Question

[Cứu em với, giải đầy đủ xíu.]

Giải phương trình:

`a, (x-1)/2012+(x-2)/2011+(x-3)/2010+...+(x-2012)/1=2012`

`b,x^4-30x^2+31x-30=0`

`c,(2x-5)^3-(x-2)^3=(x-3)^3`

Answer 1

[๖ۣۜDũ๖ۣۜN๖ۣۜG]

a)

PT <=> \(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)

<=> \(\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)

<=> \(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\right)=0\)

Mà \(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\ne0\)

<=> x - 2013 = 0

<=> x = 2013

KL: ...

b) PT <=> \(\left(x^4-5x^3\right)+\left(5x^3-25x^2\right)-\left(5x^2-25x\right)+\left(6x-30\right)=0\)

<=> \(x^3\left(x-5\right)+5x^2\left(x-5\right)-5x\left(x-5\right)+6\left(x-5\right)=0\)

<=> \(\left(x-5\right)\left(x^3+5x^2-5x+6\right)=0\)

<=> \(\left(x-5\right)\left[\left(x^3+6x^2\right)-\left(x^2+6x\right)+\left(x+6\right)\right]=0\)

<=> \(\left(x-5\right)\left[x^2\left(x+6\right)-x\left(x+6\right)+\left(x+6\right)\right]=0\)

<=> \(\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)=0\)

<=> \(\left[{}\begin{matrix}x=5\\x=-6\\x=\varnothing\end{matrix}\right.\)

KL: ...

Answer 2

[Nguyễn Lê Phước Thịnh]

a) Ta có: \(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)

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

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

\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)

\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1\right)=0\)

mà \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1>0\)

nên x-2013=0

hay x=2013

Vậy: Tập nghiệm S={2013}

b) Ta có: \(x^4-30x^2+31x-30=0\)

\(\Leftrightarrow x^4+x-30x^2+30x-30=0\)

\(\Leftrightarrow\left(x^4+x\right)-\left(30x^2-30x+30\right)=0\)

\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x+1\right)-30\right]=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+6x-5x-30\right)=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x+6\right)-5\left(x+6\right)\right]=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(x+6\right)\left(x-5\right)=0\)(1)

Ta có: \(x^2-x+1\)

\(=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)

Ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

hay \(x^2-x+1>0\forall x\)(2)

Từ (1) và (2) suy ra (x+6)(x-5)=0

\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=5\end{matrix}\right.\)

Vậy: Tập nghiệm S={-6;5}