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BT
9 tháng 3 2018
2x4-9x3+14x2-9x+2=0
<=> 2x4-2x3-7x3+7x2+7x2-7x-2x+2=0
<=> 2x3(x-1)-7x2(x-1)+7x(x-1)-2(x-1)=0
<=> (x-1)(2x3-7x2+7x-2)=0
<=> (x-1)[2x3-2x2-5x2+5x+2x-2]=0
<=> (x-1)[2x2(x-1)-5x(x-1)+2(x-1)]=0
<=> (x-1)2(2x2-5x+2)=0
<=> (x-1)2(2x2-4x-x+2)=0
<=> (x-1)2[(2x(x-2)-(x-2)]=0
<=> (x-1)2(x-2)(2x-1)=0
=> \(\hept{\begin{cases}\left(x-1\right)^2=0\\x-2=0\\2x-1=0\end{cases}}\) <=> \(\hept{\begin{cases}x_1=1\\x_2=2\\x_3=\frac{1}{2}\end{cases}}\)
16 tháng 3 2018
\(2x^4-9x^3+14x^2-9x+2=0\)
\(\Leftrightarrow2x^4-4x^3-5x^3+10x^2+4x^2-8x-x+2=0\)
\(\Leftrightarrow2x^3\left(x-2\right)-5x^2\left(x-2\right)+4x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-5x^2+4x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-2x^2-3x^2+3x+x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[2x^2\left(x-1\right)-3x\left(x-1\right)+\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
NT
2
HY
18 tháng 2 2020
\(b.6x^4+25x^3+12x^2-25x+6=0\\\Leftrightarrow 6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\\\Leftrightarrow 6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\\\Leftrightarrow \left(6x^3+13x^2-14x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left(6x^3+18x^2-5x^2-15x+x+3\right)\left(x+2\right)=0\\\Leftrightarrow \left[6x^2\left(x+3\right)-5x\left(x+3\right)+\left(x+3\right)\right]\left(x+2\right)=0\\ \Leftrightarrow\left(6x^2-5x+1\right)\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left(6x^2-3x-2x+1\right)\left(x+3\right)\left(x+2\right)=0\\\Leftrightarrow \left[3x\left(2x-1\right)-\left(2x-1\right)\right]\left(x+3\right)\left(x+2\right)=0\\\Leftrightarrow \left(3x-1\right)\left(2x-1\right)\left(x+3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\2x-1=0\\x+3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=\frac{1}{2}\\x=-3\\x=-2\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{\frac{1}{3};\frac{1}{2};-3;-2\right\}\)
HY
18 tháng 2 2020
\(2x^4-9x^3+14x^2-9x+2=0\\\Leftrightarrow 2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0\\\Leftrightarrow 2x^3\left(x-1\right)-7x^2\left(x-1\right)+7x\left(x-1\right)-2\left(x-1\right)=0\\\Leftrightarrow \left(2x^3-7x^2+7x-2\right)\left(x-1\right)=0\\\Leftrightarrow \left[2\left(x^3-1\right)-7x\left(x-1\right)\right]\left(x-1\right)=0\\\Leftrightarrow \left(x-1\right)^2\left[2\left(x^2+x+1\right)-7x\right]=0\\\Leftrightarrow \left(2x^2+2x+2-7x\right)\left(x-1\right)^2=0\\\Leftrightarrow \left(2x^2-5x+2\right)\left(x-1\right)^2=0\\\Leftrightarrow \left(2x^2-x-4x+2\right)\left(x-1\right)^2=0\\\Leftrightarrow \left[x\left(2x-1\right)-2\left(2x-1\right)\right]\left(x-1\right)^2=0\\\Leftrightarrow \left(x-2\right)\left(2x-1\right)\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-1=0\\\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\2x=1\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\\x=1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{2;\frac{1}{2};1\right\}\)
AH
Akai Haruma
Giáo viên
29 tháng 1 2020
Lời giải:
$2x^4-9x^3+14x^2-9x+2=0$
$\Leftrightarrow 2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0$
$\Leftrightarrow 2x^3(x-1)-7x^2(x-1)+7x(x-1)-2(x-1)=0$
$\Leftrightarrow (x-1)(2x^3-7x^2+7x-2)=0$
$\Leftrightarrow (x-1)[2(x^3-1)-7x(x-1)]=0$
$\Leftrightarrow (x-1)(x-1)(2x^2+2x+2-7x)=0$
$\Leftrightarrow (x-1)^2(2x^2-5x+2)=0$
$\Leftrightarrow (x-1)^2(2x^2-4x-x+2)=0$
$\Leftrightarrow (x-1)^2[2x(x-2)-(x-2)]=0$
$\Leftrightarrow (x-1)^2(2x-1)(x-2)=0$
\(\Rightarrow \left[\begin{matrix} x=1\\ x=\frac{1}{2}\\ x=2\end{matrix}\right.\)
NQ
1
23 tháng 1 2016
2x^4-9x^3+14x^2-9x+2=0
vế trái có tổng các hệ số (2-9+14-9+2)=0 nến có 1 nghiêm x=1
nên phân tích đc nhân tử là (x-1)
2x^4-9x^3+14x^2-9x+2=0 <=> (x-1)(2x^3-7x^2+7x-2)=0
<=> x=1 và 2x^3-7x^2+7x-2=0
PT: 2x^3-7x^2+7x-2=0 cũng có tổng các hệ số (2-7+7-2)=0 nên cũng có 1 nghiệm là 1 => vế trái có thể phân tích đc nhân tử (x-1)
2x^3-7x^2+7x-2=0 <=> (x-1)(2x^2-5x+2)=0
<=> x=1 và 2x^2-5x+2=0
2x^2-5x+2=0 <=> x^2 - (5/2)x + 1 =0
<=> (x-5/4)^2 - 9/16 = 0
<=> (x-5/4)^2 - (3/4)^2 = 0
PT
5
11 tháng 6 2017
\(2x^4-9x^3+14x^2-9x+2=0\)
\(\Leftrightarrow2x^4-4x^3+2x^2-5x^3+10x^2-5x+2x^2-4x+2=0\)
\(\Leftrightarrow2x^2\left(x^2-2x+1\right)-5x\left(x^2-2x+1\right)+2\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(2x^2-5x+2\right)\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(2x^2-x-4x+2\right)\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left[x\left(2x-1\right)-2\left(2x-1\right)\right]\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)^2\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\2x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Q
11 tháng 6 2017
\(2x^4-9x^3+14x^2-9x+2=0\)
\(\Leftrightarrow2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0\)
\(\Leftrightarrow2x^3\cdot\left(x-1\right)-7x^2\cdot\left(x-1\right)+7x\cdot\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^3-7x^2+7x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left[2\left(x^3-1\right)-7x\cdot\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left[2\left(x-1\right)\cdot\left(x^2+x+1\right)-7x\cdot\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left[2\left(x^2+x+1\right)-7x\right]=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2+2x+2-7x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2-5x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2-x-4x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left[x\cdot\left(2x-1\right)-2\left(2x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(x-2\right)\cdot\left(2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-2\right)\cdot\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\x-2=0\\2x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{1}{2};x_2=1;x_3=2\)
3 tháng 2 2017
a) \(x^3-7x+6=x^3+3x^2-x^2-3x-2x^2-6x+2x+6\)
=\(x^2\left(x+3\right)-x\left(x+3\right)-2x\left(x+3\right)+2\left(x+3\right)\)
=\(\left(x+3\right)\left(x^2-x-2x+2\right)\)
=\(\left(x+3\right)\left(x-2\right)\left(x-1\right)\)
=\(\left\{\begin{matrix}x+3=0=>x=-3\\x-2=0=x=2\\x-1=0=>x=1\end{matrix}\right.\)
3 tháng 2 2017
\(b...x^3-19x+30=0\)
\(=>x^3+5x^2-2x^2-10x-3x^2-15x+6x+30=0\)
=>\(x^2\left(x+5\right)-2x\left(x+5\right)-3x\left(x+5\right)+6\left(x+5\right)=0\)
=>\(\left(x+5\right)\left(x^2-2x-3x+6\right)=0\)
=>\(\left(x+5\right)\left(x-3\right)\left(x-2\right)=0\)
=>\(\left\{\begin{matrix}x-3=0=>x=3\\x-2=0=>x=2\\x+5=0=>x=-5\end{matrix}\right.\)
Vậy x=-5;2;3
\(2x^3+9x^2+14x+5=0\)
\(\Leftrightarrow\left(2x^3+x^2\right)+\left(8x^2+4x\right)+\left(10x+5\right)=0\)
\(\Leftrightarrow x^2\left(2x+1\right)+4x\left(2x+1\right)+5\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x^2+4x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x^2+4x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(tm\right)\\\left(x+2\right)^2+1=0\left(ktm\right)\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{1}{2}\right\}\)