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 \(x=\frac{4015}{2}\)

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...

Cre:Miny.vn

Bài 1: Tìm x biết: $\frac{\left(2009-x\right)^2+\left(2009-x\right..

Đặt \(2009-x=t\Rightarrow x-2010=-\left(2009-x\right)-1=-t-1\)

Suy ra:

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

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

\(\Leftrightarrow\frac{t^2-t^2-t+t^2+2t+1}{t^2+t^2+t+t^2+2t+1}=\frac{19}{49}\)

\(\Leftrightarrow\frac{t^2+t+1}{3t^2+3t+1}=\frac{19}{49}\)

\(\Leftrightarrow49t^2+49t+49=57t^2+57t+19\)

\(\Leftrightarrow8t^2+8t-30=0\)

\(\Leftrightarrow4t^2+4t-15=0\Leftrightarrow4t^2+4t+1=16\)

\(\Leftrightarrow\left(2t+1\right)^2=16\Leftrightarrow\left[{}\begin{matrix}2t+1=-4\\2t+1=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2t=-5\\2t=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=-\frac{5}{2}\\t=\frac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2009-x=-\frac{5}{2}\\2009-x=\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4023}{2}\\x=\frac{4015}{2}\end{matrix}\right.\)

Vậy \(S=\left\{\frac{4015}{2};\frac{4023}{2}\right\}\)

ĐKXĐ: \(x\ne2009\) ; \(x\ne2010\)

Đặt : a = \(2009-x\)

b = \(x-2010\)

⇒ a + b = -1 ⇒ a = - ( 1 + b )

⇒ Phương trình đã cho có dạng :

\(\dfrac{a^2+ab+b^2}{a^2-ab+b^2}=\dfrac{19}{49}\)

⇔ \(\dfrac{\left(1+b\right)^2-b\left(1+b\right)+b^2}{\left(1+b\right)^2+b\left(1+b\right)+b^2}\) \(=\dfrac{19}{49}\)

⇔ \(\dfrac{b^2+b+1}{3b^2+3b+1}\) \(=\dfrac{19}{49}\)

⇔ \(49b^2+49b+49=57b^2+57b+19\)

⇔ \(8b^2+8b-30=0\)

⇔ \(4b^2+4b-15=0\)

⇔ \(4b^2-6b+10b-15=0\)

⇔ \(2b\left(2b-3\right)+5\left(2b-3\right)=0\)

⇔ \(\left(2b-3\right)\left(2b+5\right)=0\)

⇔ \(\left[{}\begin{matrix}2b+5=0\\2b-3=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}b=\dfrac{-5}{2}\\b=\dfrac{3}{2}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-5}{2}+2010=2007,5\\x=\dfrac{3}{2}+2010=2011,5\end{matrix}\right.\)

Vậy ......

\(\left(x-2008\right)^{2010}+\left(x-2009\right)^{2010}=1\)\(1\)====>> \(\hept{\begin{cases}x-2008=1\\x-2009=0\end{cases}}< =>\hept{\begin{cases}x=20009\\x=2009\end{cases}}< =>x=20009\)                                                                                                     Vậy x=2009 thì PT có GT là 1

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