giải phuong trình
x4 + 9 = 5x(3 - x2)
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20 tháng 12 2022
1:=x^3-27-x^2-3=x^3-x^2-30
2: =x-2+125x^3+150x^2+60x+8
=125x^3+150x^2+61x+6
3: \(=2xy-5y+5y=2xy\)
4: =25x-10x^2+15x
=-10x^2+40x
NS
1
7 tháng 4 2017
Ta có:
\(x^3+5x^2+3x-9=0\)
\(\Leftrightarrow x^3+3x^2+2x^2+6x-3x-9=0\)
\(\Leftrightarrow x^2\left(x+3\right)+2x\left(x+3\right)-3\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+2x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x+3\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Vậy PT có nghiệm là \(\left\{1;-3\right\}\)
LT
1
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.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
2 tháng 11 2021
\(a,\Rightarrow\left(2x-5\right)^2+2\left(2x-5\right)\left(x+2\right)+\left(x+2\right)^2=0\\ \Rightarrow\left(2x-5+x+2\right)^2=0\\ \Rightarrow3x-3=0\\ \Rightarrow x=1\\ b,\Rightarrow9-\left(x^2-5x\right)^2=9\\ \Rightarrow x^2-5x=0\\ \Rightarrow x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
RH
18 tháng 1 2022
a) (3x + 2)2 - (3x - 2)2 = 5x + 38
<=> 6x.4 = 5x + 38 <=> 19x = 38 <=> x = 2
b) 3(x - 2)2 + 9(x - 1) = 3(x2 + x - 3)
<=> 3x2 - 12x + 12 + 9x - 9 = 3x2 + 3x - 9
<=> -6x = -12 <=> x = 2
c) (x + 3)2 - (x - 3)2 = 6x + 8
<=> 2x.6 = 6x + 8 <=> 6x = 8 <=> x = 4/3
d) (x - 1)3 - x(x + 1)2 = 5x(2 - x) - 11(x + 2)
<=> x3 - 3x2 + 3x - 1 - x3 - 2x2 - x = 10x - 5x2 - 11x - 22
<=> 3x = -21 <=> x = -7
e) (x + 1)(x2 - x + 1) - 2x = x(x - 1)(x + 1)
<=> x3 - 1 - 2x = x3 - x
<=> x = -1
ND
15 tháng 3 2018
\(x^4+9=5x\left(3-x^2\right)\)
\(\Leftrightarrow x^4+9=15x-5x^3\)
\(\Leftrightarrow x^4+5x^3-15x+9=0\)
\(\Leftrightarrow x^4-x^3+6x^3-6x^2+6x^2-6x-9x+9=0\)
\(\Leftrightarrow\left(x^4-x^3\right)+\left(6x^3-6x^2\right)+\left(6x^2-6x\right)-\left(9x-9\right)=0\)
\(\Leftrightarrow x^3\left(x-1\right)+6x^2\left(x-1\right)+6x\left(x-1\right)-9\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+6x^2+6x-9\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+3x^2+9x-3x-9\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+3\right)+3x\left(x+3\right)-3\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)\left(x^2+3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\\x^2+3x-3=0\end{matrix}\right.\)
Ta có: \(x^2+3x-3=0\)
\(\Leftrightarrow x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{21}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=\dfrac{21}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{21}}{2}\\x=\dfrac{-3-\sqrt{21}}{2}\end{matrix}\right.\)
Vậy: \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\\x=\dfrac{-3+\sqrt{21}}{2}\\x=\dfrac{-3-\sqrt{21}}{2}\end{matrix}\right.\)
MV
8 tháng 1 2018
\(x^4+2x^3-2x^2+2x-3=0\\ \Leftrightarrow x^4+3x^3-x^3-3x^2+x^2+3x-x-3=0\\ \Leftrightarrow x^3\left(x+3\right)-x^2\left(x+3\right)+x\left(x+3\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^3-x^2+x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left[x^2\left(x-1\right)+\left(x-1\right)\right]=0\\ \Leftrightarrow\left(x+3\right)\left(x-1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-1=0\\x^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\left(\text{vì }x^2+1\ge1>0\right)\)
Vậy ...
\(\left(x-1\right)\left(x^2+5x-2\right)-x^3+1=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left[\left(x^2+5x-2\right)-\left(x^2+x+1\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy ...
\(x^2+\left(x+2\right)\left(11x-7\right)=4\\ \Leftrightarrow x^2-4+\left(x+2\right)\left(11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-2\right)+\left(11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-2+11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\\ \Leftrightarrow3\left(x+2\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\4x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy ...
27 tháng 2 2018
nghiệm đâu bạn ưi...nó là phương trình vô nghiệm hay vô số nghiệm vậy m :))
Ta có PT <=> x4 + 5x3 - 15x + 9 = 0
<=> (x - 1)(x + 3)(x2 + 3x - 3) = 0
Tới đây thì đơn giản rồi