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Với x = 2014 => x + 1 = 2015
Ta có :
\(f\left(x\right)=x^{17}-2015x^{16}+2015x^{15}-2015x^{14}+....+2015x-1\)
<=> \(f\left(x\right)=x^{17}-\left(x+1\right)x^{16}+\left(x+1\right)x^{15}-\left(x+1\right)x^{14}+....+\left(x+1\right)x-1\)
<=> \(f\left(x\right)=x-1\)
<=> \(f\left(2014\right)=2014-1=2013\)
1) x (x-2016) + 2015 (2016-x) = 0
x (x-2016) - 2015 (x- 2016) = 0
(x-2015)(x-2016) =0
\(\Rightarrow\orbr{\begin{cases}x-2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}}\)
Vậy x= 2015; 2016
2) -5x (x-15) + (15-x) = 0
-5x (x-15) - (x-15) =0
(-5x -1) (x-15) =0
\(\Rightarrow\orbr{\begin{cases}-5x-1=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}-5x=1\\x=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{5}\\x=15\end{cases}}}\)
Vậy x= -1/5; 15
3) 3x (3x-7) - (7-3x) =0
3x(3x-7) + (3x -7) =0
(3x+1) (3x-7) =0
\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=-1\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{7}{3}\end{cases}}}\)
Vậy x= -1/3 ; 7/3
thay x=2014 ta có:
\(f\left(x\right)=2014^{17}-2015.2014^{16}+2015.2014^{15}-2015.2014^{14}+...+2015.2014-1 \)
=2014^17 - (2014+1).2014^16 + (2014+1).2014^15 - (2014+1).2014^14 + ... + (2014+1).2014-1
=2014^17 - 2014^17 - 2014^16 + 2014^16 + 2014^15 - 2014^15 + 2014^14 + ...-2014^3 - 2014^2 + 2014^2 + 2014 -1
=2014-1=2013
f)
\(A=\sqrt{\frac{\left(x+1\right)}{x-3}}=\sqrt{1+\frac{4}{x-3}}\)
x-3={-4)=> x=-1
\(-\frac{8}{15}< \frac{x}{15}< -\frac{2}{15}\)
=> \(-8< x< -2\)
=> \(x\in\left\{-7;-6;-5;-4;-3\right\}\)
Ta có : \(\frac{-8}{15}< \frac{x}{15}< \frac{-2}{15}\)
\(\Rightarrow-8< x< -2\)
\(\Rightarrow x\in\left\{-7;-6;-5;-4;-3\right\}\)
Vậy \(x\in\left\{-7;-6;-5;-4;-3\right\}\)
Ta có:
\(\left\{{}\begin{matrix}\left(x-15\right)^2=\left|x-15\right|^2\\\left|15-x\right|=\left|x-15\right|\end{matrix}\right.\)
\(\Leftrightarrow2015\left|x-15\right|+\left|x-15\right|^2-2014\left|x-15\right|=0\)
\(\Leftrightarrow\left|x-15\right|+\left|x-15\right|^2=0\Leftrightarrow\left|x-15\right|\left(\left|x-15\right|-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=15\\x=16\\x=14\end{matrix}\right.\)
thank bạn