Câu 1:
\(A=\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)+\dfrac{4}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)\)
\(=\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{10}{15}\right)+\dfrac{4}{9}:\left(\dfrac{2}{22}-\dfrac{5}{22}\right)\)
\(=\dfrac{4}{9}:\dfrac{-9}{15}+\dfrac{4}{9}:\dfrac{-3}{22}\)
\(=\dfrac{4}{9}:\dfrac{-3}{5}+\dfrac{4}{9}.\dfrac{-22}{3}\)
\(=\dfrac{4}{9}.\dfrac{-5}{3}+\dfrac{4}{9}.\dfrac{-22}{3}\)
\(=\dfrac{4}{9}.\left(\dfrac{-5}{3}+\dfrac{-22}{3}\right)\)
\(=\dfrac{4}{9}.\dfrac{-27}{3}\)
\(=\dfrac{4}{9}.\left(-9\right)\)
\(=-4\)
\(b,\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\left(\dfrac{-8}{5}+x\right):\dfrac{12}{13}=\dfrac{13}{6}\)
\(\dfrac{-8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\dfrac{-8}{5}+x=2\)
\(x=2-\dfrac{-8}{5}\)
\(x=\dfrac{10}{5}+\dfrac{8}{5}\)
\(x=\dfrac{18}{5}\)
Vậy .........
Câu 2:
a: 2x=3y
=>\(\dfrac{x}{3}=\dfrac{y}{2}\)
=>\(\dfrac{x}{15}=\dfrac{y}{10}\left(2\right)\)
4y=5z
=>\(\dfrac{y}{5}=\dfrac{z}{4}\)
=>\(\dfrac{y}{10}=\dfrac{z}{8}\left(1\right)\)
Từ (1),(2) suy ra \(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}\)
mà x+y+z=11
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}=\dfrac{x+y+z}{15+10+8}=\dfrac{11}{33}=\dfrac{1}{3}\)
=>\(x=\dfrac{15}{3}=5;y=\dfrac{10}{3};z=\dfrac{8}{3}\)
b: \(C=3+3^2+...+3^{2023}\)
=>\(3C=3^2+3^3+...+3^{2024}\)
=>\(3C-C=3^2+3^3+...+3^{2024}-3-3^2-...-3^{2023}\)
=>\(2C=3^{2024}-3\)
=>\(2C+3=3^{2024}\)
=>\(3^n=3^{2024}\)
=>n=2024
c: \(\sqrt{17}>\sqrt{16}=4;\sqrt{26}>\sqrt{25}=5\)
Do đó: \(\sqrt{17}+\sqrt{26}>4+5=9\)
=>\(\sqrt{17}+\sqrt{26}+1>9+1=10\)
mà \(10=\sqrt{100}>\sqrt{99}\)
nên \(\sqrt{17}+\sqrt{26}+1>\sqrt{99}\)
