Mọi người giúp mik nha  ^_^

Ta có: \(P\left(x\right)+Q\left(x\right)=2\left(1+x^2+x^4+...+x^{2010}\right)\)

\(\Rightarrow P\left(\frac{1}{2}\right)+Q\left(\frac{1}{2}\right)=2\left(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{2010}}\right)\)

Đặt \(K=\left(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{2010}}\right)\)

\(\Rightarrow\frac{1}{2^2}K=\left(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{2012}}\right)\)

\(\Rightarrow K-\frac{1}{2^2}K=1-\frac{1}{2^{2012}}\)

\(\Rightarrow\frac{3}{4}K=1-\frac{1}{2^{2012}}\)

\(\Rightarrow K=\frac{4}{3}-\frac{1}{3.2^{2010}}\)

Lúc đó \(P\left(\frac{1}{2}\right)+Q\left(\frac{1}{2}\right)=2\left(\frac{4}{3}-\frac{1}{3.2^{2010}}\right)=\frac{8}{3}-\frac{1}{3.2^{2009}}\)

\(=\frac{2^{2012}-1}{3.2^{2009}}\)

Ta thấy \(2^{2012}-1=2^{4.503}-1=\overline{...6}-1=\overline{...5}⋮5\)

Mà 3 . 2²⁰⁰⁹ không chia hết cho 5 nên khi ta rút gọn \(\frac{2^{2012}-1}{3.2^{2009}}\)đến dạng tối giản thì a vẫn chia hết cho 5.

Vậy \(a⋮5\left(đpcm\right)\)

Bài 1:

Đặt x-2009=y. Khi đó phương trình đã cho trở thành:

\(\frac{y^2-y\left(y-1\right)+\left(y-1\right)^2}{y^2+y\left(y-1\right)+\left(y-1\right)^2}=\frac{19}{49}\)

\(\Leftrightarrow4y^2-4y-15=0\)

\(\Leftrightarrow\)(2y-5).(2y+3)=0

\(\Leftrightarrow\left[\begin{matrix}y=2,5\\y=-1,5\end{matrix}\right.\)

Thay y=x-2009. Ta được: \(\left[\begin{matrix}x=2009+2,5=2011,5\\x=2009-1,5=2007,5\end{matrix}\right.\)

Vậy x=2011,5 hoặc x=2007,5

A(2010)=x^2010 - 2009x^2009 - 2009x^2008 - 2009x^2007 -...- 2009x + 1

ta có: 2010-1=2009 --> x-1=2009

thay x-1=2009 vào đa thức A(2010) ta được:

A(2010)=x^2010 - x^2009(x-1) - x^2008(x-1) - x^2007(x-1) -...- x(x-1) + 1

=x^2010 - x^2010 + x^2009 - x^2009 + x^2008 - x^2008 + x^2007 -...- x^2 + x + 1 

= x + 1 

thay x=2010 vao x+1 ta được:

2010+1=2011

vậy A(2010)=2011

1. \(\left(2x-1\right)^3+\left(x+2\right)^3=\left(3x+1\right)^3\)

\(\Rightarrow8x^3-12x^2+6x-1+x^3+6x^2+12x+8=27x^3+27x^2+9x+1\)

\(\Rightarrow-18x^3-33x^2+9x+6=0\)\(\Rightarrow\left(x+2\right)\left(-18x^2+3x+3\right)=0\)

\(\Rightarrow\left(x+2\right)\left(2x-1\right)\left(-9x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{2};x=-\frac{1}{3}\end{cases}}\)

Vậy \(x=-2;x=\frac{1}{2};x=-\frac{1}{3}\)

2. \(\frac{x-1988}{15}+\frac{x-1969}{17}+\frac{x-1946}{19}+\frac{x-1919}{21}=10\)

\(\Rightarrow\left(\frac{x-1988}{15}-1\right)+\left(\frac{x-1969}{17}-2\right)+\left(\frac{x-1946}{19}-3\right)+\left(\frac{x-1919}{21}-4\right)=0\)

\(\Rightarrow\frac{x-2003}{15}+\frac{x-2003}{17}+\frac{x-2003}{19}+\frac{x-2003}{21}=0\)

\(\Rightarrow x-2003=0\)do \(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\ne0\)

Vậy \(x=2003\)

3. Đặt \(\hept{\begin{cases}2009-x=a\\x-2010=b\end{cases}}\)

\(\Rightarrow\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\Rightarrow49a^2+49ab+49b^2=19a^2-19ab+19b^2\)

\(\Rightarrow30a^2+68ab+30b^2=0\Rightarrow\left(5a+3b\right)\left(3a+5b\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5a=-3b\\3a=-5b\end{cases}}\)

Với \(5a=-3b\Rightarrow5\left(2009-x\right)=-3\left(x-2010\right)\)

\(\Rightarrow-2x=-4015\Rightarrow x=\frac{4015}{2}\)

Với \(3a=-5b\Rightarrow3\left(2009-x\right)=-5\left(x-2010\right)\)

\(\Rightarrow2x=4023\Rightarrow x=\frac{4023}{2}\)

Vậy \(x=\frac{4023}{2}\)hoặc \(x=\frac{4015}{2}\)