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\(A=1+3^1+3^2+...+3^{2017}\)
\(3A=3+3^2+3^3+...+3^{2018}\)
\(3A-A=\left(3+3^2+3^3+...+3^{2018}\right)-\left(1+3^1+3^2+...+3^{2017}\right)\)
\(2A=3^{2018}-1\)
\(A=\frac{3^{2018}-1}{2}\)
\(\Rightarrow\)\(B-A=\frac{3^{2018}}{2}-\frac{3^{2018}-1}{2}=\frac{3^{2018}-3^{2018}+1}{2}=\frac{1}{2}\)
Vậy \(B-A=\frac{1}{2}\)
Chúc bạn học tốt ~
ta có: A = 1 + 31 + 32 + ...+ 32017
=> 3A = 31 + 32 + 33 + ....+ 32018
=> 3A - A = 32018 - 1
\(\Rightarrow A=\frac{3^{2018}-1}{2}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3^{2018-1}}{2}}{\frac{3^{2018}}{2}}=\frac{\frac{3^{2018}}{2}}{\frac{3^{2018}}{2}}-\frac{1}{\frac{3^{2018}}{2}}=1-\frac{1}{\frac{3^{2018}}{2}}\)
\(A=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{1}{2018}\)
\(A=1+\left(1+\frac{2017}{2}\right)+\left(1+\frac{2016}{3}\right)+...+\left(1+\frac{1}{2018}\right)\)
\(A=\frac{2019}{2019}+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2018}\)
\(A=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)\)
Ta có: \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)
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 \(\dfrac{a}{a+1}< 1\Leftrightarrow a+1-\dfrac{a}{a+1}>0\)
\(\Leftrightarrow A=a+1-\dfrac{a}{a+1}+\dfrac{a}{a+1}=a+1=2018\)
A = 1 + 2 + 3 + ... + 2018
= ( 1 + 2018 ) + ( 2 + 2017) + ... + ( 1009 + 1010 )
= 2019 + 2019 + ... + 2019 ( có 1009 số 2019 )
= 2019 x 1009 = 2037171
B = 1 + 3 + 5 + ... + 2017
= ( 1 + 2017 ) + ( 3 + 2015 ) + ... + ( 1007 + 1010) + 1009
= 2018 + 2018 + ... + 2018 + 1009 (có 504 số 2018)
= 2018 x 504 + 1009 = 1018081
Còn lại làm giống ý trên .
Theo bài ra, ta có: \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2017.2018.2019}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2017.2018.2019}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2018.2019}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2018.2019}\right)\)
Giải thích:
\(\frac{2}{1.2.3}=\frac{3}{1.2.3}-\frac{1}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\)
\(\frac{2}{2.3.4}=\frac{4}{2.3.4}-\frac{2}{2.3.4}=\frac{1}{1.2}-\frac{1}{3.4}\)
................................................................................
\(\frac{2}{2017.2018.2019}=\frac{2019}{2017.2018.2019}-\frac{2017}{2017.2018.2019}=\frac{1}{2017.2018}-\frac{1}{2018.2019}\)
Ta có :
a^2>hoặc=0(vì mang số mũ dương)
Tương tự => b^2 và c ^2 như a^2
mà a^2+b^2+c^2=1=>a=b=c=1
=> a^2016+b^2017+c^2018=1
Mình nghĩ \(a+b+c=1\) nữa chắc oke hơn :3
\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Rightarrow1-3abc=1-ab-bc-ca\Rightarrow3abc=ab+bc+ca\)
\(1=\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(=1+2\left(ab+bc+ca\right)\)
\(\Rightarrow ab+bc+ca=0\Rightarrow3abc=0\)
Nếu \(a=0\Rightarrow b+c=1;b^2+c^2=1;b^3+c^3=1\)
\(\Rightarrow b^2+2bc+c^2=1\Rightarrow2bc=0\Rightarrow b=0\left(h\right)c=0\)
Cứ tiếp tục thì sẽ ra nhá :))
Đáp án A
P = 3 + 2 2 2018 2 2 − 3 2017 = 3 + 2 2 2018 . 2 2 − 3 2018 . 1 2 2 − 3 = 3 + 2 2 2 2 − 3 2018 . 1 2 2 − 3 = − 2 2 − 3.