x+2.x+3.x+4.x+....+2012.x=2012.2013
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BP
1
PL
7 tháng 9 2018
Ta có : x + (x + 1) + (x + 2) + .... + (x + 2012) = 2012.2013
<=> (x + x + x + ..... + x) + (1 + 2 + .... + 2012) = 2012.2013
<=> 2013x + \(\frac{2012.2013}{2}\) = 2012.2013
<=> 2013x = 2012.2013 - \(\frac{2012.2013}{2}\)
<=> 2013x = 2025078
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
16 tháng 9 2023
\(Bài.1:\\ a,104^2-16=104^2-4^2=\left(104+4\right)\left(104-4\right)=108.100=10800\\ b,9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\\ =\left(9.2\right)^8-\left(18^8-1\right)=18^8-18^8+1=1\\ c,999^3+3.999^2+3.999+1\\ =999^3+3.999^2.1+3.999.1^2+1^3=\left(999+1\right)^3=1000^3=1000000000\\ d,42^3-6.42^2+12.42-8\\ =42^3-3.42^2.2+3.42.2^2-2^3\\ =\left(42-2\right)^3=40^3=64000\)
KV
16 tháng 9 2023
Bài 1
a) 104² - 16
= 104² - 4²
= (104 - 4)(104 + 4)
= 100.108
= 10800
b) 9⁸.2⁸ - (18⁴ - 1)(18⁴ + 1)
= 18⁸ - (18⁸ - 1)
= 18⁸ - 18⁸ + 1
= 1
c) 999³ + 3.999² + 3.999 + 1
= (999 + 1)³
= 1000³
= 1000000000
d) 42³ - 6.42² + 12.42 - 8
= (42 - 2)³
= 40³
= 64000
PA
2
AT
6 tháng 9 2018
nốt ý b:
\(\left(x-1\right)^3+1+3x\left(x-4\right)=0\)
\(\Leftrightarrow x^3-3x^2+3x-1+1+3x^2-12x=0\)
\(\Leftrightarrow x^3-9x=0\Leftrightarrow x\left(x^2-9\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy ..............
DN
6 tháng 9 2018
\(a,x\left(x-2012\right)-2013x+2012.2013=0\)
\(=x\left(x-2012\right)+2013\left(-x+2012\right)=0\)
\(\Rightarrow x\left(x-2012\right)-2013\left(x-2012\right)=0\)
\(\Rightarrow\left(x-2013\right)\left(x-2012\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2013=0\\x-2012=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2013\\x=2012\end{matrix}\right.\)
Vậy...
27 tháng 4 2016
a) x+2x+3x+4x+...+2011x = 2012.2013
\(\Rightarrow\) x(1+2+3+4+...+2011) = 4050156
\(\Rightarrow\) x.2023066 = 4050156
\(\Rightarrow\) x = 4026/2011
23 tháng 5 2020
a)
PT <=> \(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)
<=> \(\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
<=> \(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\right)=0\)
Mà \(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\ne0\)
<=> x - 2013 = 0
<=> x = 2013
KL: ...
b) PT <=> \(\left(x^4-5x^3\right)+\left(5x^3-25x^2\right)-\left(5x^2-25x\right)+\left(6x-30\right)=0\)
<=> \(x^3\left(x-5\right)+5x^2\left(x-5\right)-5x\left(x-5\right)+6\left(x-5\right)=0\)
<=> \(\left(x-5\right)\left(x^3+5x^2-5x+6\right)=0\)
<=> \(\left(x-5\right)\left[\left(x^3+6x^2\right)-\left(x^2+6x\right)+\left(x+6\right)\right]=0\)
<=> \(\left(x-5\right)\left[x^2\left(x+6\right)-x\left(x+6\right)+\left(x+6\right)\right]=0\)
<=> \(\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)=0\)
<=> \(\left[{}\begin{matrix}x=5\\x=-6\\x=\varnothing\end{matrix}\right.\)
KL: ...
23 tháng 5 2020
a) Ta có: \(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)
\(\Leftrightarrow\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}-2012=0\)
\(\Leftrightarrow\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+\frac{x-3}{2010}-1+...+\frac{x-2012}{1}-1=0\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1\right)=0\)
mà \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1>0\)
nên x-2013=0
hay x=2013
Vậy: Tập nghiệm S={2013}
b) Ta có: \(x^4-30x^2+31x-30=0\)
\(\Leftrightarrow x^4+x-30x^2+30x-30=0\)
\(\Leftrightarrow\left(x^4+x\right)-\left(30x^2-30x+30\right)=0\)
\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x+1\right)-30\right]=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+6x-5x-30\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x+6\right)-5\left(x+6\right)\right]=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x+6\right)\left(x-5\right)=0\)(1)
Ta có: \(x^2-x+1\)
\(=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
hay \(x^2-x+1>0\forall x\)(2)
Từ (1) và (2) suy ra (x+6)(x-5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=5\end{matrix}\right.\)
Vậy: Tập nghiệm S={-6;5}
ND
1
27 tháng 5 2017
x^4-2012(x^3-x^2+x-1)
mà 2012=x
suy ra h(2012)=x^4-x.x^3+x.x^2-x.x+2012
=x^4-x^4+x^3-x^2+x
=x^3-x^2+x
=2012(2012^2-2012+1)
=2012(2012.2011+1)
=2012^2.2011+2012
x+2.x+3.x+4.x+....+2012.x=2012.2013
x(1+2+3+4+...+2012)=2012.2013
x(2012.2013:2)=2012.2013
x=2012.2013:(2012.2013:2)
x=2
x+2.x+3.x+4.x+....+2012.x=2012.2013
x.(1+2+3+4+...+2012)=2012.2013
Số số hạng từ 1 đến 2012 là:
(2012-1):1+1=2012(số hạng)
Tổng từ 1 đến 2012 là:
(2012+1)x2012:2=2025078
x.(1+2+3+4+...+2012)=2012.2013
x.2025078=4050156
x=4050156:2025078
x=2
Vậy x=2