So sánh
a)\(\sqrt{35}+\sqrt{99}v\text{à}16\)
b)\(\sqrt{24}v\text{à}\sqrt{5}+\sqrt{10}\)
Những câu hỏi liên quan
Không dùng máy tính ,hãy so sánh :
1 )\(\sqrt{7-\sqrt{21}+4\sqrt{5}}v\text{à}\sqrt{5}-1\)
2 )\(\sqrt{5}+\sqrt{10}+1v\text{à}\sqrt{35}.\)
3 )\(\frac{15-2\sqrt{10}}{3}v\text{à}\sqrt{15}.\)
1) \(A=\left(\sqrt{7-\sqrt{21}+4\sqrt{5}}\right)^2=7-\sqrt{21}+4\sqrt{5}\)
\(B=\left(\sqrt{5}-1\right)^2=6-2\sqrt{5}\)
\(\Rightarrow A-B=1-\sqrt{21}+6\sqrt{5}=\left(1+\sqrt{180}\right)-\sqrt{21}>0\)
\(\Rightarrow A>B\Rightarrow\sqrt{7-\sqrt{21}+4\sqrt{5}}>\sqrt{5}-1\)
2) \(C=\left(\sqrt{5}+\sqrt{10}+1\right)^2=5+10+1+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}\)
\(=26+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}>26+10>35=\left(\sqrt{35}\right)^2\)
Vậy \(\sqrt{5}+\sqrt{10}+1>\sqrt{35}\)
3) \(\left(\frac{15-2\sqrt{10}}{3}\right)^2=\frac{225-60\sqrt{10}+40}{9}=\frac{265-60\sqrt{10}}{9}=\frac{265}{9}-\frac{20\sqrt{10}}{3}< 15\)
Vậy nên \(\frac{15-2\sqrt{10}}{3}< \sqrt{15}\)
Đúng 0
Bình luận (0)
So Sánh Các Biểu Thức Sau:
a,\(\sqrt{2}+\sqrt{11}v\text{à}\sqrt{3}+4\) 4
b, \(\sqrt{21}-\sqrt{5}v\text{à}\)\(\sqrt{20}-\sqrt{6}\)
c,\(\sqrt{24}-1v\text{à}\)\(5\)
\(a,\sqrt{2}+\sqrt{11}< \sqrt{3}+\sqrt{16}=\sqrt{3}+4\)
Đúng 0
Bình luận (0)
Câu 1: Chứng minh:
\(31.82+125.48+21.43=125.67=1500\)
Câu 2: So sánh:
1,\(\sqrt{51}-\sqrt{5}v\text{à}\sqrt{20}-\sqrt{6}\)
2,\(\sqrt{2}+\sqrt{8}v\text{à}\sqrt{3}+3\)
3,\(\sqrt{37}-\sqrt{14}v\text{à}6-\sqrt{15}\)
4,\(\sqrt{5}+\sqrt{10}v\text{à}5,3\)
So sánh các số sau:
a) \(0,5\sqrt{100}-\sqrt{\frac{4}{25}}v\text{à}\left(\sqrt{1\frac{1}{9}}-\sqrt{\frac{9}{16}}\right):5\)
b) \(\sqrt{25+9}v\text{à}\sqrt{25}+\sqrt{9}\)
so sánh
a, \(\sqrt{2}+\sqrt{3}+\sqrt{5}v\text{à}18\)
b, \(\sqrt{5}+\sqrt{7}+4v\text{à}12\)
\(\sqrt{2}+\sqrt{3}+\sqrt{5}< \sqrt{4}+\sqrt{9}+\sqrt{25}=2+3+5=10< 18\)
b) \(\sqrt{5}+\sqrt{7}+4< \sqrt{9}+\sqrt{9}+4=3+3+4=10< 12\)
Đúng 0
Bình luận (0)
Vì \(\sqrt{2}\) và các căn bậc khác đều là nhưng số thực nên ta cha nó là
\(\sqrt{2}+\sqrt{3}+\sqrt{5}=2+3+5=10\)
Mà 10<12
\(\Rightarrow dpcm\)
Đúng 0
Bình luận (3)
So sánh
\(\sqrt{2012}+\sqrt{2013}+\sqrt{2014}v\text{à}\sqrt{2009}+\sqrt{2011}+\sqrt{2019}\)
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 \(\sqrt[3]{\left(1-\sqrt{3}\right)\left(4-2\sqrt{3}\right)}\)>\(\sqrt[3]{\left(1-\sqrt{5}\right)\left(6-2\sqrt{5}\right)}\)
Đúng 0
Bình luận (0)
Không dùng máy tính, hãy so sánh: \(\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}v\text{à}\sqrt{2016}+\sqrt{2017}\)
\(\frac{2016}{\sqrt{2016}}=\sqrt{2016}\)
\(\frac{2017}{\sqrt{2017}}=\sqrt{2017}\)
=> Bằng nhau
Đúng 0
Bình luận (0)
\(\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}-\sqrt{2016}-\sqrt{2017}=\left(\frac{2016}{\sqrt{2017}}-\sqrt{2017}\right)+\left(\frac{2017}{\sqrt{2016}}-\sqrt{2016}\right)\)
\(=\frac{2016-2017}{\sqrt{2017}}+\frac{2017-2016}{\sqrt{2016}}=\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}\)
vì \(2016< 2017\Rightarrow\sqrt{2016}< \sqrt{2017}\Rightarrow\frac{1}{\sqrt{2016}}>\frac{1}{\sqrt{2017}}\Rightarrow\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}>0\)
\(\Rightarrow\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}-\sqrt{2016}-\sqrt{2017}>0\Rightarrow\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}>\sqrt{2016}+\sqrt{2017}\)
Đúng 0
Bình luận (0)
So sánh \(2+\sqrt{3}v\text{à}\sqrt{5+4\sqrt{3}}\)
Ta có
\(\left(2+\sqrt{3}\right)^2=2^2+2\cdot2\cdot\sqrt{3}+3=7+4\sqrt{3}\)
\(\Rightarrow2+\sqrt{3}=\sqrt{7+4\sqrt{3}}\)
Ta có \(7+4\sqrt{3}>5+4\sqrt{3}\)
\(\Leftrightarrow\sqrt{7+4\sqrt{3}}>\sqrt{5+4\sqrt{3}}\)
\(\Rightarrow2+\sqrt{3}>\sqrt{5+4\sqrt{3}}\)
Đúng 0
Bình luận (0)