a) 16. x= x^4
b) 2^x+ 2^x+1 = 48
c)x^2018=x^2015
Tìm x,biết:
x+2015/5 + x+2014/6 = x+2017/3 + x+2018/2
Hướng dẫn: x+2015/5+1 + x+2014/6+1 = x+2017/3+1 + x+2018/2+1
=> (x+2020)/5=(x+2020)/6=(x+2020)/3+(x+2020)/2
=>(x+2020)(1/5+1/6)=(x+2020)(1/3+1/2)
Với x+2020=0=>x=-2020
Với x+2020 khác 0=>1/5+1/6=1/3+1/2 ,vô lí
Vậy x=-2020
Tìm N(2017) biết đa thức N(x)=\(x^{2017}-2018.x^{2016}+2018.x^{2015}-2018.x^{2014}+........-2018.x^2+2018.x-1\)
Ta có: \(N\left(x\right)=x^{2017}-2018x^{2016}+2018x^{2015}-...-2018x^2+2018x-1\)
\(=x^{2017}-2018\left(x^{2016}-x^{2015}+...+x^2-x\right)-1\)
\(\Rightarrow N\left(2017\right)=2017^{2017}-2018\left(2017^{2016}-2017^{2015}+...+2017^2-2017\right)-1\)
Đặt \(A=2017^{2016}-2017^{2015}+...+2017^2-2017\)
\(\Rightarrow2017A=2017^{2017}-2017^{2016}+...+2017^3-2017^2\)
\(\Rightarrow2018A=2017^{2017}-2017\)
\(\Rightarrow A=\dfrac{2017^{2017}-2017}{2018}\)
\(\Rightarrow N\left(2017\right)=2017^{2017}-2018.\dfrac{2017^{2017}-2017}{2018}-1\)
\(=2017^{2017}-\left(2017^{2017}-2017\right)-1\)
\(=2017^{2017}-2017^{2017}+2017-1\)
\(=2016\)
Vậy N(2017) = 2016
1. Cho biểu thức B :
\(B=x^{2017}-2018.x^{2016}+2018.x^{2015}-2018.x^{2014}+...-2018.x^2+2018.x-1\)
TÍNH GIÁ TRỊ BIỂU THỨC VỚI x=2017
mik chưa dc học dạng này sr
a/ 2b -√b2−4b+4b−2
b/ |x+4| - x+4√x2+8x+16
c/√4−4a+a2−2a với -4 ≤x≤ 2
d/|x+4| - x+4√x2+8x+16
e/√4x^2-4x+1/2x-1với x<1/2
f/|x|+x√x2
với x>0
1, Cho (2x1-3y1)2018+(2x2-3y2)2018+...+(2x2015-3y2015)2018 lớn hơn bằng 0
Tính A = x1+x2+....+x2015 / y1+y2+...+y2015
S = 1 x 2 x 3 x 4 + 2 x 3 x 4 x5 + ... + 2015 x 2016 x 2017 x 2018
tìm x biết
a, (1/1x2+1/2x3+1/5x4+...+1/99x100) X=1/1x2+2x3+3x4+...+98x99
b, X/1x3+X/3x5+X/5x7+...+X/2013x2015=4/2015
c, X+1/2015+X+2/2016=X+3/2017+X+4/2018
b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)
Phương trình tương đương với:
\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)
c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)
\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)
\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
\(\left(\right. \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \hdots + \frac{1}{99 \times 100} \left.\right) \times X = 1 \times 2 + 2 \times 3 + \hdots + 98 \times 99\)
Bên trái: là dãy rút gọn:Vậy:
\(\frac{99}{100} \times X = 321860 \Rightarrow X = 321860 \times \frac{100}{99} = \boxed{325000}\)
Câu b:\(\frac{X}{1 \times 3} + \frac{X}{3 \times 5} + \hdots + \frac{X}{2013 \times 2015} = \frac{4}{2015}\)
Có 1008 phân số.Mỗi phân số rút gọn được thành:\(\frac{1}{\left(\right. 2 k - 1 \left.\right) \left(\right. 2 k + 1 \left.\right)} = \frac{1}{2} \left(\right. \frac{1}{2 k - 1} - \frac{1}{2 k + 1} \left.\right)\)
Cộng lại sẽ ra:
\(X \times \frac{1007}{2015} = \frac{4}{2015} \Rightarrow X = \boxed{\frac{4}{1007}}\)
S = 1 x 2 x 3 x 4 + 2 x 3 x 4 x 5 + ... + 2015 x 2016 x 2017 x 2018