-6\cdot\left(5-\sqrt{3}+\sqrt{5}+\sqrt{15}+\sqrt{3}-\sqrt{5}-1\right)-15\sqrt{15}\)
\(=21\cdot\left(4+\sqrt{15}\right)-6\left(4+\sqrt{15}\right)-15\sqrt{15}\)
\(=84+21\sqrt{15}-24-6\sqrt{15}-15\sqrt{15}\)
\(=60\)
a, \(\sqrt{\left(\sqrt{5}-4\right)^2}-\sqrt{5}+\sqrt{20}=4\)
\(VT=\sqrt{\left(4-\sqrt{5}\right)^2}-\sqrt{5}+\sqrt{20}=\left|4-\sqrt{5}\right|-\sqrt{5}+\sqrt{20}\)
\(=4-\sqrt{5}-\sqrt{5}+2\sqrt{5}=4\) hay \(VT=VP\)
Vậy ta có đpcm
b, Với \(x>0,x\ne4\)
\(P=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}\right):\frac{2}{x-2\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\frac{2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{2}=\frac{x}{\sqrt{x}+2}\)
1.
Giả sử điều trên là đúng ta có:
\( \left | \sqrt{5}-4 \right |-\sqrt{5}+\sqrt{20}=4\)
Ta có: \(4>\sqrt{5}\)
\(\Rightarrow 4-\sqrt{5}- \sqrt{5}+\sqrt{20}=4\)
\(\Leftrightarrow 4-\sqrt{20}+\sqrt{20}=4\)
\(\Rightarrow đpcm\)
2.
1/ \(A=\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}\) (Vì \(\sqrt{5}-\sqrt{3}>0\))
\(B=\sqrt{6+2\sqrt{5}}-\sqrt{13}+\sqrt{48}=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{13}+4\sqrt{3}=\left|\sqrt{5}+1\right|-\sqrt{13}+4\sqrt{3}=\sqrt{5}+1+\sqrt{13}+4\sqrt{5}\)
2/Ta có :
\(\left(\frac{3\sqrt{2}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right).\frac{1}{\sqrt{6}}\)
\(=\left(\frac{3\sqrt{2}}{3\sqrt{3}-3}-\frac{5\sqrt{6}}{3}\right).\frac{1}{\sqrt{6}}\)
\(=\left(\frac{3\sqrt{2}}{3\left(\sqrt{3}-1\right)}-\frac{5\sqrt{6}\left(\sqrt{3}-1\right)}{3\left(\sqrt{3}-1\right)}\right).\frac{1}{\sqrt{6}}\)
\(=\frac{3\sqrt{2}-15\sqrt{2}+5\sqrt{6}}{3\left(\sqrt{3}-1\right)}.\frac{1}{\sqrt{6}}\)
\(=\frac{-12\sqrt{2}+5\sqrt{6}}{3\left(\sqrt{3}-1\right)}.\frac{1}{\sqrt{6}}\)
\(=\frac{-7+\sqrt{3}}{6}\)
Vậy...
Bài 1:
Ta có: \(A=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5+2\cdot\sqrt{5}\cdot\sqrt{3}+3}+\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}-2\cdot\sqrt{5-2\cdot\sqrt{5}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-2\cdot\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\left|\sqrt{5}+\sqrt{3}\right|+\left|\sqrt{5}-\sqrt{3}\right|-2\cdot\left|\sqrt{5}-1\right|\)
\(=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2\)
=2
Vậy: A=2
Bài 2: Sửa đề: Chứng minh \(\left(\frac{3\sqrt{2}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right)\cdot\frac{1}{\sqrt{6}}=\frac{-7+\sqrt{3}}{6}\)
Ta có: \(\left(\frac{3\sqrt{2}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{9\sqrt{2}}{3\left(\sqrt{27}-3\right)}-\frac{\sqrt{150}\left(\sqrt{27}-3\right)}{3\cdot\left(\sqrt{27}-3\right)}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{9\sqrt{2}-45\sqrt{2}+3\sqrt{150}}{9\left(\sqrt{3}-1\right)}\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{-36\sqrt{2}+3\sqrt{150}}{9\sqrt{6}\cdot\left(\sqrt{3}-1\right)}\)
\(=\frac{\sqrt{54}\cdot\left(5-4\sqrt{3}\right)}{\sqrt{486}\cdot\left(\sqrt{3}-1\right)}\)
\(=\frac{5-4\sqrt{3}}{3\sqrt{3}-3}\)
\(=\frac{-7+\sqrt{3}}{6}\)(đpcm)
a) \(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)
b) \(\frac{1}{2\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}=\frac{2\sqrt{3}}{12}+\frac{2\sqrt{3}}{6}-\frac{6-2\sqrt{3}}{6}\)
\(=\frac{2\sqrt{3}}{12}+\frac{4\sqrt{3}}{12}-\frac{12-4\sqrt{3}}{12}=\frac{-12+10\sqrt{3}}{12}=\frac{-6+5\sqrt{3}}{6}\)
a)
\(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{8}-4\right)^2}=3-2\sqrt{2}-4+\sqrt{8}\)
\(=3-2\sqrt{2}-4+2\sqrt{2}=3-4=-1\)
b)
\(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}=\frac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{3-1}\)
\(=\frac{2\left(\sqrt{3}+1-\sqrt{3}+1\right)}{2}=\sqrt{3}+1-\sqrt{3}+1=1+1=2\)
\(\sqrt{\frac{3\sqrt{5}+1}{2\sqrt{5}-3}}\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}\)
?? :v
\(\sqrt{\frac{3\sqrt{5}+1}{2\sqrt{5}-3}}\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{\frac{\left(3\sqrt{5}+1\right)\left(2\sqrt{5}+3\right)}{\left(2\sqrt{5}-3\right)\left(2\sqrt{5}+3\right)}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{3+\sqrt{5}}.\sqrt{2}.\left(\sqrt{5}-1\right)\)
\(=\sqrt{6+2\sqrt{5}}.\left(\sqrt{5}-1\right)\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=4\)
1: \(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
2: \(=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=-\sqrt{2}\)