Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
f(x)=ax^2+bx+c
=> f(1)= a + b + c
Mà f(1)= 3 nên a + b + c = 3 /1/
f(3) = 9a + 3b + c
Mà f(3)=5 => 9a + 3b + c = 5 /2/
f(5)= 25a + 5b + c
Mà f(5)=7 nên 25a + 5b + c = 7 /3/
Lấy /2/ - /1/, ta được:
8a + 2b = 2
<=> 2(4a + b) = 2
<=> 4a + b = 1 /4/
Lấy /3/ - /1/, ta được:
24a + 4 b = 4
<=> 4(6a + b) = 4
<=> 6a + b = 1 /5/
Lấy /5/ - /4/, ta được:
2a = 0
<=> a = 0
Thay a = 0 vào /4/, ta được:
4.0 + b = 1
<=> b = 1
Thay a = 0, b = 1 vào /1/, ta được:
0 + 1 + c = 3
<=> c = 2
=> a = 0, b = 1, c = 2
Vậy f(x) = 0.x^2 + x.1 + 2 = x + 2
\(.1.\)
Ta có : \(f\left(x\right)=\frac{2}{3}-\frac{1}{2}\left|x-1\right|\)
Thay
+ \(f\left(0\right)=\frac{2}{3}-\frac{1}{2}\left|0-1\right|\)
\(\Rightarrow f\left(0\right)=\frac{2}{3}-\frac{1}{2}\)
\(\Rightarrow f\left(0\right)=\frac{4}{6}-\frac{3}{6}\)
\(\Rightarrow f\left(0\right)=\frac{1}{6}\)
+ \(f\left(-1\right)=\frac{2}{3}-\frac{1}{2}\left|-1-1\right|\)
\(\Rightarrow f\left(-1\right)=\frac{2}{3}-\frac{1}{2}.2\)
\(\Rightarrow f\left(-1\right)=\frac{2}{3}-1\)
\(\Rightarrow f\left(-1\right)=\frac{2}{3}-\frac{3}{3}\)
\(\Rightarrow f\left(-1\right)=-\frac{1}{3}\)
+ \(f\left(1\right)=\frac{2}{3}-\frac{1}{2}\left|1-1\right|\)
\(\Rightarrow f\left(1\right)=\frac{2}{3}-\frac{1}{2}.0\)
\(\Rightarrow f\left(1\right)=\frac{2}{3}-0\)
\(\Rightarrow f\left(1\right)=\frac{2}{3}\)
+ \(f\left(\frac{3}{4}\right)=\frac{2}{3}-\frac{1}{2}\left|\frac{3}{4}-1\right|\)
\(\Rightarrow f\left(\frac{3}{4}\right)=\frac{2}{3}-\frac{1}{2}.\frac{1}{4}\)
\(\Rightarrow f\left(\frac{3}{4}\right)=\frac{2}{3}-\frac{1}{8}\)
\(\Rightarrow f\left(\frac{3}{4}\right)=\frac{16}{24}-\frac{3}{24}=\frac{13}{24}\)
a/ ta có: f(0)=9*02-2=-2
f(-1/3)=9*(-1/3)2-2=-1
f(\(3\sqrt{2}\)
\(f\left(-1\right)=-a+b-c+d=2\)
\(f\left(0\right)=d=1\)
\(f\left(\frac{1}{2}\right)=\frac{1}{8}a+\frac{1}{4}b+\frac{1}{2}c+d=3\)
\(f\left(1\right)=a+b+c+d=7\)
Suy ra \(\hept{\begin{cases}-a+b-c=1\\\frac{1}{8}a+\frac{1}{4}b+\frac{1}{2}c=2\\a+b+c=6\end{cases}}\Leftrightarrow\hept{\begin{cases}2b=7\\\frac{1}{8}a+\frac{1}{4}b+\frac{1}{2}c=2\\a+b+c=6\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{1}{3}\\b=\frac{7}{2}\\c=\frac{13}{6}\end{cases}}\)
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{97\cdot99}-\frac{5}{4}\cdot\frac{13}{99}+\frac{5}{99}\cdot\frac{1}{4}\)
\(A=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right)-\frac{13}{4}\cdot\frac{5}{99}+\frac{5}{99}\cdot\frac{1}{4}\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{5}{99}\cdot\left(\frac{13}{4}-\frac{1}{4}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)-\frac{5}{99}\cdot3\)
\(A=\frac{1}{2}\cdot\frac{32}{99}-\frac{5}{33}\)
\(A=\frac{16}{99}-\frac{5}{33}=\frac{1}{99}\)