HT
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 9 2019
\(D=\sqrt{5}-\sqrt{13-4\sqrt{\left(\sqrt{5}-2\right)^2}}=\sqrt{5}-\sqrt{13-4\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}-\sqrt{21-4\sqrt{5}}=\sqrt{5}-\sqrt{\left(2\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-2\sqrt{5}+1=1-\sqrt{5}\)
\(B=10\sqrt{5}+\left|1-\sqrt{5}\right|-\frac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=10\sqrt{5}+\sqrt{5}-1-\sqrt{5}+1=10\sqrt{5}\)
\(C=\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{2+\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{12\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\sqrt{3}-1+2+\sqrt{3}+2\left(3-\sqrt{3}\right)=7\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 5 2019
Đặt \(x=\sqrt[4]{5}\Rightarrow x^4=5\Rightarrow x^4-5=0\)
\(A=\frac{2}{\sqrt{4-3x+2x^2-x^3}}=\frac{2\left(x+1\right)}{\sqrt{\left(x+1\right)^2\left(4-3x+2x^2-x^3\right)}}\)
\(=\frac{2\left(x+1\right)}{\sqrt{4+5x-x^5}}=\frac{2\left(x+1\right)}{\sqrt{4+x\left(5-x^4\right)}}=x+1=\sqrt[4]{5}+1\)
\(B=\left(\frac{-\sqrt[4]{2}\left(1-\sqrt[4]{2}\right)}{1-\sqrt[4]{2}}+\frac{1+\sqrt{2}}{\sqrt[4]{2}}\right)^2-\frac{\sqrt{1+\sqrt{2}+\frac{1}{2}}}{1+\sqrt{2}}\)
\(=\left(-\sqrt[4]{2}+\frac{1}{\sqrt[4]{2}}+\sqrt[4]{2}\right)^2-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{2}\left(\sqrt{2}+1\right)}\)
\(=\frac{1}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{2}\left(\sqrt{2}+1\right)}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0\)
YY
0
LB
0
TN
1
18 tháng 9 2019
d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)
\(\Leftrightarrow x^3=6-5x\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow x=1\)
19 tháng 9 2019
c/
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=3-1=2\)