Câu 1.
a. \(1,8.\dfrac{3}{4}+2\dfrac{1}{3}.3,5+8,2.\dfrac{3}{4}-3,5.\dfrac{1}{3}\)
\(=\dfrac{9}{5}.\dfrac{3}{4}+\dfrac{7}{3}.\dfrac{7}{2}+\dfrac{41}{5}.\dfrac{3}{4}-\dfrac{7}{2}.\dfrac{1}{3}\)
\(=\left(\dfrac{41}{5}+\dfrac{9}{5}\right).\dfrac{3}{4}+\left(\dfrac{7}{3}-\dfrac{1}{3}\right).\dfrac{7}{2}\)
\(=10.\dfrac{3}{4}+2.\dfrac{7}{2}\)
\(=\dfrac{15}{2}+7\)
\(=\dfrac{29}{2}\)
b. \(3^{10}:3^7+2.\left(-5\right)^2-4.\left(\dfrac{-1}{2011}\right)^0+41.\left(-1\right)^{11}\)
\(=3^3+2.25-4.1+41.\left(-1\right)\)
\(=27+50-4+(-41)\)
\(=32\)
Câu 2.
\(\dfrac{6}{7}+\dfrac{5}{8}:\left|x-2\right|-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{13}{56}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:\left|x-2\right|-\dfrac{3}{16}.4=\dfrac{13}{56}\)
\(\dfrac{5}{8}:\left|x-2\right|-\dfrac{3}{16}.4=\dfrac{13}{56}-\dfrac{6}{7}\)
\(\dfrac{5}{8}:\left|x-2\right|-\dfrac{3}{16}.4=\dfrac{-5}{8}\)
\(\left|x-2\right|-\dfrac{3}{16}.4=\dfrac{5}{8}:\dfrac{-5}{8}\)
\(\left|x-2\right|-\dfrac{3}{16}.4=-1\)
\(\left|x-2\right|-\dfrac{3}{4}=\dfrac{-1}{4}\)
\(\left|x-2\right|=\dfrac{-1}{4}+\dfrac{3}{4}\)
\(\left|x-2\right|=\dfrac{1}{2}\)
\(\Rightarrow x-2\in\left\{-\dfrac{1}{2};\dfrac{1}{2}\right\}\)
\(+)x-2=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+2\)
\(x=\dfrac{5}{2}\)
\(+)x-2=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}+2\)
\(x=\dfrac{3}{2}\)
Vậy x \(\in\left\{\dfrac{3}{2};\dfrac{5}{2}\right\}\)
