\(\left(x-2\right)^{2016}=\left(x-2\right)^{2014}\)
\(\Rightarrow x-2\in\left\{0;1;-1\right\}\)
Nếu x - 2 = 0 => x = 2
Nếu x - 2 = 1 => x = 3
Nếu x - 2 = -1 => x = 1
\(\left(x-2\right)^{2016}=\left(x-2\right)^{2014}\)
\(\Rightarrow x-2\in\left\{0;1;-1\right\}\)
Nếu x - 2 = 0 => x = 2
Nếu x - 2 = 1 => x = 3
Nếu x - 2 = -1 => x = 1
a)\(\left(x^2-49\right)\left(x^2-81\right)< o\)
b)\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right).x=\frac{2015}{1}+\frac{2014}{2}+\frac{2013}{3}+...+\frac{2}{2014}+\frac{1}{2015}\)
giúp mk với, mai thi rồi, mk sẽ tích
Tìm x biết \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+....\dfrac{2}{x\cdot\left(x+1\right)}=\dfrac{2014}{2016}\)
\(b,\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+...+\left(x+\frac{1}{1024}\right)=1\)
\(a,\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(c,\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{x.x+3}=\frac{34}{103}\)
\(d,\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{x.\left(x+1\right):2}=\frac{2009}{2011}\)
HỘ MÌNH VỚI . GẤP . MAI MÌNH ĐI HỌC RỒI . THANKS TRƯỚC NÀ . MÌNH SẼ TICK CHO BẠN NÀO TRẢ LỜI ĐÚNG NHẤT VÀ NHANH NHẤT NHÉ
1.So sánh:
\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2014}\) và \(4\)
2. Tính :
\(\left(1-\frac{1}{2}+\frac{1}{3}+\frac{1}{2015}-\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\)
Tìm x:
a)\(2016x+\left(\dfrac{7}{12}+\dfrac{4}{21}+\dfrac{2}{24}+\dfrac{11}{30}+\dfrac{3}{40}+\dfrac{15}{56}\right)-\left(\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}\right)=0\)
b\(\dfrac{2x-1}{x+2015}-\dfrac{4025}{x+2017}=\dfrac{x-2014}{2x-4036}-\dfrac{x-2013}{2x-4030}\) (x thuộc N)
c)\(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{x\left(x+2\right)}\right)=\dfrac{4016}{2007}\)
AI GIÚP MK VỚI MK TICK CHO
Cho đẳng thức :\(x\times\left(x+1\right)\times\left(x+2\right)\times.............\times\left(x+2016\right)=2016\)(với x>0)
Chứng tỏ rằng \(x< \dfrac{1}{2015!}\)
+Tim x :
a ) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
b )\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
c ) \(\left(x+1\right)^{x+2}=\left(x-1\right)^{x+6}\) (x ϵ Z )
d ) \(\left(2x+3\right)^{2016}=\left(2x+3\right)^{2018}\)
e ) \(\frac{3}{x+2.x+5}+\frac{5}{x+5.x+10}+\frac{7}{x+10.x+17}=\frac{x}{x+2.x+17}\) Với \(x\notin\left\{-2;-5;-10;-17\right\}\)
Tìm x:
e/ \(\left(x-5\right)^4=\left(x-5\right)^6\)
i/ \(\left(x+2\right)^5=2^{10}\)
k/ \(\left(x+1\right)^2=\left(x+1\right)^0\)
l/ \(\left(3x-4\right).\left(x-1\right)^3+0\)
Làm phép tính sau khi bỏ dấu GTTĐ :
a) \(\left|x+\frac{2}{3}\right|+\left|x-3\right|\) biết rằng x\(\ge\)3
b) \(-\left|x+\frac{2}{5}\right|+\left|\frac{4}{3}-x\right|\)biết rằng x > 2