\(\left(\dfrac{9}{2}-2x\right)\cdot1\dfrac{4}{7}=\dfrac{11}{14}\\ \\\\ \Rightarrow\left(\dfrac{9}{2}-2x\right)\cdot\dfrac{11}{7}=\dfrac{11}{14}\\ \\\\ \Rightarrow\dfrac{9}{2}-2x=\dfrac{11}{14}:\dfrac{11}{7}=\dfrac{11}{14}\cdot\dfrac{7}{11}=\dfrac{1}{2}\\ \\ \\ \Rightarrow2x=\dfrac{9}{2}-\dfrac{1}{2}=\dfrac{8}{2}=4\\ \\ \\ \Rightarrow x=4:2=2\)

(\(\dfrac{9}{2}\) - 2\(x\)) \(\times\) 1\(\dfrac{4}{7}\) = \(\dfrac{11}{14}\)

( \(\dfrac{9}{2}\) - 2\(x\))  \(\times\) \(\dfrac{11}{7}\) =   \(\dfrac{11}{14}\)

\(\dfrac{9}{2}\)  - 2\(x\)              = \(\dfrac{11}{14}\) : \(\dfrac{11}{7}\)

\(\dfrac{9}{2}\) - 2\(x\)               = \(\dfrac{1}{2}\)

       2\(x\)               =   \(\dfrac{9}{2}\) - \(\dfrac{1}{2}\)

       2\(x\)               = 4

         \(x\)                = 4 : 2

          \(x\)               = 2

a) Số đo \(\widehat{xAy}\) là: 90^o vì có kí hiệu vuông góc.

b) Số đo \(\widehat{x'Ay}\):

Vì \(\widehat{x'Ay}\) và \(\widehat{xAy}\) là hai góc kề bù nên

nên \(\widehat{x'Ax}\) = \(\widehat{x'Ay}\) + \(\widehat{xAy}\)

       180^o = \(\widehat{x'Ay}\) + 90^o

        \(\widehat{x'Ay}\) = 180^o - 90^o

        \(\widehat{x'Ay}\) = 90^o

c) Số đo \(\widehat{x'Ay'}\):

Vì  \(\widehat{xAy}\) và \(\widehat{x'Ay'}\) là hai góc đối đỉnh 

nên: \(\widehat{x'Ay'}\) = \(\widehat{xAy}\) = 90^o

d) Số đo \(\widehat{xAy'}\):

Vì \(\widehat{xAy'}\) và \(\widehat{x'Ay}\) là hai góc đối đỉnh

nên \(\widehat{xAy'}\) = \(\widehat{x'Ay}\) = 90^o

(2^10*3^10.2^10*3^9):(2^9*3^10)

(\(\dfrac{1}{9}\))² = (\(\dfrac{1^2}{3^2}\))²= ((\(\dfrac{1}{3}\))²)²=  (\(\dfrac{1}{3}\))⁴

\(-\dfrac{1}{2}x-\dfrac{9}{12}=0,25\\ \\ \Rightarrow-\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{4}\\ \\ \Rightarrow-\dfrac{1}{2}x=\dfrac{1}{4}+\dfrac{3}{4}=1\\ \\ \Rightarrow x=1:\left(-\dfrac{1}{2}\right)=-2\)

  -\(\dfrac{1}{2}\)\(x\)  - \(\dfrac{9}{12}\) = 0,25

- \(\dfrac{1}{2}\)\(x\)            = 0,25 + \(\dfrac{9}{12}\)

-\(\dfrac{1}{2}\)\(x\)             = \(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)

-\(\dfrac{1}{2}\)\(x\)            = 1

   \(x\)             = -1 : \(\dfrac{1}{2}\)

   \(x\)           =- 2

Đổi 10cm= 1dm

Cạnh đáy bể:

20:4=5(dm)

Diện tích đáy bể:

5 x 5 = 25(dm²)

Thể tích viên đá:

25 x 1= 25(dm³)

Đ.số:25 dm³

2x-x⁰=3⁵:3³=3²=9

=>2x-1=9

=>2x=9+1=10

=>x=10:2=5.

\(2x-x^0=3^5:3^3\\ 2x-1=3^{5-3}\\ 2x-1=3^2\\ 2x-1=9\\ 2x=9+1\\ 2x=10\\ x=\dfrac{10}{2}\\ x=5\)

\(\dfrac{28^7.5^8.15^9}{14^8.25^9.6^{12}}=\dfrac{2^7.14^7.5^8.5^9.3^9}{14^7.14.\left(5^2\right)^9.3^{12}.2^{12}}\\ =\dfrac{14^7.5^{17}.3^9.2^7}{14^7.14.5^{17}.5.3^9.3^3.2^7.2^5}\\ =\dfrac{1}{14.5.3^3.2^5}=\dfrac{1}{14.5.27.32}\\ =\dfrac{1}{60480}\)

\(\dfrac{28^7.5^8.15^9}{14^8.25^9.6^{12}}\\ =\dfrac{2^7.14^7.5^8.5^9.3^9}{14^8.5^9.5^9.3^{12}.2^{12}}\\ =\dfrac{1}{14.5.3^3.2^5}\\ =\dfrac{1}{60480}\)

16⁵+2¹⁵=(2⁴)⁵+2¹⁵=2²⁰+2¹⁵=2¹⁵(1+2⁵)=2¹⁵x33⋮33

=>ĐPCM

Ok e

a) \(3^{54}\)

\(2^{200}=4^{100}>3^{54}\)

\(\Rightarrow3^{54}< 2^{200}\)

b) \(15^{12}=3^{12}.5^{12}\)

\(1^3.125^3=\left(5^3\right)^3=5^9< 3^{12}.5^{12}\)

\(\Rightarrow15^{12}>1^3.125^3\)

c) \(78^{12}-78^{11}=78^{11}.\left(7-1\right)=78^{11}.6\)

\(78^{11}-78^{10}=78^{10}.\left(7-6\right)=78^{10}.6< 78^{11}.6\)

\(\Rightarrow78^{12}-78^{11}>78^{11}-78^{10}\)

d) \(72^{45}-72^{44}=72^{44}.\left(72-1\right)=72^{44}.72>27^{44}\)

\(\Rightarrow72^{45}-72^{44}>27^{44}\)

e) \(3^{39}=\left(3^3\right)^{13}=27^{13}>11^{11}\)

\(\Rightarrow3^{39}>11^{11}\)

Giải chi tiết dùm mik nhé.