\(\frac14+\frac14+\frac18-\frac14=\frac34\)
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\(P=\dfrac{\dfrac{8}{12}-\dfrac{3}{12}+\dfrac{5}{11}}{\dfrac{5}{12}+\dfrac{12}{12}-\dfrac{7}{11}}=\dfrac{\dfrac{5}{12}+\dfrac{5}{11}}{\dfrac{17}{12}-\dfrac{7}{11}}=\dfrac{115}{132}:\dfrac{103}{132}=\dfrac{115}{103}\)

\(Q=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2018}\right)\)
\(Q=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2017}{2018}\)
\(Q=\dfrac{1.2.3...2017}{2.3.4...2018}\)
\(Q=\dfrac{1}{2018}\)
Vậy \(Q=\dfrac{1}{2018}\)

\(Q=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right).........\left(1-\dfrac{1}{2018}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right).........\left(\dfrac{2018}{2018}-\dfrac{1}{2018}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.......\dfrac{2017}{2018}\)
\(=\dfrac{1}{2018}\)
Tính giá trị biểu thức
P=(x-a) (x-b) (x-c)... (x-y) (x-z) tại x=2018 câu khó hơn nhiều

a) \(\left(m-\dfrac{1}{4}\right)^3=\left(m^3-3m^2.\dfrac{1}{4}+3m\left(\dfrac{1}{4}\right)^2-\left(\dfrac{1}{4}\right)^3\right)\\ =\left(m^3-\dfrac{3}{4}m^2+\dfrac{3}{16}m-\dfrac{1}{64}\right)\)
b)\(\left(\dfrac{2}{3}-n\right)^3=\left(\dfrac{2}{3}\right)^3-3\left(\dfrac{2}{3}\right)^2n+3.\dfrac{2}{3}n^2-n^3\\ =\dfrac{8}{27}-\dfrac{4}{3}n+2n^2-n^3\)
c)\(m^3-125=m^3-5^3=\left(m-5\right)\left(m^2+5m+25\right)\)
d)\(m^3+\dfrac{1}{64}=m^3+\left(\dfrac{1}{4}\right)^3=\left(m+\dfrac{1}{4}\right)\left(m^2-\dfrac{1}{4}m+\dfrac{1}{16}\right)\)

a: \(=2\sqrt{6}\cdot3\sqrt{6}-4\sqrt{3}\cdot3\sqrt{6}+5\sqrt{2}\cdot3\sqrt{6}-\dfrac{1}{4}\cdot2\sqrt{2}\cdot3\sqrt{6}\)
\(=36-36\sqrt{2}+30\sqrt{3}-3\sqrt{3}\)
\(=36-36\sqrt{2}+27\sqrt{3}\)
b: \(=\left(-2\cdot\sqrt[3]{\dfrac{9}{5}}+4\cdot\sqrt[3]{\dfrac{1}{3}}\right):2\sqrt[3]{\dfrac{1}{3}}\)
\(=-\sqrt[3]{\dfrac{9}{5}:\dfrac{1}{3}}+2\cdot1\)
\(\simeq-1.75+2=0.25\)
3/8 đúng vì phải quy đồng mẫu số 8
\(\frac{33}{24}-\frac58\) bằng bao nhiêu vậy?