`# \text {Ryo}`

So sánh ạ?

\(2^{225}\text{ và }3^{150}\)

Ta có:

\(2^{225}=2^{3\cdot75}=\left(2^3\right)^{75}=8^{75}\\ 3^{150}=3^{2\cdot75}=\left(3^2\right)^{75}=9^{75}\)

Vì `9 > 8 \Rightarrow`\(9^{75}>8^{75}\)

\(\Rightarrow2^{225}< 3^{150}\)

2²²⁵ = (2³)⁷⁵ = 8⁷⁵

3¹⁵⁰ = (3²)⁷⁵ = 9⁷⁵

Do 8 < 9 nên 8⁷⁵ < 9⁷⁵

Vậy 2²²⁵ < 3¹⁵⁰

a) 17/20 = 323/380

18/19 = 360/380

Do 323 < 360 nên 323/380 < 360/380

Vậy 17/20 < 18/19

b) 19/18 = 1 + 1/18

2023/2022 = 1 + 1/2022

Do 18 < 2022 nên 1/18 > 1/2022

1 + 1/18 > 1 + 1/2022

Vậy 19/18 > 2023/2022

c) 13/17 = 455/595

135/175 = 27/35 = 459/595

Do 455 < 459 nên 455/595 < 459/595

Vậy 13/17 < 135/175

d) 53/63 = 11236/13356

535/636 = 11235/13356

Do 11236 > 11235 nên 11236/13356 > 11235/13356

Vậy 53/63 > 535/636

e) 13/15 = 65/75

22/25 = 66/75

Do 65 < 66 nên 65/75 < 66/75

Vậy 13/15 < 22/25

8⁵ = (2³)⁵ = 2¹⁵ = 2.2¹⁴

3.4⁷ = 3.(2²)⁷ = 3.2¹⁴

Do 2 < 3 nên 2.2¹⁴ < 3.2¹⁴

Vậy 8⁵ < 3.4⁷

\(8^5=2^{15}=2.2^{14}\)

\(3.4^7=3.2^{14}>2.2^{14}\)

\(\Rightarrow8^5< 3.4^7\)

a) \(\dfrac{17}{20}< \dfrac{18}{20}< \dfrac{18}{19}\Rightarrow\dfrac{17}{20}< \dfrac{18}{19}\)

b) \(\dfrac{19}{18}>\dfrac{19+2024}{18+2024}=\dfrac{2023}{2022}\Rightarrow\dfrac{19}{18}>\dfrac{2023}{2022}\)

c) \(\dfrac{135}{175}=\dfrac{27}{35}\)

\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)

\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)

`# \text {Kaizu DN}`

`a)`

`(3x + 6) + (7x - 14) = 0?`

\(\Rightarrow3x+6+7x-14=0\\ \Rightarrow\left(3x+7x\right)+\left(6-14\right)=0\\ \Rightarrow10x-8=0\\ \Rightarrow10x=8\Rightarrow x=\dfrac{8}{10}\\ \Rightarrow x=\dfrac{4}{5}\)

Vậy, \(x=\dfrac{4}{5}\) 

`b)`

`17y + 35 + 4x + 17 = 42`

\(\Rightarrow\left(17y+17\right)+\left(35+4x\right)=42\\ \Rightarrow17\left(y+1\right)+\left(35+4x\right)=42\)

Bạn xem lại đề ;-;.

a) \(\dfrac{5}{3a-1}=1\)

\(\Rightarrow3a-1=5\)

\(\Rightarrow3a=6\)

\(\Rightarrow a=\dfrac{6}{3}=2\)

b) \(\dfrac{5}{3a-1}=-5\)

\(\Rightarrow3a-1=5:\left(-5\right)=-1\)

\(\Rightarrow3a=-1+1=0\)

\(\Rightarrow a=0:3=0\)

\(a,2\dfrac{1}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}-\left(-\dfrac{4}{7}\right)\\ \Rightarrow\dfrac{5}{2}-x+\dfrac{4}{5}=\dfrac{26}{21}\\ \Rightarrow\dfrac{5}{2}-x=\dfrac{46}{105}\\ \Rightarrow x=\dfrac{433}{210}\\ b,-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow2x-\dfrac{4}{7}-x=\dfrac{3}{5}\\ \Rightarrow2x-x=\dfrac{41}{35}\\ \Rightarrow x=\dfrac{41}{35}\\ c,\left(\dfrac{3}{8}-\dfrac{1}{5}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\\ \Rightarrow\dfrac{7}{40}+\dfrac{5}{8}-x=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{5}-x=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{3}{5}.\)

Đồng vị: \(\widehat{C_2}=\widehat{A_2};\widehat{E_2}=\widehat{A_2};\widehat{E_2}=\widehat{C_2}\)

So le trong: \(\widehat{C_4}=\widehat{A_2};\widehat{C_3}=\widehat{A_1};\widehat{A_3}=\widehat{E_1}\)

Trong cùng phía: \(\widehat{A_3}+\widehat{E_2}=180^o;\widehat{A_4}+\widehat{E_1}=180^o;\widehat{C_3}+\widehat{A_1}=180^o\)