[x+1]+[x+2]+....+[x+2017]=10092+6051
tim x
[x+1]+[x+3]+...+[x+2017]=10092+6051
Cho \(x>y\ge0\). CMR: \(P=2017\left[\frac{x^4+y^4}{x^4-y^4}-\frac{xy}{x^2-y^2}+\frac{x+y}{2\left(x-y\right)}\right]\ge\frac{6051}{2}\)
Dat \(A=\frac{x^4+y^4}{x^4-y^4}-\frac{xy}{x^2-y^2}+\frac{x+y}{2\left(x-y\right)}\)
\(=\frac{2x^4+2y^4-2xy\left(x^2+y^2\right)+\left(x+y\right)^2\left(x^2+y^2\right)}{2x^4-2y^4}\)
\(=\frac{2x^4+2y^4+\left(x^2+y^2\right)\left[\left(x+y\right)^2-2xy\right]}{2x^4-2y^4}\)
\(=\frac{2x^4+2y^4+\left(x^2+y^2\right)^2}{2x^4-2y^4}\)
\(\Rightarrow A\ge\frac{2x^4+x^4}{2x^4}=\frac{3}{2}\)
\(\Rightarrow P=2017A\ge2017.\frac{3}{2}=\frac{6051}{2}\)
Dau '=' xay ra khi \(y=0\)
Tìm số tự nhiên x biết::8060:4<x>6051:3
2016 nha bạn hiền ơi
TUI ĐOÁN BỪA : CHỌN C
Một hình lăng trụ có 36 cạnh . Hỏi hình lăng trụ đó có bao nhiêu mặt A. 2017 B. 6051 C. 4034 D. 6045.
Bài này đáp án là 14 mặt nha
Cho (x1)2 +(x2)2 +.....+ (x2017)2 = (x1+x2+......... +x2017) / 2017
CMR x1=x2=x3=...... = x2017
cho (x1)2 + ( x2 )2 + ..... + ( x2017 )2 = ( x1+x2+...+ x2017 ) 2 /2017
CMR x1=x2=....... =x2017
tìm GTNN
B=|x-1|+2|y+2017|-2010
C=|x-2|+|x-2009|+10
D=2017|x-1|+2016-2017|x|
E=|x-1|+|x-5|+|x-7|
Ta thấy : \(\left|x-1\right|\ge0\)
\(\left|y+2007\right|\ge0\)
\(\Rightarrow B=\left|x-1\right|=2\left|y+2007\right|-2010\ge-2010\)
\(MaxB=-2010\Leftrightarrow\hept{\begin{cases}x-1=0\\y+2007=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2007\end{cases}}}\)
a)có ng` lm r`
b)Áp dụng Bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)ta có:
\(C-10\ge\left|x-2+2009-x\right|=2007\)
\(\Rightarrow C\ge2017\)
Dấu = khi x=2 hoặc x=2009
Vậy MinC=2017 khi x=2 hoặc x=2009
c)Xét từng trường hợp và ta có:
MinD=-1 khi \(x\ge1\)
d)\(\left|x-1\right|+\left|x-5\right|+\left|x-7\right|\)
\(\ge\left|x-1+0+7-x\right|=6\)
\(\Rightarrow E\ge6\)
Dấu = khi \(\hept{\begin{cases}x-1\ge0\\x-5=0\\x-7\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x=5\\x\le7\end{cases}}\Leftrightarrow x=5\)
Vậy MinE=6 khi x=5
tìm x
a) x^2 - 2x =-1
b) x^2 + 2x + 1= 0
c) 4(x-1)^2 - (x-2)^2 = 3x^2
d) x(x-2017) - x^2 ( 2017-x) = 0
a/ \(x^2-2x=-1\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\Rightarrow x=1\)
Vậy..............
b/ \(x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\Rightarrow x=-1\)
Vậy.......
c/ \(4\left(x-1\right)^2-\left(x-2\right)^2=3x^2\)
\(\Leftrightarrow4\left(x^2-2x+1\right)-\left(x^2-4x+4\right)=3x^2\)
\(\Leftrightarrow4x^2-8x+4-x^2+4x-4-3x^2=0\)
\(\Leftrightarrow-4x=0\Rightarrow x=0\)
Vậy...................
d/ \(x\left(x-2017\right)-x^2\left(2017-x\right)=0\)
\(\Leftrightarrow x^2-2017x-2017x^2+x^3=0\)
\(\Leftrightarrow x^3-2016x^2-2017x=0\)
\(\Leftrightarrow x^3+x^2-2017x^2-2017x=0\)
\(\Leftrightarrow x\left(x^2+x\right)-2017\left(x^2+x\right)=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x-2017\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\\x-2017=0\Rightarrow x=2017\end{matrix}\right.\)
Vậy pt có 3 nghiệm là.....(tự ghi ra)
\(a,x^2-2x=-1\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
\(b,x^2+2x+1=0\)
\(\Leftrightarrow\left(x+2\right)^2=0\)
\(\Rightarrow x+2=0\Rightarrow x=-2\)
\(c,4\left(x-1\right)^2-\left(x-2\right)^2=3x^2\)
\(\Leftrightarrow4\left(x^2-2x+1\right)-\left(x^2-4x+4\right)-3x^2=0\) \(\Leftrightarrow4x^2-8x+4-x^2+4x-4-3x^2=0\)
\(\Leftrightarrow-4x=0\Rightarrow x=0\)
\(d,x\left(x-2017\right)-x^2\left(2017-x\right)=0\)
\(\Leftrightarrow x^2-2017x-2017x^2+x^3=0\)
\(\Leftrightarrow x^3+x^2-2017x-2017=0\)
\(\Leftrightarrow x^2\left(x+1\right)-2017\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-2017\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x^2-2107=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x^2=2017\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\\left[{}\begin{matrix}x=\sqrt{2017}\\x=-\sqrt{2017}\end{matrix}\right.\end{matrix}\right.\)