(x + 1)12 = 20160
(x + 1)12 = 1
(x + 1)12 = 112 = (-1)12
=> x + 1 thuộc { 1 ; -1}
=> x thuộc { 0 ; -2}
vậy x thuộc { 0 ; -2}
\(\left(x+1\right)^{12}=2016^0\)
\(\Rightarrow x=2016^0-1^{12}\)
\(\Rightarrow x=0\)
(x + 1)12 = 20160
(x + 1)12 = 1
(x + 1)12 = 112 = (-1)12
=> x + 1 thuộc { 1 ; -1}
=> x thuộc { 0 ; -2}
vậy x thuộc { 0 ; -2}
\(\left(x+1\right)^{12}=2016^0\)
\(\Rightarrow x=2016^0-1^{12}\)
\(\Rightarrow x=0\)
Cho đẳng thức: \(x.\left(x+1\right).\left(x+2\right).\left(x+3\right).....\left(x+2016\right)=2016\) (với\(x>0\) )
Chứng tỏ rằng: \(x< \frac{1}{2015!}\)
tìm x
a) \(\frac{x-1}{2}+\frac{x-2}{5}=\frac{1}{4}+\frac{x-7}{10}\)
b) \(3-\frac{2}{2x-3}=\frac{2}{5}+\frac{1}{2x-3}-\frac{3}{2}\)
c)\(7\cdot\left(x-1\right)+2x\cdot\left(1-x\right)=0\)
d) \(\frac{x+1}{2008}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{x+10}{2009}+\frac{x+11}{2008}+\frac{x+12}{2007}\)
e) \(\frac{2}{\left(x-1\right)\cdot\left(x-3\right)}+\frac{5}{\left(x-3\right)\cdot\left(x-8\right)}+\frac{12}{\left(x-8\right)\cdot\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
Bài 1 Tìm \(x\in z\)
a) \(5-4^{x+2}=2016^0\)
b) \(2^{x+1}.2^{2016}=2^{2017}\)
Bài 2 Tìm \(x\)
a) \(|x^2-19|=6\)
b) \(-6.\left(x+8\right)-5.\left(4-x\right)=12\)
c) \(\left(x+1\right).\left(x^2-4\right)=0\)
d) \(x^{15}=x\)
e) \(5⋮x+1\)
Giúp mik vs ạ, mik đag cần rất gấp
R = \(\left\{2015-2016^0.\left[2^3.5-\left(-1\right)^{2016}.\frac{1}{2^{19}}.\left(2.5^2-2^4.3\right)^{20}\right]\right\}-10^3\)
a, \(\left[x+1\right]^2+\left|y+1\right|=0\)
b, \(\left|x+7\right|+\left[y-2\right]^{2016}=0\)
Tìm x,y biết:
\(\left|x+\frac{11}{7}\right|+\left|x+\frac{2}{7}\right|+\left|x+\frac{4}{7}\right|=4x\)
\(\left(x-2\right)^{2014}+\left(y-1\right)^{2016}+\left(x-y-2\right)=0\)
mk đang cần gấp giúp mk nha các bn
Tìm x,y,z biết: \(\left(x-1\right)^{2016}\)+\(\left(2y-1\right)^{2016}\)\(|^{ }x+2y-z|^{2017}_{ }_{ }\)=0
Nhanh lên chìu nay mk phải đi hk rùi nhớ giải chi tiết nha
Cho đẳng thức
\(x.\left(x+1\right).\left(x+2\right).\left(x+3\right)...\left(x+2016\right)=2016.\)
Chứng tỏ rằng x < \(\frac{1}{2015!}\)
\(\left(x-3\right)^{2016}+\left(y-4\right)^{2018}=0\)
Tim x