Câu 11:
a: \(A=\left(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\right):\dfrac{2021}{2022}+2023\)
\(=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right)\cdot\dfrac{2022}{2021}+2023\)
\(=\left(\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}\right)\cdot\dfrac{2022}{2021}+2023\)
\(=\left(\dfrac{2}{7}-\dfrac{2}{7}\right)\cdot\dfrac{2022}{2021}+2023\)
=2023
b: 4x=3y
=>\(\dfrac{x}{3}=\dfrac{y}{4}\)
5y=2z
=>\(\dfrac{y}{2}=\dfrac{z}{5}\)
=>\(\dfrac{y}{4}=\dfrac{z}{10}\)
=>\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{10}\)
mà 2x+3y-5z=8
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{10}=\dfrac{2x+3y-5z}{2\cdot3+3\cdot4-5\cdot10}=\dfrac{8}{-32}=-\dfrac{1}{4}\)
=>\(x=-\dfrac{1}{4}\cdot3=-\dfrac{3}{4};y=-\dfrac{1}{4}\cdot4=-1;z=-\dfrac{1}{4}\cdot10=-\dfrac{5}{2}\)
