x4 + x3 - 4x2 + 5x - 3 = 0
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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}
TH
2
5 tháng 3 2022
\(A=5x^3-7x^2+3x^3-4x^2+x^2-x^3+5x-1=7x^3-10x^2+5x-1\)
\(B=5x^3+3x^2-7x^4-5x^3+4x^2-x^4+3=-8x^4+7x^2+3\)
VT
0
11 tháng 10 2021
a: \(x^2-y^2-x-y\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
f: \(x^3-5x^2-5x+1\)
\(=\left(x+1\right)\left(x^2-x+1\right)-5x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-6x+1\right)\)
14 tháng 2 2022
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
14 tháng 2 2022
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
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
=>\(x\in\left\{\dfrac{7}{5};\sqrt{3};-\sqrt{3}\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 3 2020
1.
a/ \(\Leftrightarrow\left(x+1\right)\left(x^2+3x+2\right)+\left(x-1\right)\left(x^2-3x+2\right)-12=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2\right)+3x\left(x+1\right)-3x\left(x-1\right)+\left(x-1\right)\left(x^2+2\right)-12=0\)
\(\Leftrightarrow2x\left(x^2+2\right)+6x^2-12=0\)
\(\Leftrightarrow x^3+3x^2+2x-6=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+4x+6\right)=0\Rightarrow x=1\)
b/ Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)
\(x^2+\frac{1}{x^2}+3\left(x+\frac{1}{x}\right)+4=0\)
Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)
\(t^2-2+3t+4=0\Rightarrow t^2+3t+2=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=-1\\x+\frac{1}{x}=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+x+1=0\left(vn\right)\\x^2+2x+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 3 2020
1c/
\(\Leftrightarrow x^5+x^4-2x^4-2x^3+5x^3+5x^2-2x^2-2x+x+1=0\)
\(\Leftrightarrow x^4\left(x+1\right)-2x^3\left(x+1\right)+5x^2\left(x+1\right)-2x\left(x+1\right)+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4-2x^3+5x^2-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^4-2x^3+5x^2-2x+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^4-2x^3+x^2+x^2-2x+1+3x^2=0\)
\(\Leftrightarrow\left(x^2-x\right)^2+\left(x-1\right)^2+3x^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x=0\\x-1=0\\x=0\end{matrix}\right.\) \(\Rightarrow\) không tồn tại x thỏa mãn
Vậy pt có nghiệm duy nhất \(x=-1\)
AH
Akai Haruma
Giáo viên
24 tháng 8 2021
a.
$x^4-25x^3=0$
$\Leftrightarrow x^3(x-25)=0$
\(\Leftrightarrow \left[\begin{matrix} x^3=0\\ x-25=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=25\end{matrix}\right.\)
b.
$(x-5)^2-(3x-2)^2=0$
$\Leftrightarrow (x-5-3x+2)(x-5+3x-2)=0$
$\Leftrightarrow (-2x-3)(4x-7)=0$
\(\Leftrightarrow \left[\begin{matrix}
-2x-3=0\\
4x-7=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix}
x=\frac{-3}{2}\\
x=\frac{7}{4}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
24 tháng 8 2021
c.
$x^3-4x^2-9x+36=0$
$\Leftrightarrow x^2(x-4)-9(x-4)=0$
$\Leftrightarrow (x-4)(x^2-9)=0$
$\Leftrightarrow (x-4)(x-3)(x+3)=0$
\(\Leftrightarrow \left[\begin{matrix} x-4=0\\ x-3=0\\ x+3=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=4\\ x=3\\ x=-3\end{matrix}\right.\)
d. ĐK: $x\neq 0$
$(-x^3+3x^2-4x):(\frac{-1}{2}x)=0$
$\Leftrightarrow x(-x^2+3x-4):(\frac{-1}{2}x)=0$
$\Leftrightarrow -2(-x^2+3x-4)=0$
$\Leftrightarrow x^2-3x+4=0$
$\Leftrightarrow (x-1,5)^2=-1,75< 0$ (vô lý)
Vậy pt vô nghiệm.
19 tháng 6 2023
A(x)+B(x)=-2x^4+x^3+x^2+5x-5-x^4-3x^3+4x^2-6x+7
=-3x^4+4x^3+5x^2-x+2
A(x)-B(x)=-2x^4+x^3+x^2+5x-5+x^4+3x^3-4x^2+6x-7
=-x^4+4x^3-3x^2+11x-2
B(x)-C(x)
=-x^4-3x^3+4x^2-6x+7-x^3-x+2
=-x^4-4x^3+4x^2-7x+9
Ta có: \(x^4+x^3-4x^2+5x-3=0\)
\(\Leftrightarrow x^4-x^3+2x^3-2x^2-2x^2+2x+3x-3=0\)
\(\Leftrightarrow x^3.\left(x-1\right)+2x^2.\left(x-1\right)-2x.\left(x-1\right)+3.\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^3+2x^2-2x+3\right).\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^3+3x^2-x^2-3x+x+3\right).\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right).\left(x-1\right).\left(x+3\right)=0\)
Mặt khác: \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
\(\Rightarrow\)\(\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
Vậy .........
Ta có : \(x^4+x^3-4x^2+5x-3=0\left(1\right)\)
\(\Leftrightarrow\left(x^4-x^3\right)+\left(2x^3-2x^2\right)-\left(2x^2-2x\right)+\left(3x-3\right)=0\)
\(\Leftrightarrow x^3\left(x-1\right)+2x^2\left(x-1\right)-2x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2-2x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)[\left(x^3+3x^2\right)-\left(x^2+3x\right)+\left(x+3\right)]=0\)
\(\Leftrightarrow\left(x-1\right)[x^2\left(x+3\right)-x\left(x+3\right)+\left(x+3\right)]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)\left(x^2-x+1\right)=0\)
Mà \(x^2-x+1=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
Nên \(\left(1\right)\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
Vậy phương trình có tập nghiệm là : \(S=\){\(1;-3\)}