a,

\(-\frac{13}{38}=-1--\frac{25}{38}=-1+\frac{25}{38}\)

\(\frac{29}{-88}=-\frac{29}{88}=-1--\frac{59}{88}=-1+\frac{59}{88}\)

Vì \(\frac{25}{38}< \frac{59}{88}\Rightarrow-\frac{13}{38}< \frac{29}{-88}\)

b,

Ta có:

3³⁰¹ > 3³⁰⁰ = [3³]¹⁰⁰ = 27¹⁰⁰

5¹⁹⁹ < 5²⁰⁰ = [5²]¹⁰⁰ = 25¹⁰⁰

Mà 27¹⁰⁰ > 25¹⁰⁰ => 3³⁰¹ > 5¹⁹⁹

c,

\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left[2n+1\right]\left[2n+3\right]}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n+1}-\frac{1}{2n+3}\)

\(=1-\frac{1}{2n+3}< 1\)

Vậy P < 1

\(5^{199}=\left(5^{\frac{199}{301}}\right)^{301}\)

\(5^{\frac{199}{301}}< 3^1\)

\(\Leftrightarrow5^{199}< 3^{301}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n+1}-\frac{1}{2n+3}\)

\(=1-\frac{1}{2n+3}< 1\)

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

study well

chúc bạn học tốt

\(\frac{13}{38}>\frac{13}{39}=\frac{1}{3}=\frac{29}{87}>\frac{29}{88}\Rightarrow\frac{-13}{38}< \frac{29}{-88}\)

a) 13/57=13+16/57+16=29/73   ( Ghi nhớ SKG Toán 6)
-=> 13/57 < 29/73
b)  17/42 = 17-4/42-4 = 13/38
=> 17/42 > 13/38

c)7/41 = 7+6/41+6= 13/47
=> 7/41<13/47

a, \(-\frac{13}{38}\) và \(\frac{29}{-88}\)

Ta có : 

\(\left(-13\right)\cdot\left(-88\right)=1144>29\cdot38=1102\)

b, Ta có :

\(-\frac{1}{5}< 0< \frac{1}{1000}\)

\(\Rightarrow\text{ }-\frac{1}{5}< \frac{1}{1000}\)

Ta có:

\(\frac{13}{38}>\frac{13}{39}=\frac{1}{3}\)

\(\frac{-12}{-37}=\frac{12}{37}< \frac{12}{36}=\frac{1}{3}\)

Vì \(\frac{13}{38}>\frac{1}{3}>\frac{-12}{-37}\)

=> \(\frac{13}{38}>\frac{-12}{-37}\)

ta có: \(\frac{13}{38}>\frac{13}{39}=\frac{1}{3}=\frac{29}{87}>\frac{29}{88}\)

=> \(\frac{13}{88}>\frac{29}{88}\), mà đối với số âm, số nào dương lớn hơn thì nhỏ hơn

=> \(\frac{-13}{38}

Bài 1:

Ta có:

\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)

\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)

Lại có:

\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)

\(\Rightarrow\left(\frac{1}{10}\right)^{15}< \left(\frac{3}{10}\right)^{20}\)

Bài 2:

Ta có:

\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)

\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)

Mà \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)

\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)

\(\Rightarrow13A>13B\Rightarrow A>B\)