So sánh: \(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+...+\sqrt{n\left(n+1\right)};\frac{\left(n+1\right)\left(n+2\right)}{4}\)
Những câu hỏi liên quan
Chứng minh rằng:
a)\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\) là số nguyên
b)\(\left(\sqrt{3}-1\right).\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
So sánh \(A=\sqrt{20+1}+\sqrt{40+2}+\sqrt{60+3}\) và \(B=\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{20}+\sqrt{40}+\sqrt{60}\right)\)
Câu 1: Thực hiện phép tính:
a. \(\sqrt{3}\left(2\sqrt{6}-\sqrt{3}\right)-6\sqrt{2}\)
b. \(6\sqrt{12}-\sqrt{20}-2\sqrt{27}+\sqrt{125}\)
c. \(\sqrt{\left(1-\sqrt{3}\right)^2}-3\sqrt{\dfrac{1}{3}}\)
d. \(\dfrac{6}{\sqrt{6}}-\dfrac{5}{\sqrt{6}-1}\)
\(a,=6\sqrt{2}-3-6\sqrt{2}=-3\\ b,=12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}=6\sqrt{3}+3\sqrt{5}\\ c,=\sqrt{3}-1-\sqrt{3}=-1\\ d,=\sqrt{6}-\dfrac{5\left(\sqrt{6}+1\right)}{5}=\sqrt{6}-\sqrt{6}-1=-1\)
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so sánh\(\sqrt[3]{\left(1-\sqrt{3}\right)\left(4-2\sqrt{3}\right)}v\text{à}\sqrt[3]{\left(1-\sqrt{5}\right)\left(6-2\sqrt{5}\right)}\)
\(\sqrt[3]{\left(1-\sqrt{3}\right)\left(4-2\sqrt{3}\right)}=\sqrt[3]{\left(1-\sqrt{3}\right)\left(\sqrt{3}-1\right)^2}\)=\(\sqrt[3]{\left(1-\sqrt{3}\right)^3}\)=1-\(\sqrt{3}\)
\(\sqrt[3]{\left(1-\sqrt{5}\right)\left(6-2\sqrt{5}\right)}=\sqrt[3]{\left(1-\sqrt{5}\right)\left(\sqrt{5}-1\right)^2}\)=\(\sqrt[3]{\left(1-\sqrt{5}\right)^3}\)=1-\(\sqrt{5}\)
Ta thấy \(\sqrt{5}>\sqrt{3}\)nên 1-\(\sqrt{3}\)>\(1-\sqrt{5}\)
Vậy \(\sqrt[3]{\left(1-\sqrt{3}\right)\left(4-2\sqrt{3}\right)}\)>\(\sqrt[3]{\left(1-\sqrt{5}\right)\left(6-2\sqrt{5}\right)}\)
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rg: \(\sqrt{\left(\sqrt{7}-4\right)}^2\) = 3
chứng minh:
\(\left(\sqrt{8}-5\sqrt{2}+\sqrt{20}\right)\sqrt{5}-\left(3\sqrt{\dfrac{1}{10}}+10\right)=3.3\sqrt{10}\)
\(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}\left(5\sqrt{\dfrac{1}{2}}+12\right)=-14.5\sqrt{2}\)
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: \(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}\cdot\left(\dfrac{5}{2}\sqrt{2}+12\right)\)
\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\left(5\sqrt{3}+12\sqrt{6}\right)\)
\(=-60-144\sqrt{2}+30\sqrt{2}+144\)
\(=84-114\sqrt{2}\)
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a)\(\left(\sqrt{3}-\sqrt{2}+1\right).\left(\sqrt{3}-1\right).\)
b)\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
c)\(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
d)\(\left(\sqrt{8}-5\sqrt{2}+\sqrt{20}\right).\sqrt{5}-\left(3.\sqrt{\frac{1}{10}}+10\right)\)
giúp mk zới:((
Adfrac{dfrac{sqrt{2+sqrt{3}}}{2}}{dfrac{sqrt{2+sqrt{3}}}{2}-dfrac{2}{sqrt{16}}+dfrac{sqrt{2+sqrt{3}}}{2sqrt{3}}}
Bdfrac{2left(dfrac{sqrt{2}+sqrt{3}}{6sqrt{2}}right)^{^{^{-1}}}+3left(dfrac{sqrt{2}+sqrt{3}}{4sqrt{3}}right)^{^{^{-1}}}}{left(dfrac{2+sqrt{6}}{12}right)^{^{^{-1}}}+left(dfrac{3+sqrt{6}}{12}right)^{^{^{-1}}}}
Cíu em với các pro ~
P/s: Câu B em làm đc r mà k biết kết quả đúng k nữa nên up lên hỏi luôn :)))
Đọc tiếp
\(A=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{16}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(B=\dfrac{2\left(\dfrac{\sqrt{2}+\sqrt{3}}{6\sqrt{2}}\right)^{^{^{-1}}}+3\left(\dfrac{\sqrt{2}+\sqrt{3}}{4\sqrt{3}}\right)^{^{^{-1}}}}{\left(\dfrac{2+\sqrt{6}}{12}\right)^{^{^{-1}}}+\left(\dfrac{3+\sqrt{6}}{12}\right)^{^{^{-1}}}}\)
Cíu em với các pro ~
P/s: Câu B em làm đc r mà k biết kết quả đúng k nữa nên up lên hỏi luôn :)))
Câu A dưới mẫu em ghi nhầm nha các pro
Mẫu là \(\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\)
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Tính:
\(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)
\(=2\sqrt{5}+\sqrt{5}-4\sqrt{5}\)
\(=-\sqrt{5}\)
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e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
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Tính:
\(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
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a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
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