Tính F
\(F=2017.2018-\sqrt{2017^2+2018^2+2017^2.2018^2}\)
\(\frac{1+2017\sqrt{2018}\:-2018\sqrt{2017}}{\sqrt{2017\:\:}+\sqrt{2018}+\sqrt{2017}\cdot\sqrt{2018}}=\sqrt{2017.2018\:}\)
Thực hiện phép tính
a)\(\sqrt{1+\left(\dfrac{1}{2017}+\dfrac{1}{2019}\right)^2}\)
b)\(\sqrt{2017^2+2017^2.2018^2+2018^2}\)
a: Đặt a=2017
\(A=\sqrt{1+\left(\dfrac{1}{a}+\dfrac{1}{a+2}\right)^2}\)
\(=\sqrt{1+\left(\dfrac{2a+2}{a\left(a+2\right)}\right)^2}\)
\(=\sqrt{1+\dfrac{4a^2+8a+4}{a^2\cdot\left(a+2\right)^2}}=\sqrt{\dfrac{\left(a^2+a\right)^2+4a^2+8a+4}{a^2\left(a+2\right)^2}}\)
\(=\sqrt{\dfrac{\left(a^2+a\right)^2+4\left(a+1\right)^2}{a^2\left(a+2\right)^2}}\)
\(=\dfrac{\sqrt{\left(a^2+a\right)^2+4\left(a+1\right)^2}}{a\left(a+2\right)}\)
\(=\dfrac{\sqrt{\left(2017^2+2017\right)^2+4\cdot2018^2}}{2017\cdot2019}\)
b: Đặt 2017=a
\(B=\sqrt{a^2+a^2\cdot\left(a+1\right)^2+\left(a+1\right)^2}\)
\(=\sqrt{2a^2+2a+1+\left(a^2+a\right)^2}\)
\(=\sqrt{\left(a^2+a+1\right)^2}=a^2+a+1\)
\(=2017^2+2017+1=4070307\)
\(\sqrt{2017^2+2017.2018+2018^2}\)
Cho đa thức F(x)= a.x^2+b.x+c biết F(0)= 2016, F(1)= 2017, F(-1)= 2018. Tính F(2).
Xét đa thức \(F\left(x\right)=ax^2+bx+c\)
\(F\left(0\right)=c=2016\)
\(F\left(1\right)=a+b+c=2017\Rightarrow a+b=1\) (1)
\(F\left(-1\right)=a-b+c=2018\Rightarrow a-b=2\) (2)
Từ (1), (2)
\(\Rightarrow\hept{\begin{cases}a+b-a+b=-1\\a+b+a-b=3\end{cases}}\Rightarrow\hept{\begin{cases}2b=-1\\2a=3\end{cases}}\Rightarrow\hept{\begin{cases}b=-0,5\\a=1,5\end{cases}}\)
\(\Rightarrow F\left(2\right)=1,5.2^2-0,5.2+2016=2021\)
Vậy \(F\left(2\right)=2021\).
(1.2+2.3+......+2017.2018+2018.2019)-(1 mũ2+2 mũ2+.......+2017 mũ2 +2018 mũ 2
(1.2 + 2.3 + 3.4 + ... + 2018.2019) - (12 + 22 + ... + 20182)
= (1.2 + 2.3 + ... + 2018.2019) - (1.1 + 2.2 + ... + 2018.2018)
= (1.2 + 2.3 + ... + 2018.2019) - [1.(2 - 1) + 2.(3 - 1) + ... + 2018.(2019 - 1)]
= (1.2 + 2.3 + ... + 2018.2019) - (1.2 + 2.3 + ... + 2018.2019 - 1 - 2 - 3 - ... - 2018)
= (1.2 + 2.3 + ... + 2018.2019) - [1.2 + 2.3 + ... + 2018.2019 - (1 + 2 + ... + 2018)]
= (1.2 + 2.3 + ... + 2018.2019) - (1.2 + 2.3 + ... + 2018.2019) + (1 + 2 + 3 + ... + 2018)
= 1 + 2 + ... + 2018 (có : (2018 - 1) : 1 + 1 = 2018 (số))
= (2018 + 1).2018 : 2
= 2037171
cho f(x)=x^3-3x^2+3x+3 cm f(2018/2017)< f(2017/2016)
cho đa thức F(x) = ax2 + bx + c. Biết F(0) = 2017; F(1) = 2018; F(-1) = 2019. Tính F(2)?
Theo đề bài f(0)= 2017 => c= 2017
f(1)= 2018 => a + b + c = 2018 => a + b = 1 (1)
f(-1)= 2019 => a - b + c= 2019 => a - b= 2 (2)
Cộng theo vế của (1) và (2), ta được
2a = 3 => a = 3/2
=>b= -1/2
Vậy a=3/2, b=-1/2, c= 2017. Khi đó f(2)= 6 - 2 + 2017= 2021
Vậy f(2)= 2021
À nhầm, dòng thứ 2 từ dưới lên phải là f(2)= 6 - 1 + 2017= 2022 nha, mình nhấn nhầm
Tính giá trị của biểu thức: \(A=\sqrt{1+2017^2+\dfrac{2017^2}{2018^2}}+\dfrac{2017}{2018}\)
Đặt \(2017=a\)
\(A=\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1\right)^2-2a+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1\right)^2-2\left(a+1\right)\cdot\dfrac{a}{a+1}+\left(\dfrac{a}{a+1}\right)^2}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1-\dfrac{a}{a+1}\right)^2}+\dfrac{a}{a+1}\\ A=\left|a+1-\dfrac{a}{a+1}\right|+\dfrac{a}{a+1}\)
Ta có \(\dfrac{a}{a+1}< 1\Leftrightarrow a+1-\dfrac{a}{a+1}>0\)
\(\Leftrightarrow A=a+1-\dfrac{a}{a+1}+\dfrac{a}{a+1}=a+1=2018\)
Cho f(x)= x x 2 + 1 ( 2 x 2 + 1 + 2017 ) , biết F(x) là một nguyên hàm của f(x) thỏa mãn F(0)=2018. Tính F(2)
A. F(2) = 5+2017 5
B. F(2) = 4+2017 4
C. F(2) = 3+2017 3
D. F(2)= 2022