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\(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>2^2=4\left(5>4\right)\\ \Leftrightarrow\sqrt{2}+\sqrt{3}>2\)
\(\left(\sqrt{8}+\sqrt{5}\right)^2=13+2\sqrt{40};\left(\sqrt{7}-\sqrt{6}\right)^2=13-2\sqrt{42}\\ 2\sqrt{40}>0>-2\sqrt{42}\\ \Leftrightarrow13+2\sqrt{40}>13-2\sqrt{42}\\ \Leftrightarrow\left(\sqrt{8}+\sqrt{5}\right)^2>\left(\sqrt{7}-\sqrt{6}\right)^2\\ \Leftrightarrow\sqrt{8}+\sqrt{5}>\sqrt{7}-\sqrt{6}\)
a/ giả sử \(\sqrt{7}-\sqrt{2}< 1\)
\(\Leftrightarrow\sqrt{7}< 1+\sqrt{2}\)
\(\Leftrightarrow 7< 1+2\sqrt{2}+2\)
\(\Leftrightarrow4< 2\sqrt{2}\Leftrightarrow16< 8\left(sai\right)\)
vậy \(\sqrt{7}-\sqrt{2}>1\)
câu b, c bạn làm tương tụ nhé , giả sử một đẳng thức tạm, sau đó bình phương lên rồi làm theo như trên là được nha
Bài này cũng dễ
a, \(\sqrt{7}-\sqrt{2}\) lớn hơn \(1\) . Vì
\(\sqrt{7}-\sqrt{2}=1,231537749\)
\(1=1\)
b, \(\sqrt{8}+\sqrt{5}\) bé hơn \(\sqrt{7}+\sqrt{6}\) . Vì
\(\sqrt{8}+\sqrt{5}=5,064495102\)
\(\sqrt{7}+\sqrt{6}=5,095241054\)
c, \(\sqrt{2005}+\sqrt{2007}\) lớn hơn \(\sqrt{2006}\) . Vì
\(\sqrt{2005}+\sqrt{2007}=89,57677992\)
\(\sqrt{2006}=44,78839135\)
\(A=\sqrt{6+2\sqrt{5}}-\sqrt{5}=\sqrt{5}+1-\sqrt{5}=1\)
\(B=\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)
Do đó: A=B
\(\sqrt{6+2\sqrt{5}}-\sqrt{5}=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}=\left|\sqrt{5}+1\right|-\sqrt{5}=1\)
\(\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt[3]{\left(\sqrt{2}\right)^3+1^3+3.2+3\sqrt{2}}-\sqrt{2}=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)
--> Bằng nhau
Võ Đông Anh Tuấn
Áp dụng \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\)
a)
\(7=\sqrt{49}\\ 3\sqrt{5}=\sqrt{9}\cdot\sqrt{5}=\sqrt{9\cdot5}=\sqrt{45}\\ \text{Vì }\sqrt{49}>\sqrt{45}\text{ nên }7>3\sqrt{5}\)
Vậy \(7>3\sqrt{5}\)
b)
\(2\sqrt{7}+3=\sqrt{4}\cdot\sqrt{7}+3=\sqrt{4\cdot7}+3=\sqrt{28}+3\\ \sqrt{28}+3>\sqrt{25}+3=5+3=8\)
Vậy \(8< 2\sqrt{7}+3\)
c)
\(3\sqrt{6}=\sqrt{9}\cdot\sqrt{6}=\sqrt{9\cdot6}=\sqrt{54}\\ 2\sqrt{15}=\sqrt{4}\cdot\sqrt{15}=\sqrt{4\cdot15}=\sqrt{60}\\ \text{Vì } \sqrt{54}< \sqrt{60}\text{nên }3\sqrt{6}< 2\sqrt{15}\)
Vậy \(3\sqrt{6}< 2\sqrt{15}\)
a \(\left(\sqrt{5\sqrt{7}}\right)^4=\left(\left(\sqrt{5\sqrt{7}}\right)^2\right)^2=\left(5\sqrt{7}\right)^2=25\cdot7=175\)
\(=\left(\sqrt{7\sqrt{5}}\right)^4=\left(\left(\sqrt{7\sqrt{5}}\right)^2\right)^2=\left(7\sqrt{5}\right)^2=49\cdot5=240\)
vì 175<240\(\Rightarrow\left(\sqrt{5\sqrt{7}}\right)^4< \left(\sqrt{7\sqrt{5}}\right)^4\Rightarrow\sqrt{5\sqrt{7}}< \sqrt{7\sqrt{5}}\)
b \(6=\sqrt{36}\)
\(\sqrt{31}< \sqrt{36};\sqrt{19}>\sqrt{17}\Rightarrow\sqrt{31}-\sqrt{19}< \sqrt{36}-\sqrt{17}=6-\sqrt{17}\)
c \(\left(\sqrt{10}+\sqrt{17}\right)^2=10+2\sqrt{10\cdot17}+17=27+2\sqrt{170}\)
\(\left(\sqrt{61}\right)^2=61=27+34=27+2\cdot17=27+2\sqrt{289}\)
vì \(2\sqrt{170}< 2\sqrt{289}\Rightarrow27+2\sqrt{170}< 27+2\sqrt{289}\Rightarrow\left(\sqrt{10}+\sqrt{17}\right)^2< \left(\sqrt{61}\right)^2\)
\(\Rightarrow\sqrt{10}+\sqrt{17}< \sqrt{61}\)
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) \(5=\sqrt{25}< \sqrt{29}\)
\(\Rightarrow-5>-\sqrt{29}\)
\(a,\left(\sqrt{\sqrt{3}}\right)^4=3< 4=\left(\sqrt{2}\right)^4\Rightarrow\sqrt{\sqrt{3}}< \sqrt{2}\\ b,\left(\sqrt{2\sqrt{3}}\right)^4=12< 18=\left(\sqrt{3\sqrt{2}}\right)^4\Rightarrow\sqrt{2\sqrt{3}}=\sqrt{3\sqrt{2}}\\ c,\left(2+\sqrt{6}\right)^2=8+4\sqrt{6};5^2=25=8+17;\left(4\sqrt{6}\right)^2=96< 289=17^2\\ \Rightarrow4\sqrt{6}< 17\Rightarrow2+\sqrt{6}< 5\\ d,\left(7-2\sqrt{2}\right)^2=57-28\sqrt{2};4^2=16=57-41;\left(28\sqrt{2}\right)^2=1568< 41^2=1681\\ \Rightarrow28\sqrt{2}< 41\Rightarrow7-2\sqrt{2}>4\\ e,\left(\sqrt{15}+\sqrt{8}\right)^2=23+4\sqrt{30};7^2=49=23+26;\left(4\sqrt{30}\right)^2=240< 676=26^2\\ \Rightarrow4\sqrt{30}< 26\Rightarrow\sqrt{15}+\sqrt{8}< 7\)
\(f,\left(\sqrt{37}-\sqrt{14}\right)^2=51-2\sqrt{518};\left(6-\sqrt{15}\right)^2=51-12\sqrt{15};\left(2\sqrt{518}\right)^2=2072;\left(12\sqrt{15}\right)^2=2160\\ \Rightarrow2\sqrt{518}< 12\sqrt{15}\Rightarrow\sqrt{37}-\sqrt{14}>6-\sqrt{15}\)
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.
$(2\sqrt{3}+6\sqrt{2})^2=84+24\sqrt{6}< 84+24\sqrt{9}< 169$
$\Rightarrow 2\sqrt{3}+6\sqrt{2}< 13$
$\Rightarrow \frac{13-2\sqrt{3}}{6}> \sqrt{2}$