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2]\(\sqrt{3}\)+1+\(\sqrt{4-4\sqrt{3}+3}\)=\(\sqrt{3}+1+\sqrt{\left(2-\sqrt{3}\right)^2}=\sqrt{3}+1+2-\sqrt{3}=3\)
4\(\left(\dfrac{\sqrt{3}.\left(2+\sqrt{3}\right)+2.\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}\right)=\dfrac{\sqrt{3}.\left(2+\sqrt{3}\right)+2.\left(2-\sqrt{3}\right)}{1}\)
1: \(=2\sqrt{7}-12\sqrt{7}+15\sqrt{7}+27\sqrt{7}=32\sqrt{7}\)
3: \(=\sqrt{5}-2-\sqrt{14+6\sqrt{5}}\)
\(=\sqrt{5}-2-3-\sqrt{5}=-5\)
4: \(=2\sqrt{3}+3+4-2\sqrt{3}=7\)
5: \(=3-\sqrt{2}+3+\sqrt{2}+4-3=7\)
6: \(=\sqrt{\dfrac{6+2\sqrt{5}}{4}}+\sqrt{\dfrac{14-6\sqrt{5}}{4}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}}{2}=\dfrac{4}{2}=2\)
8: \(=\sqrt{5}-1+\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{4}}-\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{4}}\)
\(=\sqrt{5}-1+\dfrac{3-\sqrt{5}}{2}-\dfrac{3+\sqrt{5}}{2}\)
\(=\dfrac{2\sqrt{5}-2+3-\sqrt{5}-3-\sqrt{5}}{2}=\dfrac{-2}{2}=-1\)
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
♡
\(\dfrac{2}{1-\sqrt{2}}-\dfrac{2}{1+\sqrt{2}}\)
\(=\dfrac{2\left(1+\sqrt{2}\right)-2\left(1-\sqrt{2}\right)}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}\)
\(=\dfrac{2+2\sqrt{2}-2+2\sqrt{2}}{1-2}=-4\sqrt{2}\)
♡
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left[-\dfrac{\sqrt{2}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}-\sqrt{5}\right]\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-3\)
♡
\(\dfrac{2}{7+4\sqrt{3}}+\dfrac{2}{7-4\sqrt{3}}\)
\(=\dfrac{2\left(7-4\sqrt{3}\right)+2\left(7+4\sqrt{3}\right)}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\dfrac{14-8\sqrt{3}+14+8\sqrt{3}}{49-48}\)
= 28
♡
\(\dfrac{2}{\sqrt{5}+1}-\sqrt{\dfrac{2}{3-\sqrt{5}}}\)
\(=\dfrac{2}{\sqrt{5}+1}-\sqrt{\dfrac{4}{6-2\sqrt{5}}}\)
\(=\dfrac{2}{\sqrt{5}+1}-\dfrac{2}{\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\dfrac{2\left(\sqrt{5}-1\right)-2\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\dfrac{2\sqrt{5}-2-2\sqrt{5}-2}{5-1}\)
= - 1
♡
\(\dfrac{4}{1-\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\)
\(=\dfrac{4\left(1+\sqrt{3}\right)}{1-3}-\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)}\)
\(=-2-2\sqrt{3}-\sqrt{3}=-2-3\sqrt{3}\)
♡
\(\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)
\(=\dfrac{2}{4+\sqrt{6+2\sqrt{5}}}\) (nhân [căn 2] vào cả tử và mẫu)
\(=\dfrac{2}{4+\sqrt{\left(\sqrt{5}+1\right)^2}}\)
\(=\dfrac{2}{5+\sqrt{5}}=\dfrac{2\left(5-\sqrt{5}\right)}{25-5}=\dfrac{5-\sqrt{5}}{10}\)
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
\(1.A=\left(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{3+\sqrt{5}}\right).\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\left(\dfrac{3+\sqrt{5}}{9-5}-\dfrac{3-\sqrt{5}}{9-5}\right).\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}=\dfrac{2\sqrt{5}}{4}.\sqrt{5}=\dfrac{5}{2}\) \(2.B=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=\dfrac{\sqrt{2}-1}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{100}-\sqrt{99}}{100-99}=\sqrt{100}-1\)
\(3.C=\sqrt[3]{7+5\sqrt{2}}-\sqrt[3]{5\sqrt{2}-7}=\sqrt[3]{\left(\sqrt{2}\right)^3+3.2.1+3.\sqrt{2}.1+1}-\sqrt[3]{\left(\sqrt{2}\right)^3-3.2.1+3.\sqrt{2}.1-1}=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}=\sqrt{2}+1-\left(\sqrt{2}-1\right)=2\) \(4.Sai-đề\) ???
Sorry và cám ơn bạn.
4.\(\sqrt[3]{9+4\sqrt{5}}\) + \(\sqrt[3]{9-4\sqrt{5}}\)
1,
a,\(4\sqrt{\dfrac{9}{2}}+\sqrt{2}+\sqrt{\dfrac{1}{18}}=4\sqrt{\dfrac{18}{4}}+\sqrt{2}+\sqrt{\dfrac{1}{9.2}}=4\dfrac{\sqrt{18}}{2}+\sqrt{2}+\dfrac{1}{3}\sqrt{\dfrac{1}{2}}=2\sqrt{9.2}+\sqrt{2}+\dfrac{1}{3}\sqrt{\dfrac{2}{4}}=2.3\sqrt{2}+\sqrt{2}+\dfrac{\sqrt{2}}{6}=6\sqrt{2}+\sqrt{2}+\sqrt{2}\dfrac{1}{6}=\dfrac{43}{6}\sqrt{2}\) b,\(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}=4\sqrt{4.5}-3\sqrt{25.5}+5\sqrt{9.5}-15\dfrac{\sqrt{5}}{5}=4.2\sqrt{5}-3.5\sqrt{5}+5.3\sqrt{5}-3\sqrt{5}=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
*) Giải phương trình :
\(\sqrt{4x-8}+5\sqrt{x-2}-\sqrt{9x-18}=20\) ( ĐKXĐ : x \(\ge\) 2 )
\(\Leftrightarrow\sqrt{4\left(x-2\right)}+5\sqrt{x-2}-\sqrt{9\left(x-2\right)}=20\)
\(\Leftrightarrow2\sqrt{x-2}+5\sqrt{x-2}-3\sqrt{x-2}=20\)
\(\Leftrightarrow4\sqrt{x-2}=20\)
\(\Leftrightarrow\sqrt{x-2}=5\)
\(\Leftrightarrow x-2=25\)
\(\Leftrightarrow x=27\) ( thỏa mãn điều kiện )
Vậy phương trình có nghiệm x = 27 .
a: \(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+\dfrac{9}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=60-20\sqrt{18}+\dfrac{45}{2}\sqrt{12}\)
\(=60-60\sqrt{2}+45\sqrt{3}\)
b: \(=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}\)
\(=\dfrac{2\sqrt{5}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}=\dfrac{2\sqrt{5}+3}{9+6\sqrt{2}}\)
Bài 2:
a: \(=\sqrt{5}-2\)
b: \(=2\sqrt{3}+4\sqrt{3}-5\sqrt{3}-9\sqrt{3}=-8\sqrt{3}\)
c: \(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2\sqrt{2}}=\sqrt{16-8}=2\sqrt{2}\)
d: \(=\sqrt{2}+1-2+\sqrt{2}=2\sqrt{2}-1\)
e: \(=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\dfrac{6+2\sqrt{5}}{4}\)
\(=\dfrac{16-3-\sqrt{5}}{2}=\dfrac{13-\sqrt{5}}{2}\)
f: \(=\sqrt{5\sqrt{3+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{5\sqrt{3+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3+5\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3+25-5\sqrt{3}}}\)
\(=\sqrt{5\sqrt{28-5\sqrt{3}}}\)