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22 tháng 11 2015

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

26 tháng 12 2021

D

26 tháng 12 2021

D

18 tháng 4 2017

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


18 tháng 4 2017

y = f (x)= x2 - 2

f (2) = 22 - 2 = 4 - 2 = 2

f (1) = 12 - 2 = 1 - 2 = -1

f (0) = 02 - 2 = 0 - 2 = -2

f (-1) = (-1)2 - 2 = 1 - 2 = -1

f (-2) = (-2)2 - 2 = 4 - 2 = 2

11 tháng 5 2017

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

26 tháng 6 2017

Có: \(f\left(x\right)=2x^2-5\)

\(\Rightarrow f\left(1\right)=2.1^2-5=-3\)

\(f\left(-2\right)=2.\left(-2\right)^2-5=3\)

\(f\left(0\right)=2.0^2-5=-5\)

\(f\left(2\right)=2.2^2-5=3\)

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:\(x+1=a\Rightarrow\frac{\left(2y+1\right)a^3-20a^2-1}{a^2}=\frac{a^2-1}{a^2}\Leftrightarrow\left(2y+1\right)a^3-20a^2-1=a^2-1\)

\(\Leftrightarrow\left(2y+1\right)a^3-20a^2=a^2\Leftrightarrow\left(2ay+a\right)-20=1\left(coi:x=-1cophailanghiemko\right)\)

\(\Leftrightarrow2ay+a=21\Leftrightarrow a\left(2y+1\right)=21\Leftrightarrow\left(x+1\right)\left(2y+1\right)=21\)

30 tháng 6 2015

\(\text{1)}\)

\(\text{Thay }x=-2,\text{ ta có: }f\left(-2\right)-5f\left(-2\right)=\left(-2\right)^2\Rightarrow f\left(-2\right)=-1\)

\(\Rightarrow f\left(x\right)=x^2+5f\left(-2\right)=x^2-5\)

\(f\left(3\right)=3^2-5\)

\(\text{2)}\)

\(\text{Thay }x=1,\text{ ta có: }f\left(1\right)+f\left(1\right)+f\left(1\right)=6\Rightarrow f\left(1\right)=2\)

\(\text{Thay }x=-1,\text{ ta có: }f\left(-1\right)+f\left(-1\right)+2=6\Rightarrow f\left(-1\right)=2\)

\(\text{3)}\)

\(\text{Thay }x=2,\text{ ta có: }f\left(2\right)+3f\left(\frac{1}{2}\right)=2^2\text{ (1)}\)

\(\text{Thay }x=\frac{1}{2},\text{ ta có: }f\left(\frac{1}{2}\right)+3f\left(2\right)=\left(\frac{1}{2}\right)^2\text{ (2)}\)

\(\text{(1) - 3}\times\text{(2) }\Rightarrow f\left(2\right)+3f\left(\frac{1}{2}\right)-3f\left(\frac{1}{2}\right)-9f\left(2\right)=4-\frac{1}{4}\)

\(\Rightarrow-8f\left(2\right)=\frac{15}{4}\Rightarrow f\left(2\right)=-\frac{15}{32}\)

17 tháng 4 2016

sai 1 chút chỗ cÂU 3

nhân vs 3 thì phải là 1/12

28 tháng 8 2017

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)\(y=5-2x\Rightarrow x=\dfrac{5y}{2}\)

\(y=5\Rightarrow x=\dfrac{5-5}{2}=0\)

\(y=3\Rightarrow x=\dfrac{5-3}{2}=1\)

\(y=-1\Rightarrow x=\dfrac{5-\left(-1\right)}{2}=\dfrac{5+1}{2}=3\)


17 tháng 1 2021

f(0) = 1

\(\Rightarrow\) a.02 + b.0 + c = 1 

\(\Rightarrow\) c = 1

Vậy hệ số a = 0; b = 0; c = 1

f(1) = 2

\(\Rightarrow\) a.12 + b.1 + c = 2

\(\Rightarrow\) a + b + c = 2

Vậy hệ số a = 1; b = 1; c = 1

f(2) = 4

\(\Rightarrow\) a.22 + b.2 + c = 4

\(\Rightarrow\) 4a + 2b + c = 4

Vậy hệ số a = 4; b = 2; c = 1

Chúc bn học tốt! (chắc vậy :D)

 

4 tháng 2 2021

\(f\left(-1\right)=2\Rightarrow-a+b-c+d=2\\ f\left(0\right)=1\Rightarrow d=1\\ f\left(1\right)=7\Rightarrow a+b+c+d=7\\ f\left(\dfrac{1}{2}\right)=3\Rightarrow\dfrac{1}{8}a+\dfrac{1}{4}b+\dfrac{1}{2}c+d=3\)

\(d=1\Rightarrow-a+b-c=1;a+b+c=6\\ \Rightarrow2b=7\\ \Rightarrow b=\dfrac{7}{2}\\ \Rightarrow\dfrac{1}{8}a+\dfrac{7}{8}+\dfrac{1}{2}c=2\\ \Rightarrow\dfrac{1}{2}\left(\dfrac{1}{4}a+\dfrac{7}{4}+c\right)=2\\ \Rightarrow\dfrac{1}{4}a+\dfrac{7}{4}+c=4\\ \Rightarrow a+7+4c=16\\ \Rightarrow a+4c=9;a+c=6-\dfrac{7}{2}=\dfrac{5}{2}\\ \Rightarrow3c=\dfrac{13}{2}\Rightarrow c=\dfrac{13}{6}\\ \Rightarrow a=\dfrac{5}{2}-\dfrac{13}{6}=\dfrac{1}{3}\)

Vậy \(\left(a;b;c;d\right)=\left(\dfrac{1}{3};\dfrac{7}{2};\dfrac{13}{6};1\right)\)