\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2010}{1}=\left(-2010\right)\)

\(\Rightarrow\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+2}{2009}+1\right)+...+\left(\frac{x+2010}{1}+1\right)=-2010+2010\)

\(\Rightarrow\frac{x+2011}{2010}+\frac{x+2011}{2009}+...+\frac{x+2011}{1}=0\)

\(\Rightarrow\left(x+2011\right)\left(1+\frac{1}{2}+...+\frac{1}{2009}+\frac{1}{2010}\right)=0\)

\(\Rightarrow x+2011=0\Leftrightarrow x=-2011\)

x=-2011

cách bạn giải bài này là gì vậy ?

bài này khá khó chịu tui làm bên h r` thì phải mà giờ lật lại có toi bn rảnh thì vô đây tìm nhé h.vn/vip/thangbnsh

đặt x+1=y sau đó xét các trường hợp của y \\\\\ y >1 \\\y =1\\ 0< y <1\\\ y =0\\y <0

nham nha dat x+2 =y

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 \(x=\frac{4015}{2}\)

\(\frac{x}{2008}+\frac{x+1}{2009}+...+\frac{x+4}{2012}=5\)

\(\Leftrightarrow\left(\frac{x}{2008}-1\right)+\left(\frac{x+1}{2009}-1\right)+...+\left(\frac{x+4}{2012}-1\right)=0\)

\(\Leftrightarrow\frac{x-2008}{2008}+\frac{x-2008}{2009}+...+\frac{x-2008}{2012}=0\)

\(\Leftrightarrow\left(x-2008\right)\left(\frac{1}{2008}+\frac{1}{2009}+..+\frac{1}{2012}\right)=0\)

Mà \(\left(\frac{1}{2008}+\frac{1}{2009}+..+\frac{1}{2012}\right)\ne0\)

Nên \(x-2008=0\)

\(\Leftrightarrow x=2008\)

Vậy : \(x=2008\)

\(\frac{x}{2008}+\frac{x+1}{2009}+\frac{x+2}{2010}+\frac{x+3}{2011}+\frac{x+4}{2012}=5\)

\(\Leftrightarrow\frac{x}{2008}+\frac{x+1}{2009}+\frac{x+2}{2010}+\frac{x+3}{2011}+\frac{x+4}{2012}-5=0\)

\(\Leftrightarrow\left(\frac{x}{2008}-1\right)+\left(\frac{x+1}{2009}-1\right)+\left(\frac{x+2}{2010}-1\right)+\left(\frac{x+3}{2011}-1\right)+\left(\frac{x+4}{2012}-1\right)=0\)

\(\Leftrightarrow\frac{x-2008}{2008}+\frac{x-2008}{2009}+\frac{x-2008}{2010}+\frac{x-2008}{2011}+\frac{x-2008}{2012}=0\)

\(\Leftrightarrow\left(x-2008\right)\left(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)=0\)

Vì \(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\ne0\)

\(\Rightarrow x-2008=0\)\(\Leftrightarrow x=2008\)

Vậy \(x=2008\)

Đặt \(x-2009=y\) khi đó phương trình trở thành:
\(\dfrac{y^2-y\left(y-1\right)+\left(y-1\right)^2}{y^2+y\left(y-1\right)+\left(y-1\right)^2}=\dfrac{19}{49}\)

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

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

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

Đổi lại:\(y=x-2009\) ,ta được:

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

Vậy...

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