đáp án 4 ảnh

\(\text{Ta có :}2^{91}=(2^{13})^7=8192^7\)

\(5^{35}=(5^5)^7=3125^7\)

Mà 8192 > 3125 nên 8192⁷ > 3125⁷ hoặc 2⁹¹ > 5³⁵

Vậy : 2⁹¹ > 5³⁵

TA có :

2⁹¹ = ( 2¹³ )⁷ = 8192⁷ ( 1 )

5³⁵ = ( 5⁵ )⁷ = 3125⁷ ( 2 )

Từ ( 1 ) và ( 2 ) => 2⁹¹ > 5³⁵

\(25.\left(-\frac{1}{5}\right)^3+\frac{1}{5}-2.\left(-\frac{1}{2}\right)^2-\frac{1}{3}\)

\(\Leftrightarrow25.\left(-\frac{1}{5}\right)^3-2.\left(-\frac{1}{2}\right)^2+\frac{1}{5}-\frac{1}{2}\)

\(\Leftrightarrow-\frac{1}{5}-2.\left(-\frac{1}{2}\right)^2+\frac{1}{5}-\frac{1}{2}\)

\(\Leftrightarrow-\frac{1}{5}+\frac{1}{5}-2\left(-\frac{1}{2}\right)^2-\frac{1}{2}\)

\(\Leftrightarrow-2.\left(-\frac{1}{2}\right)^2-\frac{1}{2}\)

\(\Leftrightarrow-\frac{1}{2}-\frac{1}{2}\)

\(\Leftrightarrow\frac{-1-1}{2}\)

\(\Rightarrow-1\)

\(\frac{1}{7}.\frac{-4}{9}+\frac{1}{7}.\frac{-14}{9}\)

\(=\frac{1}{7}.\orbr{ }\frac{-4}{9}+\frac{-14}{9}\)

\(=\frac{1}{7}.-2\)

\(=\frac{2}{7}\)

\(\frac{1}{7}\cdot-\frac{4}{9}+\frac{1}{7}\cdot-\frac{14}{9}\)

\(=\frac{1}{7}\cdot(-\frac{4}{9}+-\frac{14}{9})\)

\(=\frac{1}{7}\cdot-\frac{18}{9}\)

\(=\frac{1}{7}\cdot-2\)

\(=-\frac{2}{7}\)

Chúc bạn hok tốt ~

\(\frac{32}{2^x}=2\)

\(\Rightarrow\frac{2^5}{2^x}=2^1\)

\(\Rightarrow5-x=1\)

\(\Rightarrow x=5-1\)

\(\Rightarrow x=4\)

\(\frac{32}{2^x}=2\)

=> 2^x = 32 : 2

=> 2^x = 16

=> 2^x = 2⁴

=> x = 4

Năm mới vui vẻ nha!

Hk tốt

Mn giải giúp mình với ạ ! Mình đang cần gấp lắm ! =.=

Giải : Xét \(\Delta\)AMB và \(\Delta\)AMC

có AB = AC (gt)

    AM : chung

   MB = MC (gt)

=> \(\Delta AMB=\Delta AMC\)(c.c.c)

=> \(\widehat{CAM}=\widehat{MAB};\widehat{C}=\widehat{B};\widehat{CMA}=\widehat{AMB}\)(các cặp góc tương ứng)

Mà \(\widehat{CAM}+\widehat{MAB}=40^0\)(gt)

hay \(2.\widehat{CAM}=40^0\)

=> \(\widehat{CAM}=40^0:2\)

=> \(\widehat{CAM}=20^0\)=> \(\widehat{MAB}=20^0\)

Ta có : \(\widehat{CMA}+\widehat{BMA}=180^0\)(kề bù)

hay \(2.\widehat{CAM}=180^0\)

=> \(\widehat{CAM}=180^0:2\)

=> \(\widehat{CAM}=90^0\)

Xét \(\Delta\)AMB có \(\widehat{AMB}=90^0\)=> \(\widehat{C}+\widehat{CAM}=90^0\)(t/c của 1 tam giác)

=> \(\widehat{C}=90^0-\widehat{CAM}=90^0-20^0=70^0\)

Vì \(\widehat{C}=\widehat{B}\)=> \(\widehat{B}=70^0\)

Vậy ....