x-1/2019+x-2/2018=x-3/2017+x-4/2016
\(\dfrac{x-1}{2019}+\dfrac{x-2}{2018}=\dfrac{x-3}{2017}+\dfrac{x-4}{2016}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2019}-1\right)+\left(\dfrac{x-2}{2018}-1\right)=\left(\dfrac{x-3}{2017}-1\right)+\left(\dfrac{x-4}{2016}-1\right)\)
\(\Leftrightarrow\dfrac{x-2020}{2019}+\dfrac{x-2020}{2018}=\dfrac{x-2020}{2017}+\dfrac{x-2020}{2016}\)
\(\Leftrightarrow\dfrac{x-2020}{2019}+\dfrac{x-2020}{2018}-\dfrac{x-2020}{2017}-\dfrac{x-2020}{2016}\)
\(\Leftrightarrow\left(x-2020\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\right)=0\)
Mà \(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}e0\)
\(\Leftrightarrow x-2020=0\)
\(\Leftrightarrow x=2020\)
Tính nhanh:
a)\(\left(-\dfrac{1}{2}\right)\)-\(\left(-\dfrac{3}{5}\right)\)+\(\left(-\dfrac{1}{5}\right)\)+\(\dfrac{1}{131}\)-\(\left(-\dfrac{2}{7}\right)\)+\(\dfrac{4}{35}\)-\(\dfrac{7}{18}\)
Cho
\(\dfrac{4^x}{2^{x+y}}=8\) và \(\dfrac{9^{x+y}}{3^{5y}}=243\) (x,y \(\in\)N)
Tính xy
B= \(\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)\) - \(\left(5+\dfrac{1}{3}-\dfrac{6}{5}\right)\) - \(\left(6-\dfrac{7}{4}+\dfrac{3}{2}\right)\)
Tính giá trị cưa biểu thức có nhiều dấu ngoặc:
A= \(\left(6-\dfrac{2}{3}+\dfrac{1}{3}\right)\) - \(\left(5+\dfrac{5}{7}-\dfrac{3}{2}\right)\) - \(\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
A = \(\dfrac{1}{1x2}+\dfrac{1}{3x4}+\dfrac{1}{5x6}+-+\dfrac{1}{99x100}\)
CM \(\dfrac{7}{12}< A< \dfrac{5}{6}\)
Tính B=\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{n+1}\right)\) với \(n\in N\)
Tìm x biết : a) \(x^2-3=1\) b) \(2x^3+12=-4\) c) \(\left(2x-3\right)^2=16\) d) \(x:\left(\dfrac{-1}{2}\right)^1=\dfrac{-1}{2}\) h) \(\left(\dfrac{1}{4}-x\right)^2=\dfrac{25}{16}\) i) \(\left(x+\dfrac{1}{3}\right)^3=\dfrac{-1}{8}\) k) \(\left(\dfrac{1}{2}-x\right)^2=1\)
Tìm x,y thỏa mãn:
\(\left|x+y-7\right|+\left|x.y-10\right|\le0\)
Rút gọn biểu thức sau:
A= 42.252+32.125/23.52
Chứng minh rằng n3 - 3n2-n+3 chia hết cho 48 với mọi số nguyên lẻ n
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