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a) \(VT=2\sqrt{6}-4\sqrt{2}+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)
\(=2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}\)
\(=-4\sqrt{2}+1+4\sqrt{2}+8\)
\(=1+8\)
\(=9\)
\(\Rightarrow VT=VP\) (đpcm).
b) \(VT=\left(3\sqrt{10}-3\sqrt{2}+\sqrt{50}-\sqrt{10}\right)\sqrt{3-\sqrt{5}}\)
\(=\left(3\sqrt{10}-3\sqrt{2}+5\sqrt{2}-\sqrt{10}\right)\sqrt{3-\sqrt{5}}\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)
\(=\sqrt{\left(2\sqrt{10}+2\sqrt{2}\right)^2\cdot\left(3-\sqrt{5}\right)}\)
\(=\sqrt{\left(40+8\sqrt{20}+8\right)\left(3-\sqrt{5}\right)}\)
\(=\sqrt{\left(48+16\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(=\sqrt{16\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(=\sqrt{16\left(9-5\right)}\)
\(=\sqrt{64}\)
\(=8\)
\(\Rightarrow VT=VP\) (đpcm).
c) \(VT=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{2+\sqrt{5}}\)
\(=2\left(\sqrt{5}+2\right)-\dfrac{2\left(2-\sqrt{5}\right)}{-1}\)
\(=2\sqrt{5}+4+2\left(2-\sqrt{5}\right)\)
\(=2\sqrt{5}+4+4-2\sqrt{5}\)
\(=4+4\)
\(=8\)
\(\Rightarrow VT=VP\) (đpcm).
2: \(=\sqrt{2}-1-\sqrt{2}=-1\)
3: \(=\dfrac{2+\sqrt{3}}{2-\sqrt{3}}-\dfrac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\dfrac{7+4\sqrt{3}-7+4\sqrt{3}}{1}=8\sqrt{3}\)
4: \(=1+\dfrac{2-\sqrt{3}}{2-\sqrt{3}}=1+1=2\)
\(a.\dfrac{\sqrt{7}-5}{2}-\dfrac{6}{\sqrt{7}-2}+\dfrac{1}{3+\sqrt{7}}+\dfrac{3}{5+2\sqrt{7}}=\dfrac{\sqrt{7}-5}{2}+\dfrac{3-\sqrt{7}}{2}+\dfrac{6\sqrt{7}-15}{3}-\dfrac{6\sqrt{7}+12}{3}=-10\)
\(b.\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}=\left(\sqrt{5}+1\right)\left(6-2\sqrt{5}\right)\sqrt{5+2\sqrt{5}+1}=\left(\sqrt{5}+1\right)^2\left(6-2\sqrt{5}\right)=\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)=36-20=16\)
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
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: \(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}\cdot\left(\dfrac{5}{2}\sqrt{2}+12\right)\)
\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\left(5\sqrt{3}+12\sqrt{6}\right)\)
\(=-60-144\sqrt{2}+30\sqrt{2}+144\)
\(=84-114\sqrt{2}\)
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: \(=\left(-\sqrt{5}-\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=-2
b: \(=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}\)
c: \(=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-2-\sqrt{10}+3\sqrt{7}+2\)
\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)
1) \(=2\sqrt{5}-3+5-2\sqrt{5}=2\)
2) \(=\dfrac{2\sqrt{3}-2-2\sqrt{3}-2}{3-1}=\dfrac{-4}{2}=-2\)
3) \(=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}=2\sqrt{2}\)
bạn ơi sao câu 3 lại ra là \(\sqrt{\left(\sqrt{5+\sqrt{2}}\right)^2}\) vậy ạ, bạn giải thích giúp mình được không