Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có \(\sqrt{8}+3< \sqrt{9}+3=3+3=6\)
=> \(\sqrt{8}+3< 6\)
Ta có \(\sqrt{48}< \sqrt{49};\sqrt{35}< \sqrt{36}\)
=> \(\sqrt{48}+\sqrt{35}< \sqrt{49}+\sqrt{46}\)
=> \(\sqrt{48}+\sqrt{35}< 13\)
=> \(\sqrt{48}< 13-\sqrt{35}\)
c) Ta có \(-\sqrt{19}< -\sqrt{17}\)
=> \(\sqrt{31}-\sqrt{19}< \sqrt{31}-\sqrt{17}\)
=> \(\sqrt{31}-\sqrt{19}< \sqrt{36}-17=6-\sqrt{17}\)
d) Ta có \(9=\sqrt{81}\Leftrightarrow\sqrt{81}>\sqrt{80}\);
\(-\sqrt{58}>-\sqrt{59}\)
=> \(\sqrt{81}-\sqrt{58}>\sqrt{80}-\sqrt{59}\)
<=> \(9-\sqrt{58}>\sqrt{80}-\sqrt{59}\)
a: \(\sqrt{17}+\sqrt{26}=\dfrac{9}{\sqrt{26}-\sqrt{17}}>9\)
e: \(\sqrt{13}-\sqrt{12}=\dfrac{1}{\sqrt{13}+\sqrt{12}}\)
\(\sqrt{12}-\sqrt{11}=\dfrac{1}{\sqrt{12}+\sqrt{11}}\)
mà \(\sqrt{13}+\sqrt{12}>\sqrt{11}+\sqrt{12}\)
nên \(\sqrt{13}-\sqrt{12}< \sqrt{12}-\sqrt{11}\)
d: \(9-\sqrt{58}=\sqrt{49}-\sqrt{58}< 0< \sqrt{80}-\sqrt{59}\)
a: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)
\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)
mà 112<117
nên \(4\sqrt{7}< 3\sqrt{13}\)
b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)
\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)
mà 108>64
nên \(3\sqrt{12}>2\sqrt{16}\)
c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)
\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)
mà \(\dfrac{21}{4}>\dfrac{36}{7}\)
nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)
d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)
\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)
mà 108>64
nên \(3\sqrt{12}>2\sqrt{16}\)
Ta có: \(12>9\)
\(6\sqrt{3}>4\sqrt{5}\)
Do đó: \(12+6\sqrt{3}>9+4\sqrt{5}\)
\(\Leftrightarrow\sqrt{12+6\sqrt{3}}>\sqrt{9+4\sqrt{5}}\)
a) \(9=6+3=6+\sqrt{9}\)
\(6+2\sqrt{2}=6+\sqrt{8}\)
\(\sqrt{8}< \sqrt{9}\) nên \(6+\sqrt{8}=6+2\sqrt{2}< 6+\sqrt{9}=9\)
b) \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}=5+\sqrt{24}\)
\(3^2=9=5+4=5+\sqrt{16}\)
\(\sqrt{16}< \sqrt{24}\Rightarrow3^2< \left(\sqrt{2}+\sqrt{3}\right)^2\Rightarrow3< \sqrt{2}+\sqrt{3}\)
c) \(9+4\sqrt{5}=\left(2+\sqrt{5}\right)^2\)
\(16=\left(2+2\right)^2=\left(2+\sqrt{4}\right)^2\)
\(\sqrt{4}< \sqrt{5}\Rightarrow2+\sqrt{4}< 2+\sqrt{5}\Rightarrow\left(2+\sqrt{4}\right)^2=16< \left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)
d) \(\left(\sqrt{11}-\sqrt{3}\right)^2=14-2\sqrt{33}=14-\sqrt{132}\)
\(2^2=14-10=14-\sqrt{100}\)
\(\sqrt{100}< \sqrt{132}\Leftrightarrow-\sqrt{100}>-\sqrt{132}\Leftrightarrow14-\sqrt{100}>14-\sqrt{132}\)
\(\Rightarrow2>\sqrt{11}-\sqrt{3}\)
Câu 1:
\(\sqrt{7}>0;\sqrt{11}>0\\ =>\sqrt{7}+\sqrt{11}>0\)
Ta có: \(8< 12\\ \Rightarrow\sqrt{8}< \sqrt{12}\\ \Rightarrow\sqrt{8}-\sqrt{12}< 0\)
=> \(\sqrt{7}+\sqrt{11}>0>\sqrt{8}-\sqrt{12}\)
=> \(\sqrt{7}+\sqrt{11}>\sqrt{8}-\sqrt{12}\)
Tiếp sức cho anh Đạt !
Bài 2 : Ta có : \(\left\{{}\begin{matrix}\sqrt{81}>\sqrt{80}\\-\sqrt{58}>-\sqrt{59}\end{matrix}\right.\Rightarrow\sqrt{81}+\left(-\sqrt{58}\right)>\sqrt{80}+\left(-\sqrt{59}\right)\Rightarrow9-\sqrt{58}>\sqrt{80}-\sqrt{59}\)
a) \(1=\sqrt{1}< \sqrt{2}\)
b) \(2=\sqrt{4}>\sqrt{3}\)
c) \(6=\sqrt{36}< \sqrt{41}\)
d) \(7=\sqrt{49}>\sqrt{47}\)
e) \(2=1+1=\sqrt{1}+1< \sqrt{2}+1\)
f) \(1=2-1=\sqrt{4}-1>\sqrt{3}-1\)
g) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\)
h) \(\sqrt{3}>0>-\sqrt{12}\)
i) \(5=\sqrt{25}< \sqrt{29}\)
\(\Rightarrow-5>-\sqrt{29}\)
1) \(\sqrt{17}>\sqrt{16}=4\)
\(\sqrt{26}>\sqrt{25}=5\)
Vế cộng vế ta có: \(\sqrt{17}+\sqrt{26}>9\)
2) Ta có: \(13-\sqrt{35}>13-\sqrt{36}=13-6=7\left(1\right)\)
\(\sqrt{48}< \sqrt{49}=7\left(2\right)\)
Từ (1);(2), Suy ra: \(13-\sqrt{35}>\sqrt{48}\)