Sửa đề: căn 9-4căn 5
\(VT=\sqrt{5}-2-\sqrt{5}+5=3\)=VP
Đúng 0
Bình luận (0)
Các câu hỏi tương tự
Rút gọn bt a,Afrac{left(5+2sqrt{6}right)left(49-20sqrt{6}right)left(sqrt{5-2sqrt{6}}right)}{9sqrt{3}-11sqrt{2}}b,Csqrt{4+sqrt{5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}}c,left(sqrt{3}-1right)sqrt{6+2sqrt{2}sqrt{3-sqrt{sqrt{2}+sqrt{12}+sqrt{18-sqrt{128}}}}}d,sqrt[3]{3+sqrt{9+frac{125}{27}}}-sqrt[3]{-3+sqrt{9+frac{125}{27}}}
Đọc tiếp
Rút gọn bt
\(a,A=\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\)
\(b,C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(c,\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(d,\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
a)\(\left(\sqrt{5}+2\right).\left(17-4\sqrt{9+4\sqrt{5}}\right)?\)
b)\(\left(\sqrt{3-1}\right).\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
c) \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}\)
d) \(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
\(\left(6\right)\dfrac{3\sqrt{x}}{5\sqrt{x}-1}\le-3\)
\(\left(7\right)\dfrac{8\sqrt{x}+8}{6\sqrt{x}+9}>\dfrac{8}{3}\)
\(\left(8\right)\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}< -4\)
\(\left(9\right)\dfrac{4\sqrt{x}+6}{5\sqrt{x}+7}\le-\dfrac{2}{3}\)
\(\left(10\right)\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}>-6\)
Chứng minha,sqrt{9-4sqrt{5}}-sqrt{5}-2b,frac{sqrt{2}+1}{sqrt{2}-1}3+2sqrt{2}c,2sqrt{2}left(3-sqrt{2}right)+left(1+2sqrt{2}right)^2-2sqrt{6}9d,sqrt{frac{4}{left(2-sqrt{5}right)^2}}-sqrt{frac{4}{left(2+sqrt{5}right)^2}}8e,left(3+sqrt{5}right)left(sqrt{10}-sqrt{2}right)sqrt{3-sqrt{5}}8f,sqrt{sqrt{2}+1}-sqrt{sqrt{2}-1}sqrt{2left(sqrt{2}-1right)}
Đọc tiếp
Chứng minh
\(a,\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(b,\frac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}\)
\(c,2\sqrt{2}\left(3-\sqrt{2}\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)
\(d,\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}=8\)
\(e,\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
\(f,\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
a) \(\left(2+\sqrt{3}\right)\cdot\sqrt{7-4\sqrt{3}}\)
b) \(\sqrt{4+2\sqrt{3}}-\sqrt{5+2\sqrt{6}}+\sqrt{2}+\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
c) \(\sqrt{3-\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
Chứng minha,sqrt{9-4sqrt{5}}-sqrt{5}-2b,frac{sqrt{2}+1}{sqrt{2}-1}3+2sqrt{2}c,2sqrt{2}left(sqrt{3}-2right)+left(1+2sqrt{2}right)^2-2sqrt{6}9d,sqrt{frac{4}{left(2-sqrt{5}right)^2}}-sqrt{frac{4}{left(2+sqrt{5}right)^2}}8e,left(3+sqrt{5}right)left(10-sqrt{2}right)sqrt{3-sqrt{5}}8f,sqrt{sqrt{2}+1}-sqrt{sqrt{2}-1}sqrt{2left(sqrt{2}-1right)}
Đọc tiếp
Chứng minh
\(a,\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(b,\frac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}\)
\(c,2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)
\(d,\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}=8\)
\(e,\left(3+\sqrt{5}\right)\left(10-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
\(f,\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
1. sqrt{x-2sqrt{x-1}}+sqrt{x+3-4sqrt{x-1}}left(2 x 5right)2. frac{6}{1-sqrt{3}}-frac{3sqrt{3}-1}{sqrt{3}+1}+sqrt{3}3. sqrt{29-12sqrt{5}+sqrt{24-8sqrt{3}}}4. sqrt{frac{4}{9-4sqrt{5}}}-sqrt{frac{4}{9+4sqrt{5}}}5. 5sqrt{frac{1}{5}}+frac{1}{2}sqrt{x}-frac{5}{4}sqrt{frac{4}{5}+sqrt{5}}6. frac{6-sqrt{6}}{sqrt{6}-1}-9sqrt{frac{2}{3}}-frac{4}{2-sqrt{6}}7. left(frac{sqrt{x}-2}{x-1}-frac{sqrt{x}+2}{x+2sqrt{x}+1}right)frac{left(sqrt{x}-1right)^2}{2}left(xge0,xne1right)
Đọc tiếp
1. \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+3-4\sqrt{x-1}}\left(2< x< 5\right)\)
2. \(\frac{6}{1-\sqrt{3}}-\frac{3\sqrt{3}-1}{\sqrt{3}+1}+\sqrt{3}\)
3. \(\sqrt{29-12\sqrt{5}+\sqrt{24-8\sqrt{3}}}\)
4. \(\sqrt{\frac{4}{9-4\sqrt{5}}}-\sqrt{\frac{4}{9+4\sqrt{5}}}\)
5. \(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{x}-\frac{5}{4}\sqrt{\frac{4}{5}+\sqrt{5}}\)
6. \(\frac{6-\sqrt{6}}{\sqrt{6}-1}-9\sqrt{\frac{2}{3}}-\frac{4}{2-\sqrt{6}}\)
7. \(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\frac{\left(\sqrt{x}-1\right)^2}{2}\left(x\ge0,x\ne1\right)\)
RÚT GỌN
D=\(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\)
N=\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)
C=\(\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{8-2\sqrt{15}}\right)\)
M=\(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}\)
21 tháng 12 2023 lúc 15:18
Bài 1: Tínha) 5sqrt{8}-4sqrt{27}-2sqrt{75}+sqrt{108} b) 1sqrt{left(3-sqrt{6}right)^2}+sqrt{left(1-sqrt{6}right)^2} c) dfrac{5sqrt{3}-3sqrt{5}}{sqrt{5}-sqrt{3}}+dfrac{1}{4+sqrt{15}} d) dfrac{2sqrt{3-sqrt{5}}left(3+sqrt{5}right)}{sqrt{10}-sqrt{2}}-dfrac{sqrt{15}+sqrt{5}}{sqrt{12}+2} Bài 2: Cho (d₁): y dfrac{1}{2}x-4 và (d₂): y -3x+3 . Vẽ (d₁) và (d₂) trên cùng một hệ trục tọa độ. Tìm tọa độ giao điểm A của 2 đường thẳng trên. Helpp!!
Đọc tiếp
Bài 1: Tính
a) \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
b) \(1\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
c) \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
d) \(\dfrac{2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
Bài 2: Cho (d₁): y = \(\dfrac{1}{2}x-4\) và (d₂): y = \(-3x+3\) . Vẽ (d₁) và (d₂) trên cùng một hệ trục tọa độ. Tìm tọa độ giao điểm A của 2 đường thẳng trên.
Helpp!!