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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: \(13-\sqrt{35}>\sqrt{48}\)
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) \(5=\sqrt{25}< \sqrt{29}\)
\(\Rightarrow-5>-\sqrt{29}\)
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) \(\left(\sqrt{11}-\sqrt{3}\right)^2=14-2\sqrt{33}=14-\sqrt{132}\)
\(2^2=14-10=14-\sqrt{100}\)
\(\sqrt{100}< \sqrt{132}\Leftrightarrow-\sqrt{100}>-\sqrt{132}\Leftrightarrow14-\sqrt{100}>14-\sqrt{132}\)
\(\Rightarrow2>\sqrt{11}-\sqrt{3}\)
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: \(9-\sqrt{58}=\sqrt{49}-\sqrt{58}< 0< \sqrt{80}-\sqrt{59}\)
Lời giải:
a.
$\sqrt{8}+\sqrt{15}+1<\sqrt{9}+\sqrt{16}+1=3+4+1=8=\sqrt{64}< \sqrt{65}$
$\Rightarrow \sqrt{8}+\sqrt{15}< \sqrt{65}-1$
b.
$(2\sqrt{3}+6\sqrt{2})^2=84+24\sqrt{6}< 84+24\sqrt{9}< 169$
$\Rightarrow 2\sqrt{3}+6\sqrt{2}< 13$
$\Rightarrow \frac{13-2\sqrt{3}}{6}> \sqrt{2}$
\(a,\left(\sqrt{\sqrt{3}}\right)^4=3< 4=\left(\sqrt{2}\right)^4\Rightarrow\sqrt{\sqrt{3}}< \sqrt{2}\\ b,\left(\sqrt{2\sqrt{3}}\right)^4=12< 18=\left(\sqrt{3\sqrt{2}}\right)^4\Rightarrow\sqrt{2\sqrt{3}}=\sqrt{3\sqrt{2}}\\ c,\left(2+\sqrt{6}\right)^2=8+4\sqrt{6};5^2=25=8+17;\left(4\sqrt{6}\right)^2=96< 289=17^2\\ \Rightarrow4\sqrt{6}< 17\Rightarrow2+\sqrt{6}< 5\\ d,\left(7-2\sqrt{2}\right)^2=57-28\sqrt{2};4^2=16=57-41;\left(28\sqrt{2}\right)^2=1568< 41^2=1681\\ \Rightarrow28\sqrt{2}< 41\Rightarrow7-2\sqrt{2}>4\\ e,\left(\sqrt{15}+\sqrt{8}\right)^2=23+4\sqrt{30};7^2=49=23+26;\left(4\sqrt{30}\right)^2=240< 676=26^2\\ \Rightarrow4\sqrt{30}< 26\Rightarrow\sqrt{15}+\sqrt{8}< 7\)
\(f,\left(\sqrt{37}-\sqrt{14}\right)^2=51-2\sqrt{518};\left(6-\sqrt{15}\right)^2=51-12\sqrt{15};\left(2\sqrt{518}\right)^2=2072;\left(12\sqrt{15}\right)^2=2160\\ \Rightarrow2\sqrt{518}< 12\sqrt{15}\Rightarrow\sqrt{37}-\sqrt{14}>6-\sqrt{15}\)
\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}=6\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\sqrt{\left(5-2\sqrt{6}\right)^2}+\sqrt{\left(5+2\sqrt{6}\right)^2}\)
\(=5-2\sqrt{6}+5+2\sqrt{6}=10\)
\(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}+\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)
\(=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}=2\sqrt{5}+4\sqrt{2}\)
a: \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
b: \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}\)
=6
c: Ta có: \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)
\(=5-2\sqrt{6}+5+2\sqrt{6}\)
=10
d: Ta có: \(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-4\sqrt{10}}+\sqrt{53+4\sqrt{90}}\)
\(=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}\)
\(=2\sqrt{5}+4\sqrt{2}\)
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 : \(\left\{{}\begin{matrix}\sqrt{81}>\sqrt{80}\\-\sqrt{58}>-\sqrt{59}\end{matrix}\right.\Rightarrow\sqrt{81}+\left(-\sqrt{58}\right)>\sqrt{80}+\left(-\sqrt{59}\right)\Rightarrow9-\sqrt{58}>\sqrt{80}-\sqrt{59}\)
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 \(9=\sqrt{81}\Leftrightarrow\sqrt{81}>\sqrt{80}\);
\(-\sqrt{58}>-\sqrt{59}\)
=> \(\sqrt{81}-\sqrt{58}>\sqrt{80}-\sqrt{59}\)
<=> \(9-\sqrt{58}>\sqrt{80}-\sqrt{59}\)