Chứng minh các bất đẳng thức sau:
\(\frac{\sqrt{\sqrt{2}+\sqrt{2}}+\sqrt{3}\sqrt{2-\sqrt{2}}}{4}< 0,8\)
(\(\sqrt{\sqrt{3}}+\sqrt{\sqrt{5}}+\sqrt{\sqrt{7}}\))-(\(\sqrt{3}+\sqrt{5}+\sqrt{7}\)) < 3
\(\sqrt{\sqrt{17+12\sqrt{2}}-\sqrt{2}}>\sqrt{3}-1\)
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TN
17 tháng 8 2015
a)\(\frac{3.\sqrt{6}}{2}+\frac{2.\sqrt{2}}{\sqrt{3}}-\frac{4.\sqrt{3}}{\sqrt{2}}=\frac{3\sqrt{6}}{2}+\frac{2\sqrt{2}.\sqrt{3}}{\sqrt{3}.\sqrt{3}}-\frac{4.\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{3\sqrt{6}}{2}+\frac{2\sqrt{6}}{3}-\frac{4\sqrt{6}}{2}=\frac{2\sqrt{6}}{3}-\frac{\sqrt{6}}{2}=\frac{4\sqrt{6}-3\sqrt{6}}{6}=\frac{\sqrt{6}}{6}\)
--> dpcm
b) \(\left(\frac{-\sqrt{7}.\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}.\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right).\frac{\sqrt{7}-\sqrt{5}}{1}\)
=\(\left(-\sqrt{7}-\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
=\(-1.\left(\sqrt{7}+\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
=\(-1.\left(7-5\right)\)
=-1.2
=-2
11 tháng 8 2020
Câu b, c tương tự câu a. Mình làm câu a coi như tượng trưng nha !!!!!!
a) Đặt: \(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
<=> \(A^3=2+\sqrt{5}+2-\sqrt{5}+3\sqrt[3]{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}.\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
<=> \(A^3=4+3\sqrt[3]{4-5}.A\)
<=> \(A^3=4-3A\)
<=> \(A^3+3A-4=0\)
<=> \(\left(A-1\right)\left(A^2+A+4\right)=0\)
Có: \(A^2+A+4=\left(A+\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\)
=> \(A-1=0\)
<=> \(A=1\)
=> \(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1\)
VẬY TA CÓ ĐPCM
AT
4 tháng 7 2021
\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)
\(=\sqrt{x-1-2\sqrt{x-1+1}}+\sqrt{x-1+2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|\)
\(=\sqrt{x-1}-1+\sqrt{x-1}+1\left(x\ge2\right)=2\sqrt{x-1}\)
a) \(\dfrac{1}{\sqrt{5}+\sqrt{7}}=\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)}=\dfrac{\sqrt{7}-\sqrt{5}}{2}\)
c) \(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{5}}=\dfrac{7}{2\sqrt{5}-\sqrt{3}}=\dfrac{7\left(2\sqrt{5}+\sqrt{3}\right)}{\left(2\sqrt{5}+\sqrt{3}\right)\left(2\sqrt{5}-\sqrt{3}\right)}\)
\(=\dfrac{14\sqrt{5}+7\sqrt{3}}{17}\)
10 tháng 10 2020
Giúp bn bài 1 thôi
Bài 1:
a, \(\sqrt{7-2\sqrt{10}}=\sqrt{5-2\sqrt{10}+2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{2}\right|=\sqrt{5}-\sqrt{2}\) (\(\sqrt{5}>\sqrt{2}\)) (đpcm)
b, \(\sqrt{4+2\sqrt{3}}-\sqrt{3}=\sqrt{3+2\sqrt{3}+1}-\sqrt{3}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{3}=\sqrt{3}+1-\sqrt{3}=1\) (đpcm)
Chúc bn học tốt!
20 tháng 10 2021
\(VT=\left[\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right]\cdot\left(\sqrt{7}-\sqrt{5}\right)\\ =\left(-\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\\ =-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2=VP\)
20 tháng 10 2021
\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}=\left(-\sqrt{7}-\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 9 2019
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}=\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\sqrt{2}-1\)
Ai giúp vs 😭😭😭