Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 2:
a) \(2\sqrt{125}+\dfrac{3}{2}\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\)
\(=2\sqrt{5^2\cdot5}+\dfrac{3}{2}\sqrt{4^2\cdot5}-\sqrt{6^2\cdot5}-\dfrac{2}{7}\sqrt{7^2\cdot5}\)
\(=10\sqrt{5}+\dfrac{3\cdot4}{2}\sqrt{5}-6\sqrt{5}-\dfrac{2\cdot7}{7}\sqrt{5}\)
\(=10\sqrt{5}+6\sqrt{6}-6\sqrt{5}-2\sqrt{5}\)
\(=8\sqrt{5}\)
b) \(\sqrt{11-4\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}\right)^2-2\cdot2\cdot\sqrt{7}+2^2}-\sqrt{\left(\sqrt{7}\right)^2+2\cdot3\cdot\sqrt{7}+3^2}\)
\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\sqrt{\left(\sqrt{7}+3\right)^2}\)
\(=\sqrt{7}-2-\sqrt{7}-3\)
\(=-5\)
\(2a,\\ 2\sqrt{125}+\dfrac{3}{2}.\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\\ =2\sqrt{5^2.5}+\dfrac{3}{2}.\sqrt{4^2.5}-\sqrt{6^2.5}-\dfrac{2}{7}.\sqrt{7^2.5}\\ =2.5.\sqrt{5}+\dfrac{3}{2}.4.\sqrt{5}-6\sqrt{5}-\dfrac{2}{7}.7\sqrt{5}\\ =10\sqrt{5}+6\sqrt{5}-6\sqrt{5}-2\sqrt{5}=8\sqrt{5}\)
\(R=\dfrac{2}{\sqrt{3-\sqrt{5}-\left(\sqrt[4]{5}-1\right)^3}}=\dfrac{2}{\sqrt{\dfrac{\left(\sqrt{5}-1\right)^2}{2}-\left(\sqrt[4]{5}-1\right)^3}}\)
\(=\dfrac{2}{\sqrt{\dfrac{\left(\sqrt[4]{5}-1\right)^2\left(\sqrt[4]{5}+1\right)^2}{2}-\left(\sqrt[4]{5}-1\right)^3}}\)
\(=\dfrac{2\sqrt{2}}{\sqrt{\left(\sqrt[4]{5}-1\right)^2\left[\left(\sqrt[4]{5}+1\right)^2-2\left(\sqrt[4]{5}-1\right)\right]}}\)
\(=\dfrac{2\sqrt{2}}{\left(\sqrt[4]{5}-1\right)\sqrt{\sqrt{5}+2\sqrt[4]{5}+1-2\sqrt[4]{5}+2}}\)
\(=\dfrac{2\sqrt{2}}{\left(\sqrt[4]{5}-1\right)\sqrt{3+\sqrt{5}}}=\dfrac{2\sqrt{2}}{\left(\sqrt[4]{5}-1\right)\sqrt{\dfrac{\left(\sqrt{5}+1\right)^2}{2}}}\)
\(=\dfrac{4}{\left(\sqrt[4]{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{4\left(\sqrt[4]{5}+1\right)}{\left(\sqrt[4]{5}+1\right)\left(\sqrt[4]{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(\)\(=\dfrac{4\left(\sqrt[4]{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{4\left(\sqrt[4]{5}+1\right)}{4}\)
\(=\sqrt[4]{5}+1\)
T = \(\dfrac{\sqrt{5}\left(\sqrt{16}-\sqrt{9}\right)}{4-5}-5\sqrt{5}+\dfrac{1}{\sqrt{5}-2}+2\sqrt{5}\)
= \(-\sqrt{5}-5\sqrt{5}+2\sqrt{5}+\dfrac{1}{\sqrt{5}-2}\)
= \(-4\sqrt{5}+\dfrac{1}{\sqrt{5}-2}\)
= \(\dfrac{-4\sqrt{5}\left(\sqrt{5}-2\right)+1}{\sqrt{5}-2}\)
= \(\dfrac{-20+8\sqrt{5}+1}{\sqrt{5}-2}\)
= \(\dfrac{-19+8\sqrt{5}}{\sqrt{5}-2}\)
= \(\dfrac{19-8\sqrt{5}}{2-\sqrt{5}}\)
= \(\dfrac{\left(-2+3\sqrt{5}\right)\left(\sqrt{5}-2\right)}{-\left(\sqrt{5}-2\right)}=2-3\sqrt{5}\)
Đặt \(x=\sqrt[4]{5}\Rightarrow x^4=5\Rightarrow x^4-5=0\)
\(A=\frac{2}{\sqrt{4-3x+2x^2-x^3}}=\frac{2\left(x+1\right)}{\sqrt{\left(x+1\right)^2\left(4-3x+2x^2-x^3\right)}}\)
\(=\frac{2\left(x+1\right)}{\sqrt{4+5x-x^5}}=\frac{2\left(x+1\right)}{\sqrt{4+x\left(5-x^4\right)}}=x+1=\sqrt[4]{5}+1\)
\(B=\left(\frac{-\sqrt[4]{2}\left(1-\sqrt[4]{2}\right)}{1-\sqrt[4]{2}}+\frac{1+\sqrt{2}}{\sqrt[4]{2}}\right)^2-\frac{\sqrt{1+\sqrt{2}+\frac{1}{2}}}{1+\sqrt{2}}\)
\(=\left(-\sqrt[4]{2}+\frac{1}{\sqrt[4]{2}}+\sqrt[4]{2}\right)^2-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{2}\left(\sqrt{2}+1\right)}\)
\(=\frac{1}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{2}\left(\sqrt{2}+1\right)}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0\)
Ta có:
\(R=\)\(\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\)\(\dfrac{\sqrt{10}+3\sqrt{2}}{5+\sqrt{5}}+\dfrac{\sqrt{10}-3\sqrt{2}}{5-\sqrt{5}}\)
\(=\dfrac{4\sqrt{2}}{\sqrt{5}\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\dfrac{4\sqrt{2}}{4\sqrt{5}}=\sqrt{\dfrac{2}{5}}\)
Làm câu S tương tự như này rồi đối chiếu kết quả nha
\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-....-\frac{1}{\sqrt{24}-\sqrt{25}}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}-...-\frac{\sqrt{24}+\sqrt{25}}{(\sqrt{24}-\sqrt{25})(\sqrt{24}+\sqrt{25})}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{-1}-\frac{\sqrt{2}+\sqrt{3}}{-1}+\frac{\sqrt{3}+\sqrt{4}}{-1}-...-\frac{\sqrt{24}+\sqrt{25}}{-1}\)
\(=\frac{(1+\sqrt{2})-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-...-(\sqrt{24}+\sqrt{25})}{-1}\)
\(=\frac{1-\sqrt{25}}{-1}=4\)
\(B=\frac{5}{4+\sqrt{11}}+\frac{11-3\sqrt{11}}{\sqrt{11}-3}-\frac{4}{\sqrt{5}-1}+\sqrt{(\sqrt{5}-2)^2}\)
\(=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{\sqrt{11}(\sqrt{11}-3)}{\sqrt{11}-3}-\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\sqrt{5}-2\)
\(=\frac{5(4-\sqrt{11})}{5}+\sqrt{11}-\frac{4(\sqrt{5}+1)}{4}+\sqrt{5}-2\)
\(=4-\sqrt{11}+\sqrt{11}-(\sqrt{5}+1)+\sqrt{5}-2\)
\(=1\)
Bài này nằm trong cuốn nâng cao và phát triển của Vũ Hữu Bình, và lời giải của nó thực sự rất "ảo". Có lẽ trừ tác giả ra, khó ai mà nghĩ được ra cách giải: