chịu

\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}=\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{4}{7}\)

a: \(=\dfrac{\left(5^3\right)^3\cdot4^4}{5^{12}}=\dfrac{1}{5^3}\cdot4^4=\dfrac{4^4}{5^3}\)

b: \(=\dfrac{3^6}{\left[3^3\cdot2\right]^2}=\dfrac{1}{2^2}=\dfrac{1}{4}\)

c: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot5}=\dfrac{4}{5}\)

a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)

\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=5\left(1-\dfrac{1}{100}\right)\)

\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)

b, \(C=1.2.3+2.3.4+...+8.9.10\)

\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)

\(C=\dfrac{8.9.10.11}{4}=1980.\)

c, https://hoc24.vn/hoi-dap/question/384591.html

Câu này bạn vào đây mình đã giải câu tương tự nhé.

\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)

\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)

\(\Leftrightarrow A=\dfrac{99}{20}\)

\(1.a)\dfrac{2^3+3.26-4^3}{2^3.3^2}\)

\(=\dfrac{2^3.3.2.13-\left(2^2\right)^3}{2^3.3^2}\)

\(=\dfrac{2^4.3.13-2^6}{2^3.3^2}\)

\(=\dfrac{2^3\left(2.3.13-2^3\right)}{2^3.3^2}\)

\(=\dfrac{78-8}{9}\)

\(=\dfrac{70}{9}\)

\(b)\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)

\(=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^4.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)

\(=\dfrac{2^{12}.3^{10}+2^{13}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

\(=\dfrac{2^{12}.3^{10}\left(1+2.5\right)}{2^{11}.3^{11}\left(2.3\right)}\)

\(=\dfrac{2.11}{3.6}\)

\(=\dfrac{11}{9}\)

\(2.3^{x-1}-3^{x+1}=90\)

\(\Leftrightarrow3^x:3-3^x.3=90\)

\(\Leftrightarrow3^x\left(\dfrac{1}{3}-3\right)=90\)

\(\Leftrightarrow3^x.\dfrac{-8}{3}=90\)

\(\Leftrightarrow3^x=\dfrac{-135}{4}\)

\(\Leftrightarrow\) \(x\) không có giá trị nào để thỏa mãn đề bài.

Vậy \(x\in\varnothing\)

2:

\(3^{x-1}-3^{x+1}=90\)

\(\Rightarrow3^{x-2}-3^x=30\)

\(\Rightarrow3^x.3^{-2}-3^x=30\)

\(\Rightarrow3^x.\left(\dfrac{1}{9}-1\right)=30\)

\(\Rightarrow3^x.\dfrac{-8}{9}=30\)

\(\Rightarrow3^x=\dfrac{-135}{4}\)

\(\Rightarrow x=log_3\dfrac{-135}{4}\)

câu 2) là \(3^{x+1}-3^{x-1}=90\)

\(A=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{2^{12}\cdot3^4\cdot2}{2^{12}\cdot3^5\cdot4}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\cdot9}\)

\(=\dfrac{1}{6}-\dfrac{5\cdot\left(-6\right)}{9}=\dfrac{1}{6}+\dfrac{10}{3}=\dfrac{21}{6}=\dfrac{7}{2}\)

Sửa đề: \(C=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\right)^6\cdot3^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)

\(C=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{2^{12}\cdot3^4\cdot\left(3-1\right)}{2^{12}\cdot3^5\left(3+1\right)}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}\)

\(=\dfrac{2}{3\cdot4}-\dfrac{5\cdot\left(-6\right)}{9}\)

\(=\dfrac{2}{12}+\dfrac{30}{9}=\dfrac{1}{6}+\dfrac{10}{3}=\dfrac{1}{6}+\dfrac{20}{6}=\dfrac{21}{6}=\dfrac{7}{2}\)