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a: \(=\left(2\sqrt{2}-5\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=\left(-3\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=\left(-3\sqrt{10}+10\right)\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=-9-30\sqrt{10}+3\sqrt{10}+100=91-27\sqrt{10}\)
b: \(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}\cdot\left(\dfrac{5}{2}\sqrt{2}+12\right)\)
\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\left(5\sqrt{3}+12\sqrt{6}\right)\)
\(=-60-144\sqrt{2}+30\sqrt{2}+144\)
\(=84-114\sqrt{2}\)
a) \(\Leftrightarrow\left(\sqrt{3+\sqrt{5}}\right)^2\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
\(\Leftrightarrow\sqrt{3+\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\sqrt{\left(3-\sqrt{5}\right)\cdot\left(3+\sqrt{5}\right)}=8\)
\(\Leftrightarrow\sqrt{\frac{6+2\sqrt{5}}{2}}\cdot\left(\sqrt{5}\sqrt{2}-\sqrt{2}\right)\sqrt{3^2-5}=8\).
\(\Leftrightarrow\sqrt{\frac{5+2\sqrt{5}+1}{2}}\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{4}=8\)
\(\Leftrightarrow\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\cdot2=8\)
\(\Leftrightarrow\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=4\Leftrightarrow\left(\sqrt{5}\right)^2-1=4\Leftrightarrow5-1=4\)Đúng -ĐPCM.
Biến đổi vế trái
\(\left(3+\sqrt{5}\right).\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3-\sqrt{5}}\)=\(\left(\sqrt{3+\sqrt{5}}\right)^2.\sqrt{3-\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right)\)
=\(\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{4}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\sqrt{10\left(3+\sqrt{5}\right)}-2\sqrt{2\left(3+\sqrt{5}\right)}\)
\(=2\sqrt{30+10\sqrt{5}}-2\sqrt{6+2\sqrt{5}}\)
\(=2\sqrt{\left(5+\sqrt{5}\right)^2}-2\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=2\left(5+\sqrt{5}\right)-2\left(\sqrt{5}+1\right)\)
\(=10+2\sqrt{5}-2\sqrt{5}-2=8\)
Sau khi biến đổi ta thấy vế trái bằng vế phải. Vậy đẳng thức đã được chứng minh
a. Sửa đề: \(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
biến đổi vế trái :
ta có :\(\left(3+\sqrt{5}\right)\left(\sqrt{10}+\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)
=\(\sqrt{3+\sqrt{5}}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3-\sqrt{5}}\)
=\(\sqrt{3^2-\left(\sqrt{5}\right)^2}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right)\)
=2(\(\sqrt{30+10\sqrt{5}}-\sqrt{6+2\sqrt{5}}\))
=2(\(\sqrt{5}+5-\sqrt{5}-1\))
=2.4=8=VP
=> đpcm
b. Đặt vế trái là A
ta có \(A^2=\sqrt{2}+1-2\sqrt{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\sqrt{2}-1\)
=\(2\sqrt{2}-2\)
=2\(\left(\sqrt{2}-1\right)\)
=> A=\(\sqrt{2\left(\sqrt{2}-1\right)}\)
vậy VT=VP =>đpcm
\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\cdot\sqrt{3-\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{2}\cdot\sqrt{3-\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\cdot\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\left(3+\sqrt{5}\right)\cdot\left(\sqrt{5}-1\right)^2\)
\(=\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)\)
\(=2\cdot\left(3+\sqrt{5}\right)\cdot\left(3-\sqrt{5}\right)\)
\(=2\cdot\left(9-5\right)\)
\(=2-4=8\)
\(\left(\sqrt{8}-5\sqrt{2}+\sqrt{20}\right)\sqrt{5}-\left(3\sqrt{\frac{1}{10}}+10\right)=\left(2\sqrt{2}-5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\frac{3\sqrt{10}}{10}-10\)
\(=-3\sqrt{10}+10-\frac{3\sqrt{10}}{10}-10=-3\sqrt{10}-\frac{3\sqrt{10}}{10}=-3\sqrt{10}\left(1+\frac{1}{10}\right)=\frac{-33\sqrt{10}}{10}=-3,3\sqrt{10}\)
\(a,\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
Ta có
:\(VT=\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)
\(=|2-\sqrt{5}|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}\)
\(=-2=VP\left(đpcm\right)\)
\(b,\frac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}\)
Ta có:
\(VT=\frac{\sqrt{2}+1}{\sqrt{2}-1}\)
\(=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(=\frac{2+\sqrt{2}+\sqrt{2}+1}{\sqrt{2}^2-1^2}\)
\(=\frac{3+2\sqrt{2}}{2-1}\)
\(=3+2\sqrt{2}=VP\left(đpcm\right)\)
c,Bạn xem lại đề
\(d,\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}=8\)
Ta có:
\(VT=\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)
\(=\sqrt{\frac{2^2}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{2^2}{\left(2+\sqrt{5}\right)^2}}\)
\(=\frac{2}{|2-\sqrt{5}|}-\frac{2}{|2+\sqrt{5}|}\)
\(=\frac{2\left(2+\sqrt{5}\right)}{\left(\sqrt{5}-2\right)\left(2+\sqrt{5}\right)}-\frac{2\left(\sqrt{5}-2\right)}{\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)}\)
\(=\frac{4+2\sqrt{5}-2\sqrt{5}+4}{\sqrt{5}^2-2^2}\)
\(=\frac{8}{5-4}\)
\(=8=VP\left(đpcm\right)\)
\(\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}=8\)
\(\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)^2=8\)
\(\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)=8\)
\(18-6\sqrt{5}+6\sqrt{5}-10=8\)
8=8 ( luôn đúng )