Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(VT=\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{20}-2\right)}{2}\)
\(=\frac{\sqrt{5-2\sqrt{5}+1}\left(3+\sqrt{5}\right)\left(2\sqrt{5}-2\right)}{2}\)
\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)2\left(\sqrt{5}-1\right)}{2}\)
\(=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=18-6\sqrt{5}+6\sqrt{5}-10=8=VP\)
b, \(VT=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5-2\sqrt{5}\sqrt{3}+3}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)
\(=2\left(16-15\right)=2=VP\)
Tính
a) Ta có: \(A=\left(\sqrt{6}+\sqrt{10}\right)-\sqrt{4-\sqrt{15}}\)
\(=\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{4-\sqrt{15}}\)
\(=\sqrt{3}+\sqrt{5}-\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{3}+\sqrt{5}-\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\)
\(=\sqrt{3}+\sqrt{5}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\sqrt{3}+\sqrt{5}-\left|\sqrt{5}-\sqrt{3}\right|\)
\(=\sqrt{3}+\sqrt{5}-\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{3}+\sqrt{5}-\sqrt{5}+\sqrt{3}\)
\(=2\sqrt{3}\)
c) Ta có: \(C=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\left|\sqrt{5}-\sqrt{3}\right|\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\cdot\left(8-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)
\(=2\left[4^2-\left(\sqrt{15}\right)^2\right]\)
\(=2\cdot\left[16-15\right]=2\cdot1=2\)
\(f,\sqrt{\dfrac{3-\sqrt{5}}{2-\sqrt{3}}}\\ =\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}{4-3}}\\ =\sqrt{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}\\ =\sqrt{\dfrac{\left(6-2\sqrt{5}\right)\left(4+2\sqrt{3}\right)}{4}}\\ =\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{3}+1\right)}{2}\)
\(a,\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\\ =\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\\ =\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\\ =\sqrt{9-5}.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\\ =2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\\ =2.4\\ =8\)
câu E dễ nhất nên mình làm trước , các câu còn lại làm tương tự ( biến đổi thành hằng đẳng thức rồi rút gọn ) :
\(E=\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vì \(3-\sqrt{6}>0;2\sqrt{6}-3>0\) )
\(=\sqrt{6}\)
1/Em không chắc nha, nhất là câu c ý, nó sai sai hay là em làm sai nhỉ?
a) ĐK \(x\ge0\). Bình phương hai vế:
\(x+5=x+2\sqrt{x}+1\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4\) (TMĐK)
b)ĐK \(0\le x\le1\) . Bình phương hai vế:
\(2\sqrt{x\left(1-x\right)}=0\Leftrightarrow x\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\left(TMĐK\right)\)
c) ĐK: \(\left\{{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\Leftrightarrow5\le x\le3\) (vô lí))
Vậy không tồn tại x thỏa mãn đề bài.
\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(A=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(A=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\)
\(B=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(B=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(B=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)\)
\(B=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)=2\)
ai giúp vs