\(S=1010+1010^2+1010^3+...+1010^{1011}\)

Suy ra \(1010.S=1010^2+1010^3+1010^4+....+1010^{1012}\)

Nên\(1010.S-S=1010^{1012}-1010\)hay\(1009.S=1010^{1012}-1010\)

Khi đó \(S=\frac{1010^{1012}-1010}{1009}\)

S=1011+1010^2+1010^3+...+1010^1011

S=1+1010+1010^2+1010^3+...+1010^1011

1010.S=1010+1010^2+1010^3+1010^4+...+1010^1012

1010 S - S=1010^1012-1

1009 S=1010^1012-1

S=(1010^1012-1):1009

mình có đáp án giống bạn hùng

dễ lắm hải ơi

em vừa hỏi cô Huyền xong , đơn giản cực

B

B

B

\(2a^{2020}+2b^{2020}+2c^{2020}-2\left(ab\right)^{1010}-2\left(bc\right)^{1010}-2\left(ca\right)^{1010}=0\)

\(\Leftrightarrow\left(a^{1010}-b^{1010}\right)^2+\left(b^{1010}-c^{1010}\right)^2+\left(c^{1010}-a^{1010}\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^{1010}-b^{1010}=0\\b^{1010}-c^{1010}=0\\c^{1010}-a^{1010}=0\end{matrix}\right.\)

\(\Rightarrow\left|a\right|=\left|b\right|=\left|c\right|\)

Nếu đề không cho a;b;c dương thì không tính được cụ thể giá trị A

Nếu a;b;c dương thì \(a=b=c\Rightarrow A=0\)

122210

tick nha !!!!!!!!!!!!!!!!!!!!!

\(B=\frac{1010+1007+\frac{2017}{113}+\frac{2017}{117}-\frac{1010}{119}-\frac{1007}{119}}{1010+1008+\frac{2018}{113}+\frac{2018}{117}-\frac{1010}{119}-\frac{1008}{119}}\)

\(B=\frac{2017+\frac{2017}{113}+\frac{2017}{117}-\frac{2017}{119}}{2018+\frac{2018}{113}+\frac{2018}{117}-\frac{2018}{119}}\)

\(B=\frac{2017.\left(1+\frac{1}{113}+\frac{1}{117}-\frac{1}{119}\right)}{2018.\left(1+\frac{1}{113}+\frac{1}{117}-\frac{1}{119}\right)}\)

\(B=\frac{2017}{2018}\)

Vậy \(B=\frac{2017}{2018}\)

Chúc bạn học tốt !!! 

dễ lắm thử nghỉ đi banj

\(bx^2=ay^2\Leftrightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}\Leftrightarrow\left(\dfrac{x^2}{a}\right)^{1010}=\left(\dfrac{y^2}{b}\right)^{1010}\\ \Leftrightarrow\dfrac{x^{2020}}{a^{1010}}=\dfrac{y^{2020}}{a^{1010}}\)

Áp dụng t/c dtsbn:

\(\dfrac{x^{2020}}{a^{1010}}=\dfrac{y^{2020}}{b^{1010}}=\dfrac{x^{2020}+y^{2020}}{a^{1010}+b^{1010}}\left(3\right)\)

Đặt \(\dfrac{x^2}{a}=\dfrac{y^2}{b}=k\Leftrightarrow x^2=ak;y^2=bk\)

\(x^2+y^2=1\Leftrightarrow ak+bk=1\Leftrightarrow k\left(a+b\right)=1\Leftrightarrow a+b=\dfrac{1}{k}\)

\(\Leftrightarrow\dfrac{2}{\left(a+b\right)^{1010}}=\dfrac{2}{\left(\dfrac{1}{k}\right)^{1010}}=2:\dfrac{1}{k^{1010}}=k^{1010}\left(1\right)\)

Mà \(\dfrac{x^{2020}}{a^{1010}}=\dfrac{\left(x^2\right)^{1010}}{a^{1010}}=\dfrac{a^{1010}k^{1010}}{a^{1010}}=k^{1010}\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\left(3\right)\) ta được đpcm