minh nghi ban dang chep sai de bai 

the de bai cua minh thi giai nhu sau

A=1/2015.2016+1/2016.2017+......+1/2029.2030

A=1/2015-1/2016+1/2016-1/2017+.....+1/2029-1/2030

A=1/2015-1/2030=3/818090

\(2029-\left\{\left[39-\left(2^3.3-21\right)^2:3+2018^0\right]\right\}\)

\(=2029-\left\{\left[39-\left(8.3-21\right)^2:3+2018^0\right]\right\}\)

\(=2029-\left\{\left[39-\left(24-21\right)^2:3+2018^0\right]\right\}\)

\(=2029-\left\{\left[39-3^2:3+2018^0\right]\right\}\)

\(=2029-\left\{\left[39-9:3+1\right]\right\}\)

\(=2029-\left\{\left[39-3+1\right]\right\}\)

\(=2029-37\)

\(=2022\)

Đáp án sai rồi kìa Pé Mun, bằng 1992 chứ ? Sao lại lấy 2029 - 7 =)))?

A = 1 + 2 + 2² +...........+ 2²⁰²⁹

A = ( 1 + 2 + 2² + 2³ + 2⁴ ) +...........+( 2²⁰²⁵ + 2²⁰²⁶ + 2²⁰²⁷ + 2²⁰²⁸ + 2²⁰⁰²⁹)

A = 1(1 + 2 + 2² + 2³ + 2⁴) +............+ 2²⁰²⁵( 1 + 2 + 2² + 2³ + 2⁴ )

A = 1 . 31 +.........+ 2²⁰²⁵ . 31

A = 31( 1 + .......... + 2²⁰²⁵)

Vì 31 chia hết cho 31 => 31( 1+...........+2²⁰²⁵) chia hết cho 31

                                       Hay A chia hết cho 21.                  ( Tính chất 1)

\(A=1+2+2^2+....+2^{2029}\)

\(A=\left(1+2+2^2+2^3+2^4\right)+.....+\left(2^{2025}+2^{2026}+2^{2027}+2^{2028}+2^{2029}\right)\)

\(A=31.1+....+2^{2025}.31\)

\(A=31.\left(1+....+2^{2025}\right)\)

\(\Rightarrow A⋮31\left(đpcm\right)\)

\(P=x^2+\left(2xy-6x\right)+2y^2-8y+2029\)

\(P=x^2+2x\left(y-3\right)+\left(y-3\right)^2-\left(y-3\right)^2+2y^2-8y+2029\)

\(P=\left(x+y-3\right)^2-\left(y^2-6y+9\right)+2y^2-8y+2029\)

\(P=\left(x+y-3\right)^2+y^2-2y+1+2019\)

\(P=\left(x+y-3\right)^2+\left(y-1\right)^2+2019\) \(\ge2019\forall x,y\)

\(P=2019\Leftrightarrow\left\{{}\begin{matrix}x+y-3=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy Min P = 2019 \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

1.\(\Leftrightarrow a^2+b^2-ab-a-b+3>0\)

\(\Leftrightarrow2a^2+2b^2-2ab-2a-2b+6>0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+4>0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2+4>0\) ( luôn đúng )

Do đó suy ra đpcm

\(\frac{x-2}{2012}+\frac{x-3}{2011}+\frac{x-4}{2010}+\frac{x-2029}{5}=0\)

\(\Leftrightarrow\frac{x-2}{2012}-1+\frac{x-3}{2011}-1+\frac{x-4}{2010}-1+\frac{x-2029}{5}+3=0\)

\(\Leftrightarrow\frac{x-2014}{2012}+\frac{x-2014}{2011}+\frac{x-2014}{2010}+\frac{x-2014}{5}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+\frac{1}{5}\right)=0\)

\(\Leftrightarrow x-2014=0\).Do \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+\frac{1}{5}\ne0\)

\(\Leftrightarrow x=2014\)

= (1-2015).(2-2015)....(2015-2015)....(2029-2015).(2030-2015)

= (1-2015).(2-2015)....0....(2029-2015).(2030-2015)

= 0

\(=\dfrac{97}{60};=\dfrac{5}{12}\)

\(\dfrac{97}{80}\)

\(\dfrac{5}{12}\)