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\(\dfrac{x-13}{87}-1+\dfrac{x-27}{73}-1+\dfrac{x-67}{33}-1+\dfrac{x-73}{27}-1=0\)
\(\Leftrightarrow\dfrac{x-100}{87}+\dfrac{x-100}{73}+\dfrac{x-100}{33}+\dfrac{x-100}{27}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{87}+\dfrac{1}{73}+\dfrac{1}{33}+\dfrac{1}{27}\ne0\right)=0\Leftrightarrow x=100\)
\(3,8276< \overline{3,8ab5}< 3,836\)
=>\(276< \overline{ab5}< 360\)
=>\(\left(a,b\right)\in\left\{\left(2;8\right);\left(2;9\right);\left(3;0\right);\left(3;1\right);\left(3;2\right);\left(3;3\right);\left(3;4\right);\left(3;5\right)\right\}\)
\(5,2863< \overline{5,a8b1}< 5,3841\)
=>\(2863< \overline{a8b1}< 3841\)
=>\(\left(a,b\right)\in\left\{\left(2;7\right);\left(2;8\right);\left(2;9\right);\left(3;0\right);\left(3;1\right);\left(3;2\right);\left(3;3\right)\right\}\)
Ta có \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{2022^2}< \dfrac{1}{2021.2022}\)
cộng vế với vế
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2022^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
Vậy ta có đpcm
\(\dfrac{x+4}{3}=\dfrac{x-11}{-6}\)
\(\dfrac{2x+8}{6}=\dfrac{-x+11}{6}\)
\(\Leftrightarrow2x+8=-x+11\)
\(\Leftrightarrow3x=3\)
\(\Leftrightarrow x=1\)
=>0,25x+2020=50+2020
=>0,25x=50
hay x=200