\(a,\dfrac{13}{38}\) và \(\dfrac{1}{3}.\)

Ta có: \(\dfrac{13}{38}>\dfrac{13}{39}=\dfrac{1}{3}.\)

\(\Rightarrow\dfrac{13}{38}>\dfrac{1}{3}.\)

\(b,\sqrt{235}\) và \(15.\)

Ta có: \(\sqrt{235};15=\sqrt{225}.\)

Vì \(\sqrt{235}>\sqrt{225}\) (do \(235>225\))

nên \(\sqrt{235}>15.\)

2⁹⁰=(2⁵)¹⁸=32¹⁸

5³⁶=(5²)¹⁸=25¹⁸

Vì 32>25 nên 32¹⁸>25¹⁸

=>2⁹⁰>5³⁶

b,15=\(\sqrt{225}\) <\(\sqrt{235}\)

=> 15<\(\sqrt{235}\)

c, Ta có: \(\dfrac{1}{3}=\dfrac{13}{39}\)

vì 38<39

nên \(\dfrac{13}{38}>\dfrac{13}{39}\)

a) 2⁹⁰= (2¹⁰)⁹mà 2¹⁰=(2⁵)²

5³⁶= (5⁴)⁹mà 5⁴=(5²)²

Do 2⁵>5²nên 2⁹⁰>5³⁶

a) \(2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}\)

Do : \(8^{100}< 9^{100}\)

=> \(2^{300}< 3^{200}\)

b) Do \(\dfrac{13}{38}>\dfrac{13}{39}\)

Mà : \(\dfrac{13}{39}=\dfrac{1}{3}\)

=> \(\dfrac{13}{38}>\dfrac{1}{3}\)

c)Do : \(\sqrt{235}>\sqrt{225}\)

Mà : \(\sqrt{225}=15\)

=> \(\sqrt{235}>15\)

a) Ta có:

\(2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}\)

Ta thấy 8<9 suy ra \(8^{100}< 9^{100}\)

Vậy \(2^{300}< 3^{200}\)

b) Ta có:

38<39 suy ra \(\dfrac{13}{38}>\dfrac{13}{39}=\dfrac{1}{3}\)

suy ra \(\dfrac{13}{38}>\dfrac{1}{3}\)

\(ta có A=\dfrac{13^{15}+1}{13^{16}+1}=\dfrac{13^{15}}{13^{16}}+1\)=\(\dfrac{1}{13}+1\)

B=\(\dfrac{13^{16}+1}{13^{17}+1}=\dfrac{13^{16}}{13^{17}}+1\)=\(\dfrac{1}{13}+1\)

vậy A=B

\(A=\dfrac{13^{15}+1}{13^{16}+1}vàB=\dfrac{13^{16}+1}{13^{17}+1}\)

ta có

\(\dfrac{13^{16}+1}{13^{17}+1}< 1\Rightarrow\dfrac{13^{16}+1+12}{13^{17}+1+12}=\dfrac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}=\dfrac{13^{15}+1}{13^{16}+1}=A\)

vậy B<A

\(A=\dfrac{13^{15}+1}{13^{16}+1}vàB=\dfrac{13^{16}+1}{13^{17}+1}\)

ta có B<1 nên

\(\dfrac{13^{16}+1}{13^{17}+1}< \dfrac{13^{16}+1+12}{13^{17}+1+12}=\dfrac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}=\dfrac{13^{15}+1}{13^{16}+1}=A\)

Vậy B<A

a)=

b)>

c)=

Ta có: \(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{120}+\sqrt{121}}\)

\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{120}+11\)

=10

Ta có: \(B=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{35}}\)

\(=\dfrac{2}{\sqrt{1}+\sqrt{1}}+\dfrac{2}{\sqrt{2}+\sqrt{2}}+...+\dfrac{2}{\sqrt{35}+\sqrt{35}}\)

\(\Leftrightarrow B< 2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{35}+\sqrt{36}}\right)\)

\(\Leftrightarrow B< 2\cdot\left(-\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}-...-\dfrac{1}{\sqrt{35}}+\dfrac{1}{\sqrt{36}}\right)\)

\(\Leftrightarrow B< 2\cdot\left(-\dfrac{1}{1}+\dfrac{1}{6}\right)\)

\(\Leftrightarrow B< -\dfrac{5}{3}< 10=A\)

a)\(\frac{13}{38}\)>\(\frac{1}{3}\)                              b)\(\sqrt{235}\)<15

study well

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