Giải pt sau:
1)x4+2x3-4x2-5x-6=0
2)2x3-7x2+7x-2=0
Giải hộ nka!!Mk đang cần gấp...
Đọc tiếp
Giải pt sau:
1)x4+2x3-4x2-5x-6=0
2)2x3-7x2+7x-2=0
Giải hộ nka!!Mk đang cần gấp gấp lắm!!
K
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Những câu hỏi liên quan
20 tháng 1 2017
a, 3x2 - 8x2 - 2x+3=0
2x(3-8) - 2x+3=0
2x5 - 2x+3=0
2x5 - 2x=0-3=
2x5 - 2x=-3
2x(5-x)=-3
5-x=-3/2
5-x=1,5
x=5-1,5
x=3,5
N
1
29 tháng 8 2023
b: 4x^2-20x+25=(x-3)^2
=>(2x-5)^2=(x-3)^2
=>(2x-5)^2-(x-3)^2=0
=>(2x-5-x+3)(2x-5+x-3)=0
=>(3x-8)(x-2)=0
=>x=8/3 hoặc x=2
c: x+x^2-x^3-x^4=0
=>x(x+1)-x^3(x+1)=0
=>(x+1)(x-x^3)=0
=>(x^3-x)(x+1)=0
=>x(x-1)(x+1)^2=0
=>\(x\in\left\{0;1;-1\right\}\)
d: 2x^3+3x^2+2x+3=0
=>x^2(2x+3)+(2x+3)=0
=>(2x+3)(x^2+1)=0
=>2x+3=0
=>x=-3/2
a: =>x^2(5x-7)-3(5x-7)=0
=>(5x-7)(x^2-3)=0
=>\(x\in\left\{\dfrac{7}{5};\sqrt{3};-\sqrt{3}\right\}\)
YY
2
10 tháng 2 2022
a, \(x^4-x^2-2=0\Leftrightarrow x^4-2x^2+x^2-2=0\)
\(\Leftrightarrow x^2\left(x^2-2\right)+\left(x^2-2\right)=0\Leftrightarrow\left(x^2+1>0\right)\left(x^2-2\right)=0\Leftrightarrow x=\pm\sqrt{2}\)
b, \(\Leftrightarrow x^2\left(x^2+2x+1\right)=0\Leftrightarrow x^2\left(x+1\right)^2=0\Leftrightarrow x=0;x=-1\)
c, \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1>0\right)=0\Leftrightarrow x=1\)
d, \(\Leftrightarrow6x^2-3x-4x+2=0\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\Leftrightarrow x=\dfrac{2}{3};x=\dfrac{1}{2}\)
R
10 tháng 2 2022
a)
/ \(x^4+x^2-2=0\)
\(\Leftrightarrow\left(x^2\right)^2-x^2+2x^2-2=0\\ \Leftrightarrow x^2\left(x^2-1\right)+2\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+2=0\\x+1=0\\x-1-0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
NT
2
17 tháng 2 2020
2x5 - 7x4 + 5x3 + 5x2 - 7x + 2 = 0
<=> 2x5-4x4-3x4+6x3-x3+2x2+3x2-6x-x+2=0
<=> 2x4(x-2)-3x3(x-2)-x2(x-2)+3x(x-2)-(x-2)=0
<=>(x-2)(2x4-3x3-x2+3x-1)=0
<=>(x-2)(2x4-x3-2x3+x2-2x2+x+2x-1)=0
<=>(x-2)[x3(2x-1)-x2(2x-1)-x(2x-1)+2x-1]=0
<=>(x-2)(2x-1)(x3-x2-x+1)=0
<=>(x-2)(2x-1)[x2(x-1)-(x-1)]=0
<=>(x-2)(2x-1)(x-1)(x2-1)=0
<=>(x-2)(2x-1)(x-1)2(x+1)=0
=> x-2=0 => x=2
hoặc 2x-1=0=>x=1/2
hoặc x-1=0=>x=1
hoặc x+1=0=>x=-1
Vậy...
MN
17 tháng 2 2020
\(2x^5-7x^4+5x^3+5x^2-7x+2=0\)
\(\Leftrightarrow\left(2x^5-4x^4+2x^3\right)-\left(3x^4-6x^3+3x^2\right)-\left(3x^3-6x^2+3x\right)+\left(2x^2-4x+2\right)=0\)
\(\Leftrightarrow2x^3\left(x^2-2x+1\right)-3x^2\left(x^2-2x+1\right)-3x\left(x^2-2x+1\right)+2\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)\left(2x^3-3x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(2x^3+2x^2-5x^2-5x+2x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left[2x^2\left(x+1\right)-5x\left(x+1\right)+2\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)\left(2x^2-5x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)\left(2x^2-4x-x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)\left[2x\left(x-2\right)-\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)\left(x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\)\(x-1=0\)
hoặc \(x+1=0\)
hoặc \(x-2=0\)
hoặc \(2x-1=0\)
\(\Leftrightarrow\)\(x=1\)
hoặc \(x=-1\)
hoặc \(x=2\)
hoặc \(x=\frac{1}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{1;-1;2;\frac{1}{2}\right\}\)
8 tháng 1 2021
a) Ta có: \(x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)
mà \(x^2+1>0\forall x\)
nên x+1=0
hay x=-1
Vậy: S={-1}
b) Ta có: \(x^3-6x^2+11x-6=0\)
\(\Leftrightarrow x^3-x^2-5x^2+5x+6x-6=0\)
\(\Leftrightarrow x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)
Vậy: S={1;2;3}
c) Ta có: \(x^3-x^2-21x+45=0\)
\(\Leftrightarrow x^3-3x^2+2x^2-6x-15x+45=0\)
\(\Leftrightarrow x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+2x-15\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+5x-3x-15\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2\cdot\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy: S={3;-5}
d) Ta có: \(x^4+2x^3-4x^2-5x-6=0\)
\(\Leftrightarrow x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6=0\)
\(\Leftrightarrow x^3\left(x-2\right)+4x^2\cdot\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+4x^2+4x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+3x^2+x^2+4x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+3\right)+\left(x+1\right)\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)=0\)
mà \(x^2+x+1>0\forall x\)
nên (x-2)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy: S={2;-3}
NT
5
BA
29 tháng 7 2019
\(5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x^2-5x=x-1\)
\(\Leftrightarrow5x^2-5x-x+1=0\)
\(\Leftrightarrow5x^2-6x+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-\frac{1}{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-\frac{1}{5}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
BA
29 tháng 7 2019
\(2\left(x-7\right)-x^2+7x=0\)
\(2\left(x-7\right)-x\left(x-7\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2-x=0\\x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)
TL
2 tháng 2 2016
(2/2 + 3/6 + 4/12 + 6/30) nhân 10 - x=0
(9/6+5/4) nhân 10 -x=0
11/4 nhân 10 - x= 0
2,75 nhân 10 -x=0
27.5 - x=0
x=27.5 - 0= 27.5
dùm mình nha...
2 tháng 2 2016
1/5x6 sửa thành 1/4x5 nhé!
(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5)x10-x=0
(1-1/5)x10-x=0
4/5x10-x=0
8-x=0
x=8-0=8
1, \(x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6=0\)
\(\Leftrightarrow x^3\left(x-2\right)+4x^2\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^3+4x^2+4x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^3+3x^2+x^2+3x+x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2+x+1>0\right)\left(x+3\right)\left(x-2\right)=0\Leftrightarrow x=-3;x=2\)
2, \(2\left(x^3-1\right)-7x\left(x-1\right)=0\)
\(\Leftrightarrow2\left(x-1\right)\left(x^2+x+1\right)-7x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^2-5x+2\right)=0\Leftrightarrow x=1;x=\dfrac{1}{2};x=2\)