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a, \(\sqrt{21}>\sqrt{20}\)
\(-\sqrt{5}>-\sqrt{6}\)
\(\Rightarrow\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)
b, \(\sqrt{2}< \sqrt{3}\)
\(\sqrt{8}< \sqrt{9}=3\)
\(\Rightarrow\sqrt{2}+\sqrt{8}< \sqrt{3}+3\)
\(\text{c) }\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 30\)
Ta có : \(6< 6.25\Rightarrow\sqrt{6}< \sqrt{6.25}\Rightarrow\sqrt{6}< 2.5\)
\(12< 12.25\Rightarrow\sqrt{12}< \sqrt{12.25}\Rightarrow\sqrt{12}< 3.5\)
\(20< 20.25\Rightarrow\sqrt{20}< \sqrt{20.25}\Rightarrow\sqrt{20}< 4.5\)
\(30< 30.25\Rightarrow\sqrt{30}< \sqrt{30.25}\Rightarrow\sqrt{30}< 5.5\)
\(42< 42.25\Rightarrow\sqrt{42}< \sqrt{42.25}\Rightarrow\sqrt{42}< 6.5\)
\(50< 56.5\Rightarrow\sqrt{50}< \sqrt{56.25}\Rightarrow\sqrt{50}< 7.5\) \(\left(1\right)\)
Từ \(\left(1\right)\) suy ra :
\(\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 2.5+3.5+4.5+5.5+6.5+7.5\)
\(\Rightarrow\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 30\) \(\left(ĐPCM\right)\)
Vậy \(\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 30\)
\(\)\(\text{a) }\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 24\)
Ta có : \(1< 9\Rightarrow\sqrt{1}< \sqrt{9}\Rightarrow\sqrt{1}< 3\)
\(2< 9\Rightarrow\sqrt{2}< \sqrt{9}\Rightarrow\sqrt{2}< 3\)
\(3< 9\Rightarrow\sqrt{3}< \sqrt{9}\Rightarrow\sqrt{3}< 3\)
\(...\)
\(8< 9\Rightarrow\sqrt{8}< \sqrt{9}\Rightarrow\sqrt{8}< 3\) \(\left(1\right)\)
Từ \(\left(1\right)\) suy ra :
\(\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 3+3+...+3_{\left(\text{8 số hạng 3}\right)}\) \(\) \(\)
\(\) \(\Rightarrow\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 3\cdot8\)
\(\Rightarrow\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 24\) \(\left(ĐPCM\right)\)
Vậy \(\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 24\)
\(\text{b) }\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>10\)
Ta có : \(1< 100\Rightarrow\sqrt{1}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{1}}< \dfrac{1}{\sqrt{100}}\)
\(2< 100\Rightarrow\sqrt{2}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{2}}< \dfrac{1}{\sqrt{100}}\)
\(...\)
\(100=100\Rightarrow\sqrt{100}=\sqrt{100}\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}\) \(\left(1\right)\)
Từ \(\left(1\right)\) suy ra :
\(\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}_{\left(\text{100 số hạng}\dfrac{1}{\sqrt{100}}\right)}\)
\(\Rightarrow\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}\cdot100\)
\(\Rightarrow\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>\dfrac{10}{\sqrt{100}}\)
\(\Rightarrow\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>10\) \(\left(ĐPCM\right)\)
Vậy \(\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>10\)
\(\)
b, \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
.............................................
Cộng với vế 99 của BĐT trên, ta được:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}>99.\frac{1}{10}=\frac{99}{10}\)
\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{99}{10}=\frac{1}{10}=\frac{100}{10}=10\)
Wrecking Ball đã làm đúng
to ra kết quả giống cậu : Wrecking Ball
là đáp án đúng
tk nha ( chúc bn học gioi )
a, Vì
\(\sqrt{21}-\sqrt{5}=2346507717\)
\(\sqrt{20}-\sqrt{6}=2022646212\)
b, Vì
\(\sqrt{2}+\sqrt{8}=4242640687\)
\(\sqrt{3}+3=4732050808\)
c, Vì
\(\sqrt{5}+\sqrt{10}=5398345638\)
\(5,3=5,3\)
P/s; Ủa tôi tưởng lớp 8 mới học về Căn thức chứ
a)\(\sqrt{1}\)+\(\sqrt{9}\)+\(\sqrt{25}\)+\(\sqrt{49}\)+\(\sqrt{81}\)
=1+3+5+7+9
=25
b)=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{4}\)
=\(\dfrac{6}{12}\)+\(\dfrac{4}{12}\)+\(\dfrac{2}{12}\)+\(\dfrac{3}{12}\)
=\(\dfrac{15}{12}\)
c) =0,2+0.3+0,4
= 0.9
d) =9-8+7
=8
j) =1,2-1,3+1.4
= (-0,1)+1,4
=1,4
g) \(\dfrac{2}{5}\)+\(\dfrac{5}{2}\)+\(\dfrac{9}{10}\)+\(\dfrac{3}{4}\)
= (\(\dfrac{4}{10}\)+\(\dfrac{15}{10}\)+\(\dfrac{9}{10}\))+\(\dfrac{3}{4}\)
= \(\dfrac{14}{5}\)+\(\dfrac{3}{4}\)
=\(\dfrac{56}{20}\)+\(\dfrac{15}{20}\)
= \(\dfrac{71}{20}\)
Nhớ tick cho mk nha~