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

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

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

6 tháng 1 2019

\(f\left(\frac{1}{3}\right)+2f\left(\frac{1}{\frac{1}{3}}\right)=\left(\frac{1}{3}\right)^2\Rightarrow f\left(\frac{1}{3}\right)+2f\left(3\right)=\frac{1}{9}\)(1)

\(f\left(3\right)+2f\left(\frac{1}{3}\right)=3^2\Rightarrow2f\left(3\right)+4f\left(\frac{1}{3}\right)=18\)(2)

Từ (1) và (2) \(\Rightarrow2f\left(3\right)+4f\left(\frac{1}{3}\right)-f\left(\frac{1}{3}\right)-2f\left(3\right)=18-\frac{1}{9}\)

\(\Rightarrow3f\left(\frac{1}{3}\right)=\frac{161}{9}\Rightarrow f\left(\frac{1}{3}\right)=\frac{161}{27}\)

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

20 tháng 11 2018

\(f\left(2.f\left(2015\right)\right)=2015.5-1\)

\(\Rightarrow f\left(2.50\right)=10074\Rightarrow f\left(100\right)=10074\)

20 tháng 11 2018

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: \(f\left(0\right)=7-3f\left(1\right)=7-3.\frac{12}{7}=\frac{13}{7}\)

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