Tìm x biết \(|x+1|+|x+2|+|x+3|+...+|x+2016|=2017x\)
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a/ Với \(x=2016\Rightarrow2017=x+1\)
\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+2025\)
\(A=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+2025\)
\(A=2025-x=9\)
b/ Với \(x=-1\Rightarrow\left\{{}\begin{matrix}x^{2k}=1\\x^{2k+1}=-1\end{matrix}\right.\) ta có:
\(Q=2017-2016+2015-2014+...+3-2+1\)
\(Q=1+1+1+...+1+1\) (có \(\frac{2016}{2}+1=1009\) số 1)
\(Q=1009\)
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x=2016 nên x+1=2017
\(f\left(x\right)=x^{99}-x^{98}\left(x+1\right)+x^{97}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)-1\)
\(=x^{99}-x^{99}-x^{98}+x^{98}+x^{97}-x^{97}+...-x^3-x^2+x^2+x-1\)
=x-1=2015
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\(f\left(x\right)=x^{99}-2017x^{98}+2017x^{97}-...+2017x-1\)
\(=x^{99}-2016x^{98}-x^{98}+2016x^{97}+...-x^2+2016x+x-2016+2015\)
\(=x^{98}\left(x-2016\right)-x^{97}\left(x-2016\right)+...-x\left(x-2016\right)+\left(x-2016\right)+2015\)
\(=\left(x^{98}-x^{97}+...-x+1\right)\left(x-2016\right)+2015\)
\(\Rightarrow f\left(2016\right)=2015\)
Vậy...
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\(f\left(x\right)=x^{99}-2017^{x^{98}}+2017^{x^{97}}-...+2017x-1\)
\(f\left(2016\right)=2016^{99}-2017.2016^{98}+2017.2016^{97}-...+2017.2016-1\)
\(f\left(2016\right)=2016^{99}-\left(2016+1\right).2016^{98}+\left(2016+1\right).2016^{97}-...+\left(2016+1\right).2016-1\)
\(f\left(2016\right)=2016^{99}-2016^{99}-2016^{98}+2016^{98}+2016^{97}-2016^{97}-2016^{96}+...+2016^2+2016-1\)
\(f\left(2016\right)=2016-1\)
\(f\left(2016\right)=2015\)
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Lời giải:
Tại $x=2016$ thì $x-2016=0$
Khi đó:
$A=x^{2016}(x-2016)-x^{2015}(x-2016)+x^{2014}(x-2016)-x^{2013}(x-2016)+.....-x(x-2016)+x-2017$
$=x^{2016}.0-x^{2015}.0+......-x.0+2016-2017=2016-2017=-1$
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\(A=x^3+2x^2+3x\\ =x\left(x^2+2x+1\right)\\ =x\left(x+1\right)^2\\ =1999.\left(1999+1\right)=1999.2000\\ =3998000\)
\(B=x^4-2017x^3+2017x^2-2017x+2018\\ =x^4-2016x^3-x^3+2016x^2+x^2-2016x-x+2016+2\\ =x^3\left(x-2016\right)-x^3\left(x-2016\right)+x\left(x-2016\right)-\left(x-2016\right)+2\\ =\left(x-2016\right)\left(x^3+x-1\right)+2=0+2=0\)
Bạn xem lại đề câu a nhé , theo mk thì phải là 2 thì tính ms nhanh đc, 3 thì cũng giải đc nhưng ko hợp lí lắm
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Ta có: \(x^2-2017x+2016=0\)
\(\Rightarrow x^2-x-2016x+2016=0\)
\(\Rightarrow x\left(x-1\right)-2016\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x-2016\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=2016\end{cases}}}\)
Vậy \(x=\left\{1;2016\right\}\)
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Đk : \(2017x\ge0\Rightarrow x\ge0\)
Với x \(\ge0\)=> |x + 1| > 0 ; |x + 2| > 0 ; .... |x + 2016| > 0
Khi đó |x + 1| + |x + 2| + |x + 3| + ... + |x + 2016| (2016 hạng tử)
= x + 1 + x + 2 + x + 3 + ... + x + 2016
= 2016x + 2016.(2016 + 1):2
= 2016x + 2033136
Khi đó |x + 1| + |x + 2| + |x + 3| + ... + |x + 2016| = 2017x
<=> 2016x + 2033136 = 2017x
<=> x = 2033136 (tm)
Vậy x = 2033136
\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+.....+\left|x+2016\right|=2017x\)
Vì \(VT\ge0\)\(\Rightarrow VP\ge0\)\(\Rightarrow2017x\ge0\)\(\Rightarrow x\ge0\)
Vì \(x\ge0\)\(\Rightarrow\left|x+1\right|=x+1\); \(\left|x+2\right|=x+2\);..........; \(\left|x+2016\right|=x+2016\)
\(\Rightarrow x+1+x+2+x+3+.....+x+2016=2017x\)
\(\Leftrightarrow2016x+\left(1+2+3+.....+2016\right)=2017x\)
\(\Leftrightarrow x=1+2+3+........+2016=\frac{2016.\left(2016+1\right)}{2}=\frac{2016.2017}{2}=2033136\)
Vậy \(x=2033136\)