\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

   =  \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

    = \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)

     = \(\frac{1}{4}+\frac{1}{2}\)

      =  \(\frac{3}{4}\)

b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)

    =\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)

   = \(-\frac{35}{27}+\frac{47}{21}\)

   =        \(\frac{178}{189}\)

c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)

  = \(\frac{117}{13}-\frac{311}{65}\)

 =       \(\frac{274}{65}\)

d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)

= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)

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

=     \(\frac{1}{3}+\frac{5}{2}\)

=         \(\frac{17}{6}\)

Tìm X

a) \(2x+\dfrac{3}{24}=3x-\dfrac{1}{32}\)

\(\Leftrightarrow\left(2x+\dfrac{3}{24}\right)-\left(3x-\dfrac{1}{32}\right)=0\)

\(\Leftrightarrow2x+\dfrac{3}{24}-3x+\dfrac{1}{32}=0\)

\(\Leftrightarrow\left(\dfrac{3}{24}+\dfrac{1}{32}\right)+\left(2x-3x\right)=0\)

\(\Leftrightarrow\dfrac{5}{32}-x=0\)

\(\Leftrightarrow x=\dfrac{5}{32}\)

1. So sánh

a) \(25^{50}\) và \(2^{300}\)

\(25^{50}=25^{1.50}=\left(25^1\right)^{50}=25^{50}\)

\(2^{300}=2^{6.50}=\left(2^6\right)^{50}=64^{50}\)

Vì \(25< 64\) nên \(25^{50}< 64^{50}\)

Vậy \(25^{50}< 2^{300}\)

b) \(625^{15}\) và \(12^{45}\)

\(625^{15}=625^{1.15}=\left(625^1\right)^{15}=625^{15}\)

\(12^{45}=12^{3.15}=\left(12^3\right)^{15}=1728^{15}\)

Vì \(625< 1728\) nên \(625^{15}< 1728^{15}\)

Vậy \(625^{15}< 12^{45}\)

1.So sánh

a)\(25^{50}\) và \(2^{300}\)

Ta có : \(2^{300}=\left(2^6\right)^{50}=64^{50}\)

Vì \(25^{50}< 64^{50}\) nên \(25^{50}< 2^{300}\)

b)\(625^{15}\) và \(12^{45}\)

Ta có : \(12^{45}=\left(12^3\right)^{15}=1728^{15}\)

Vì \(625^{15}< 1728^{15}\) nên \(625^{15}< 12^{45}\)

1) \(\frac{25}{12}.x+\frac{11}{15}=\frac{9}{10}\)

=> \(\frac{25}{12}.x=\frac{9}{10}-\frac{11}{15}\)

=> \(\frac{25}{12}.x=\frac{1}{6}\)

=> \(x=\frac{1}{6}:\frac{25}{12}\)

=> \(x=\frac{2}{25}\)

Vậy \(x=\frac{2}{25}\).

3) \(\frac{29}{12}.\left[x\right]-\frac{5}{6}=\frac{3}{8}\)

=> \(\frac{29}{12}.\left[x\right]=\frac{3}{8}+\frac{5}{6}\)

=> \(\frac{29}{12}.x=\frac{29}{24}\)

=> \(x=\frac{29}{24}:\frac{29}{12}\)

=> \(x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\).

4) \(\left[4x+\frac{3}{4}\right]-\frac{5}{4}=2\)

=> \(\left[4x+\frac{3}{4}\right]=2+\frac{5}{4}\)

=> \(4x+\frac{3}{4}=\frac{13}{4}\)

=> \(4x=\frac{13}{4}-\frac{3}{4}\)

=> \(4x=\frac{5}{2}\)

=> \(x=\frac{5}{2}:4\)

=> \(x=\frac{5}{8}\)

Vậy \(x=\frac{5}{8}\).

5) 2^x + 2^x+3 = 144

⇔ 2^x + 2^x . 2³ = 144

⇔ 2^x . (1 + 2³) = 144

⇔ 2^x . 9 = 144

⇔ 2^x = 144 : 9

⇔ 2^x = 16

⇔ 2^x = 2⁴

=> x = 4

Vậy x = 4.

Chúc bạn học tốt!

1)

a. \(\left(3x^2-50\right)^2=5^4\)

\(\Leftrightarrow3x^4-50=625\)

\(\Leftrightarrow3x^4=675\)

\(\Leftrightarrow x^4=225\)

\(\Leftrightarrow x=\sqrt{15}\) 

2)

a. \(\frac{\left(3^4-3^3\right)^4}{27^3}=\frac{3^{16}-3^{12}}{\left(3^3\right)^3}=\frac{3^{12}.3^4-3^{12}}{3^9}=\frac{3^{12}\left(3^4-1\right)}{3^9}\)

\(=\frac{3^{12}.80}{3^9}=3^3.80=27.80=2160\)

b. \(\frac{25^3}{\left(5^5-5^3\right)^2}=\frac{\left(5^2\right)^3}{5^{10}-5^6}=\frac{5^6}{5^6.5^4-5^6}=\frac{5^6}{5^6\left(5^4-1\right)}\)

\(=\frac{5^6}{5^6.624}=\frac{1}{624}\)

a) (2^x)⁵ : 4³ = 8¹⁵ => 2^5x = 8¹⁵.4³ = (2³)¹⁵.(2²)³ = 2⁴⁵.2⁶ = 2⁵¹ => 5x = 51 => x = 10,2 

b) (3²)^x .9³ = 243⁹ => 3^2x = 243⁹ : 9³ = (3⁵)⁹ : (3²)³ = 3⁴⁵ : 3⁶ = 3³⁹ => 2x = 39 => x = 19,5

c) (1/125)³.5^x = 25⁵ => 5^x = 25⁵ : (1/125)³ = (5²)⁵ : (1/5³)³ = 5¹⁰ : (5^-3)³ = 5¹⁰ : 5^-9 = 5¹⁹ => x = 19

d) 1/81 : 3^x = 1/729 => 3^x = 1/81 : 1/729 = 1/3⁴.729 = 3^-4.3⁶ = 3² => x = 2

e) (5x - 2)⁴ = 16⁸ = (16²)⁴ = 256⁴

=> 5x - 2 = -256 ; 256 => 5x = -254 ; 258 => x = -50,8 ; 51,6

P/S : Thay x = 10,2 vào câu a , x = 19,5 vào câu b sẽ thấy điều hư cấu : 2^10,2 và 9^19,5.Ko thể tính được giá trị của 2 lũy thừa này.

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Bài 1 :

a) \(\frac{12}{21}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{4}{7}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{1}{7}-\frac{2}{3}=-\frac{11}{21}\)

b) \(\left(-\frac{25}{13}\right)+\left(-\frac{9}{17}\right)+\frac{12}{13}+\left(-\frac{25}{17}\right)\)

\(=\left[\left(-\frac{25}{13}\right)+\frac{12}{13}\right]+\left[\left(-\frac{9}{17}\right)+\left(-\frac{25}{17}\right)\right]\)

\(=-1+\left(-2\right)=-1-2=-3\)

c) \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\cdot1=\frac{5}{9}\)

Bài 2 :

a)  \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)

=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}=-\frac{29}{70}\)

=> \(x=\left(-\frac{29}{70}\right):\frac{2}{3}=\left(-\frac{29}{70}\right)\cdot\frac{3}{2}=-\frac{87}{140}\)

b) \(x:\frac{5}{2}-\frac{1}{2}=-\frac{2}{3}\)

=> \(x:\frac{5}{2}=-\frac{2}{3}+\frac{1}{2}=-\frac{1}{6}\)

=> \(x=\left(-\frac{1}{16}\right)\cdot\frac{5}{2}=-\frac{5}{32}\)

c) Bạn chỉ cần xét hai trường hợp âm và dương thôi :>