\(\sqrt{361=19}\)

\(\sqrt{121^2=121}\)

Vaayj \(\sqrt{361}+\sqrt{121^2=}19+121=140\)

\(\sqrt{361}+\sqrt{121^2}\)

\(=19+121\)

\(=140\)

\(k\)\(nha\)

Giải:

Mình không ghi căn được nên mình không ghi lại đề nha

(4+10-11):căn bậc hai của 3

=2

\(\sqrt{81}=9\)

\(\sqrt{0,64}=0,8\)

\(\sqrt{\frac{49}{100}}=\frac{7}{10}\)

\(\sqrt{8100}=90\)

\(\sqrt{100=}10\)

\(\sqrt{0,01}=0,1\)

\(\sqrt{\frac{4}{25}}=\frac{2}{5}\)

\(\sqrt{\frac{0,09}{121}}=\frac{0,3}{11}\)

\(\sqrt{81}=9\);\(\sqrt{0,64}=0,8\);\(\sqrt{\frac{49}{100}}=\frac{7}{10}\);\(\sqrt{8100}=90\); \(\sqrt{100}=10\); \(\sqrt{0,01}=0,1\); \(\sqrt{\frac{4}{25}}=\frac{2}{5}\); \(\sqrt{\frac{0,09}{121}}=\frac{0,3}{11}=\frac{3}{110}\)

\(a,\sqrt{81}=9\)

\(b.\sqrt{8100}=90\)

\(c,\sqrt{64}=8\)

\(d,\sqrt{25}=5\)

\(e,\sqrt{0,64}=0,8\)

\(f,\sqrt{10000}=100\)

\(g,\sqrt{0,01}=0,1\)

\(h,\sqrt{\frac{49}{100}}=\frac{7}{10}\)

\(i,\sqrt{\frac{0,09}{121}}=\frac{0,3}{11}\)

\(j,\sqrt{\frac{4}{25}}=\frac{2}{5}\)

~Study well~

#JDW

a) 9

b) 90 

c) 8

d) 5

e) 0,8

f) 100

g) 0,1

h) \(\frac{7}{10}\)

i) \(\frac{0,3}{11}\)

j) 0,4.

bạn đang hỏi hay đang trả lời vậy???????????

6 

-4

3 phần 5

3

11

Câu a)
\(A=\sqrt{20+1}+\sqrt{40+2}+\sqrt{60+3}\)
\(=\sqrt{1\left(20+1\right)}+\sqrt{2\left(20+1\right)}+\sqrt{3\left(20+1\right)}\)
\(=\sqrt{20+1}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)

\(B=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{20}+\sqrt{40}+\sqrt{60}\)
\(=1\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{1}\cdot\sqrt{20}+\sqrt{2}\cdot\sqrt{20}+\sqrt{3}\cdot\sqrt{20}\right)\)
\(=\sqrt{1}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\sqrt{20}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)
\(=\left(\sqrt{20}+\sqrt{1}\right)\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)

Ta thấy: \(\hept{\begin{cases}\left(\sqrt{20+1}\right)^2=20+1\\\left(\sqrt{20}+\sqrt{1}\right)^2=20+1+2\sqrt{20}\end{cases}}\)
\(\Rightarrow\left(\sqrt{20+1}\right)^2< \left(\sqrt{20}+\sqrt{1}\right)^2\Rightarrow\sqrt{20+1}< \sqrt{20}+\sqrt{1}\)
Vậy A < B.

a) A<B

\(\frac{1}{\sqrt{1}}< \frac{1}{\sqrt{121}}\)

\(\frac{1}{\sqrt{2}}< \frac{1}{\sqrt{121}}\)

................

\(\frac{1}{\sqrt{121}}=\frac{1}{\sqrt{121}}\)

Suy ra \(\frac{1}{\sqrt{1}}\)+\(\frac{1}{\sqrt{2}}\)+.............+\(\frac{1}{\sqrt{121}}\)<\(\frac{1}{\sqrt{121}}+\frac{1}{\sqrt{121}}+\frac{1}{\sqrt{121}}+......\frac{1}{\sqrt{121}}\)=\(\frac{121}{11}\)=11(đpcm)(vì có 121 chữ số)\(\frac{1}{\sqrt{121}}\))

Khuyển Dạ Xoa : \(\sqrt{1}< \sqrt{121}\Rightarrow\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{121}}\)  chứ?