Bài 3: Phương trình đưa được về dạng ax + b = 0

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Phạm Thùy Linh

Mọi người giúp mk bài này nha:

\(2x^4-9x^3+14x^2-9x+2=0\)

Lightning Farron
11 tháng 6 2017 lúc 21:15

\(2x^4-9x^3+14x^2-9x+2=0\)

\(\Leftrightarrow2x^4-4x^3+2x^2-5x^3+10x^2-5x+2x^2-4x+2=0\)

\(\Leftrightarrow2x^2\left(x^2-2x+1\right)-5x\left(x^2-2x+1\right)+2\left(x^2-2x+1\right)=0\)

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

\(\Leftrightarrow\left(2x^2-x-4x+2\right)\left(x^2-2x+1\right)=0\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\2x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)

qwerty
11 tháng 6 2017 lúc 21:17

\(2x^4-9x^3+14x^2-9x+2=0\)

\(\Leftrightarrow2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0\)

\(\Leftrightarrow2x^3\cdot\left(x-1\right)-7x^2\cdot\left(x-1\right)+7x\cdot\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x^3-7x^2+7x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left[2\left(x^3-1\right)-7x\cdot\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left[2\left(x-1\right)\cdot\left(x^2+x+1\right)-7x\cdot\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left[2\left(x^2+x+1\right)-7x\right]=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2+2x+2-7x\right)=0\)

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

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2-x-4x+2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left[x\cdot\left(2x-1\right)-2\left(2x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(x-2\right)\cdot\left(2x-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\x-2=0\\2x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(x_1=\dfrac{1}{2};x_2=1;x_3=2\)

Như Khương Nguyễn
11 tháng 6 2017 lúc 21:26

\(2x^4-9x^3+14x^2-9x+2=0\)

\(\Leftrightarrow2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0\)

\(\Leftrightarrow\)\(2x^3\left(x-1\right)-7x^2\left(x-1\right)+7x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x^3-7x^2+7x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x^3-4x^2-3x^2+6x+x-2\right)=0\)

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

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

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

Vậy ...................

Nguyễn Tấn Dũng
11 tháng 6 2017 lúc 21:45

\(2x^4-9x^3+14x^2-9x+2=0\)

\(\Leftrightarrow\) \(2x^4-x^3-8x^3+4x^2+10x^2-5x-4x+2=0\)

\(\Leftrightarrow\) \(x^3\left(2x-1\right)-4x^2\left(2x-1\right)+5x\left(2x-1\right)-2\left(2x-1\right)=0\)

\(\Leftrightarrow\) \(\left(2x-1\right)\left(x^3-4x^2+5x-2\right)=0\)

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

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

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

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

\(\Leftrightarrow\) \(\left[\begin{array}{} 2x-1=0\\ (x-1)^2=0\\ x-2=0 \end{array} \right.\) \(\Leftrightarrow\) \(\left[\begin{array}{} 2x-1=0\\ x-1=0\\ x-2=0 \end{array} \right.\) \(\Leftrightarrow\) \(\left[\begin{array}{} x=\dfrac{1}2\\ x=1\\ x=2 \end{array} \right.\)

Vậy S={\(\dfrac{1}{2}\) ; 1; 2}

Phạm Thùy Linh
12 tháng 6 2017 lúc 15:29

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