Cho hàm số f(x)\(=\frac{x}{2^x}\).Tìm \(x\inℕ^∗\)biết \(f\left(1\right)+f\left(2\right)+f\left(3\right)+...+f\left(x\right)=\frac{2^{x+1}-x}{2^x}-\frac{1}{512}\)
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S
4 tháng 9 2019
\(f\left(x\right)=\frac{x^2+2x+1-x^2}{x^2\left(x+1\right)^2}=\frac{\left(x+1\right)^2-x^2}{x^2\left(x+1\right)^2}=\frac{1}{x^2}-\frac{1}{\left(x+1\right)^2}\)
\(\Rightarrow f\left(1\right)+f\left(2\right)+....+f\left(x\right)=1-\frac{1}{2^2}+\frac{1}{2^2}-....-\frac{1}{\left(x+1\right)^2}\)
\(\Rightarrow\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x=\frac{x\left(x+2\right)}{\left(x+1\right)^2}\)
\(\Leftrightarrow\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x=\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-20+\left(x+1\right)=\frac{x\left(x+2\right)}{\left(x+1\right)^2}\)
Dat:\(x+1=a\Rightarrow\frac{\left(2y+1\right)a^3-20a^2-1}{a^2}=\frac{a^2-1}{a^2}\Leftrightarrow\left(2y+1\right)a^3-20a^2-1=a^2-1\)
\(\Leftrightarrow\left(2y+1\right)a^3-20a^2=a^2\Leftrightarrow\left(2ay+a\right)-20=1\left(coi:x=-1cophailanghiemko\right)\)
\(\Leftrightarrow2ay+a=21\Leftrightarrow a\left(2y+1\right)=21\Leftrightarrow\left(x+1\right)\left(2y+1\right)=21\)
16 tháng 4 2019
Ta có:
f(x)=\(\frac{x^2+2x+1-x^2}{x^2\left(x+1\right)^2}=\frac{\left(x+1\right)^2-x^2}{x^2\left(x+1\right)^2}=\frac{1}{x^2}-\frac{1}{\left(x+1\right)^2}\)
\(\Rightarrow f\left(1\right)=1-\frac{1}{2^2};f\left(2\right)=\frac{1}{2^2}-\frac{1}{3^2};...;f\left(x\right)=\frac{1}{x^2}-\frac{1}{\left(x-1\right)^2}\)
=> \(S=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{x^2}-\frac{1}{\left(x+1\right)^2}=1-\frac{1}{\left(x+1\right)^2}\)
Theo bài ra ta có :
\(1-\frac{1}{\left(x+1\right)^2}=\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x\)
<=> \(1-\frac{1}{\left(x+1\right)^2}=2y\left(x+1\right)-\frac{1}{\left(x+1\right)^2}-19+x\)
<=> 1=2y(x+1)-19+x
<=> (2y+1)(x+1)=21
x, y thuộc N => 2y+1, x+1 thuộc N
Ta có bảng
| x+1 | 3 | 1 | 7 | 21 |
| 2y+1 | 7 | 21 | 3 | 1 |
| x | 2 | 0 | 6 | 20 |
| y | 3 | 10 | 1 | 0 |
Vậy....
17 tháng 4 2019
Cô Linh Chi:
phần bảng x không có giá trị bằng 0
Nếu x = 0 thì hàm số f (x) có giá trị bằng 0
ML
30 tháng 6 2015
\(\text{1)}\)
\(\text{Thay }x=-2,\text{ ta có: }f\left(-2\right)-5f\left(-2\right)=\left(-2\right)^2\Rightarrow f\left(-2\right)=-1\)
\(\Rightarrow f\left(x\right)=x^2+5f\left(-2\right)=x^2-5\)
\(f\left(3\right)=3^2-5\)
\(\text{2)}\)
\(\text{Thay }x=1,\text{ ta có: }f\left(1\right)+f\left(1\right)+f\left(1\right)=6\Rightarrow f\left(1\right)=2\)
\(\text{Thay }x=-1,\text{ ta có: }f\left(-1\right)+f\left(-1\right)+2=6\Rightarrow f\left(-1\right)=2\)
\(\text{3)}\)
\(\text{Thay }x=2,\text{ ta có: }f\left(2\right)+3f\left(\frac{1}{2}\right)=2^2\text{ (1)}\)
\(\text{Thay }x=\frac{1}{2},\text{ ta có: }f\left(\frac{1}{2}\right)+3f\left(2\right)=\left(\frac{1}{2}\right)^2\text{ (2)}\)
\(\text{(1) - 3}\times\text{(2) }\Rightarrow f\left(2\right)+3f\left(\frac{1}{2}\right)-3f\left(\frac{1}{2}\right)-9f\left(2\right)=4-\frac{1}{4}\)
\(\Rightarrow-8f\left(2\right)=\frac{15}{4}\Rightarrow f\left(2\right)=-\frac{15}{32}\)
15 tháng 10 2020
Hướng dẫn:
Đặt: \(\frac{1}{x}=t\)( t khác 0; 1)
=> \(f\left(1-t\right)+2f\left(t\right)=\frac{1}{t}\)=> \(2f\left(1-t\right)+4f\left(t\right)=\frac{2}{t}\)(1)
Đặt: \(\frac{1}{x}=1-t\)
=> \(f\left(t\right)+2f\left(1-t\right)=\frac{1}{1-t}\)(2)
Lấy (1) - (2) => \(f\left(t\right)=\frac{1}{3}\left(\frac{2}{t}-\frac{1}{1-t}\right)\)
Vậy \(f\left(x\right)=\frac{1}{3}\left(\frac{2}{x}-\frac{1}{1-x}\right)\)
P/s: Chú ý điều kiện