\(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)\)

= \(\left(\sqrt{19}\right)^2-3^2\)

= \(19-9\)

= 10

\(\sqrt{2019}-\sqrt{2017}\approx0,022\) \(< \sqrt{19}-\sqrt{17}\approx0,23\)

chắc vậy :))

\(6-\sqrt{17}=\sqrt{36}-\sqrt{17}>\sqrt{31}-\sqrt{17}>\sqrt{31}-\sqrt{19}\)

HAY  \(6-\sqrt{17}>\sqrt{31}-\sqrt{19}\)

1/ bình phương hai vế được (căn11)^2+(căn5)^2=11+5   4^2=16 vậy căn 11+căn 5=4

2/ tương tự (3 căn3 )^2=27   (căn19)^2-(căn 2)^2=19-2=17  vậy 3 căn 3 >căn 19-căn2

a,√x+11=11

x+11=11\(^2\)

x+11=121

x=121-11

x=110

b,\(\sqrt{19-x}=19\)

\(\Rightarrow19-x=19^2\)

\(19-x=\)\(361\)

\(x=19-361\)

\(x=-342\)

Ý bạn là \(18< \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}< 19\) ?

Ta có:

\(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}=\frac{2}{2\sqrt{1}}+\frac{2}{2\sqrt{2}}+...+\frac{2}{2\sqrt{100}}\)

\(\Rightarrow A>\frac{2}{1+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{100}+\sqrt{101}}\)

\(\Rightarrow A>\frac{2\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\frac{2\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{2\left(\sqrt{101}-\sqrt{100}\right)}{\left(\sqrt{101}-\sqrt{100}\right)\left(\sqrt{101}+\sqrt{100}\right)}\)

\(\Rightarrow A>2\left(\sqrt{2}-1+\sqrt{3}-2+...+\sqrt{101}-\sqrt{100}\right)\)

\(\Rightarrow A>2\left(\sqrt{101}-1\right)>2\left(\sqrt{100}-1\right)=18\)

Tương tự:

\(A=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}=1+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{2}{2\sqrt{100}}\)

\(\Rightarrow A< 1+\frac{2}{1+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{100}}\)

Nhân liên hợp tử mẫu và rút gọn ta được (giống chứng minh >18 bên trên):

\(A< 1+2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\)

\(\Rightarrow A< 1+2\left(\sqrt{100}-1\right)=1+18=19\)

\(\Rightarrow18< A< 19\) (đpcm)

giup minh vs

mình chịu lun