Có phải sai đề ko sao 2 phân số có mẫu 19 zữ

đề sở sao sai

A ) 2³⁰⁰ = ( 2³)¹⁰⁰=8¹⁰⁰

3²⁰⁰ = ( 3²)¹⁰⁰= 9¹⁰⁰

Nên 2³⁰⁰<3²⁰⁰

B)4²¹ = (4³)⁷=64⁷

8¹⁴= ( 8²)⁷=64⁷

Nên 8¹⁴=4²¹

a) \(2^{300}=2^{3\cdot100}=8^{100}\)

\(3^{200}=3^{2\cdot100}=9^{100}\)

vì \(8^{100}< 9^{100}\Rightarrow2^{300}< 3^{200}\)

a, ta có :\(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)

lại có : \(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)

vì \(9^{100}>8^{100}=>2^{300}< 3^{200}\)

b,ta có : \(4^{21}=\left(2^2\right)^{21}=2^{42}\)

lại có :\(8^{14}=\left(2^3\right)^{14}=2^{42}\)

vì \(2^{42}=2^{42}=>4^{21}=8^{14}\)

c, ta có :\(31^{25}< 32^{25}=\left(2^5\right)^{25}=2^{75}\)

lại có : \(17^{19}>16^{19}=\left(2^4\right)^{19}=2^{76}\)

vì \(31^{25}< 2^{75}< 2^{76}< 17^{19}=>31^{25}< 17^{19}\)

Lời giải:
PT $\Leftrightarrow \frac{x-342}{15}-1+\frac{x-323}{17}-2+\frac{x-300}{19}-3+\frac{x-273}{21}-4=0$

$\Leftrightarrow \frac{x-357}{15}+\frac{x-357}{17}+\frac{x-357}{19}+\frac{x-357}{21}=0$

$(x-357)\left(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\right)=0$

Dễ thấy: $\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\neq 0$

$\Rightarrow x-357=0$

$\Rightarrow x=357$

A) =(5+(-15))+(9+(-19))+(-11)+(13+17)+(21+(-23))

=(-10)+(-10)+(-11)+20+(-2)

=((-10)+(-10))+(-11)+(-2)+20

=(-20)+(-11)+20+(-2)

=((-20)+20)+((-11)+(-2))

=0+(-13)=-13

B) doi ti !!!!! Tich cho mik nha

\(\dfrac{25^5+25^7+25^9}{5^{11}+5^{13}+5^{15}+5^{17}+5^{19}+5^{21}}\)

\(=\dfrac{5^{10}+5^{14}+5^{18}}{5^{11}+5^{13}+5^{17}+5^{19}+5^{21}}\)

\(=\dfrac{5^{10}\left(1+5^4+5^8\right)}{5^{11}\left(1+5^2+5^4+5^6+5^8+5^{10}\right)}\)

\(=\dfrac{1+5^4+5^8}{5+5^3+5^5+5^7+5^9+5^{11}}\)

\(=\dfrac{1+5^4+5^8}{\left(5+5^5+5^9\right)+\left(5^3+5^7+5^{11}\right)}\)

\(=\dfrac{1+5^4+5^8}{\left(1+5^4+5^8\right)\left(5+5^3\right)}\)

\(=\dfrac{1}{5+5^3}\)

\(=\dfrac{1}{130}\)

\(\)