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f (1) = 2 . 12 - 5 = -3
f (-2) = 2 . (-2)2 - 5 = 3
f (0) = 2 . 02 - 5 = -5
f (2) = 2 . 22 - 5 = 3
Có: \(f\left(x\right)=2x^2-5\)
\(\Rightarrow f\left(1\right)=2.1^2-5=-3\)
\(f\left(-2\right)=2.\left(-2\right)^2-5=3\)
\(f\left(0\right)=2.0^2-5=-5\)
\(f\left(2\right)=2.2^2-5=3\)
Ta có: y=f(x)=x2−2y=f(x)=x2−2
Thay f(2); f(1); f(0); f(-1); f(-2) vào hàm số:
f(2)=22−2=4−2=2f(2)=22−2=4−2=2
f(1)=12−2=1−2=−1f(1)=12−2=1−2=−1
f(0)=02−2=−2f(0)=02−2=−2
f(−1)=(−1)2−2=1−2=−1f(−1)=(−1)2−2=1−2=−1
f(−2)=(−2)2−2=4−2=2
a) f(-2)=5 – 2. (-2) = 5 + 4 = 9;
f(-1) = 5 – 2.(-1) = 5 + 2 = 7;
f(0) = 5 – 2.0 = 5;
f(3) = 5 – 2.3 = 5 – 6 = -1.
b)\(y=5-2x\Rightarrow x=\dfrac{5y}{2}\)
\(y=5\Rightarrow x=\dfrac{5-5}{2}=0\)
\(y=3\Rightarrow x=\dfrac{5-3}{2}=1\)
\(y=-1\Rightarrow x=\dfrac{5-\left(-1\right)}{2}=\dfrac{5+1}{2}=3\)
\(\text{1)}\)
\(\text{Thay }x=-2,\text{ ta có: }f\left(-2\right)-5f\left(-2\right)=\left(-2\right)^2\Rightarrow f\left(-2\right)=-1\)
\(\Rightarrow f\left(x\right)=x^2+5f\left(-2\right)=x^2-5\)
\(f\left(3\right)=3^2-5\)
\(\text{2)}\)
\(\text{Thay }x=1,\text{ ta có: }f\left(1\right)+f\left(1\right)+f\left(1\right)=6\Rightarrow f\left(1\right)=2\)
\(\text{Thay }x=-1,\text{ ta có: }f\left(-1\right)+f\left(-1\right)+2=6\Rightarrow f\left(-1\right)=2\)
\(\text{3)}\)
\(\text{Thay }x=2,\text{ ta có: }f\left(2\right)+3f\left(\frac{1}{2}\right)=2^2\text{ (1)}\)
\(\text{Thay }x=\frac{1}{2},\text{ ta có: }f\left(\frac{1}{2}\right)+3f\left(2\right)=\left(\frac{1}{2}\right)^2\text{ (2)}\)
\(\text{(1) - 3}\times\text{(2) }\Rightarrow f\left(2\right)+3f\left(\frac{1}{2}\right)-3f\left(\frac{1}{2}\right)-9f\left(2\right)=4-\frac{1}{4}\)
\(\Rightarrow-8f\left(2\right)=\frac{15}{4}\Rightarrow f\left(2\right)=-\frac{15}{32}\)
\(f\left(x\right)=\frac{x^2+2x+1-x^2}{x^2\left(x+1\right)^2}=\frac{\left(x+1\right)^2-x^2}{x^2\left(x+1\right)^2}=\frac{1}{x^2}-\frac{1}{\left(x+1\right)^2}\)
\(\Rightarrow f\left(1\right)+f\left(2\right)+....+f\left(x\right)=1-\frac{1}{2^2}+\frac{1}{2^2}-....-\frac{1}{\left(x+1\right)^2}\)
\(\Rightarrow\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x=\frac{x\left(x+2\right)}{\left(x+1\right)^2}\)
\(\Leftrightarrow\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x=\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-20+\left(x+1\right)=\frac{x\left(x+2\right)}{\left(x+1\right)^2}\)
Dat:\(x+1=a\Rightarrow\frac{\left(2y+1\right)a^3-20a^2-1}{a^2}=\frac{a^2-1}{a^2}\Leftrightarrow\left(2y+1\right)a^3-20a^2-1=a^2-1\)
\(\Leftrightarrow\left(2y+1\right)a^3-20a^2=a^2\Leftrightarrow\left(2ay+a\right)-20=1\left(coi:x=-1cophailanghiemko\right)\)
\(\Leftrightarrow2ay+a=21\Leftrightarrow a\left(2y+1\right)=21\Leftrightarrow\left(x+1\right)\left(2y+1\right)=21\)
Ta có : \(y=f\left(x\right)=2x^2-3x+1\)
\(f\left(-1\right)=2\left(-1\right)^2-3.\left(-1\right)+1=2.1-\left(-3\right)+1=2+3+1=6\)
\(f\left(2\right)=2.2^2-3.2+1=2.4-6+1=8-6+1=3\)
\(f\left(\frac{-1}{2}\right)=2\left(\frac{1}{2}\right)^2-3.\frac{1}{2}+1=2.\frac{1}{4}-\frac{3}{2}+1=\frac{1}{2}-\frac{3}{2}+\frac{2}{2}=0\)
Ta có: \(\left(0+1\right).f\left(0\right)+3f\left(1-0\right)=2.0+7\)
\(\Rightarrow f\left(0\right)+3f\left(1\right)=7\Rightarrow3f\left(0\right)+9f\left(1\right)=21\) (1)
\(\left(1+1\right)f\left(1\right)+3f\left(1-1\right)=2.1+7\)
\(\Rightarrow2f\left(1\right)+3f\left(0\right)=9\)(2)
Từ (1) và (2) ta được: \(3f\left(0\right)+9f\left(1\right)-2f\left(1\right)-3f\left(0\right)=21-9\)
\(\Rightarrow7f\left(1\right)=12\Rightarrow f\left(1\right)=\frac{12}{7}\)
Khi đó: \(f\left(0\right)=7-3f\left(1\right)=7-3.\frac{12}{7}=\frac{13}{7}\)
a) theo tính chất ta có: f(0+0)= f(0)+f(0)
=> f(0)=f(0)+f(0)
=> f(0)-f(0)=f(0)+f(0)-f(0)
=> 0=f(0)
hay f(0)=0
b) f(0)=f(-x+x)=f(-x)+f(x)
=>0=f(-x)+f(x)
=> f(-x)=0-f(x)=-f(x)
c) \(f\left(x_1-x_2\right)=f\left(x_1+\left(-x_2\right)\right)=f\left(x_1\right)+f\left(-x_2\right)=f\left(x_1\right)-f\left(x_2\right)\)
f(0) = 2.0 - 1 = 0 - 1 = -1
f(-1) = 2.(-1) - 1 = -2 - 1 = -3
f(1) = 2.1 - 1 = 2 - 1 = 1
f(10) = 2.10 - 1 = 20 - 1 = 19