\(5x^3-2x^2-7x=0\)
Giải pt
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NP
1
TN
18 tháng 12 2016
a)\(9x^2+5x+2=0\)
\(\Delta=5^2-4\cdot9\cdot2=-47< 0\)
Vô nghiệm
b)\(5x^2+4x-2=0\)
\(\Delta=4^2-4\cdot5\cdot\left(-2\right)=56\)
\(x_{1,2}=\frac{-4\pm\sqrt{56}}{10}\)
c)\(2x^3+7x^2+7x+2=0\)
\(\Rightarrow2x^3+6x^2+4x+x^2+3x+2=0\)
\(\Rightarrow2x\left(x^2+3x+2\right)+\left(x^2+3x+2\right)=0\)
\(\Rightarrow\left(x^2+3x+2\right)\left(2x+1\right)=0\)
\(\Rightarrow\left(x^2+2x+x+2\right)\left(2x+1\right)=0\)
\(\Rightarrow\left[x\left(x+2\right)+\left(x+2\right)\right]\left(2x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(2x+1\right)=0\)
=>x=-1 hoặc x=-2 hoặc \(x=-\frac{1}{2}\)
31 tháng 5 2018
MK làm lại câu b hồi nãy mk chép nhầm đề :))
b) / 2x + 1/ - / 5x - 2/ = 3 ( 1)
Lập bảng xét dấu , ta có :
+) Với : x < \(\dfrac{-1}{2}\) , ta có :
( 1) ⇔ - 2x - 1 + 5x - 2 = 3
⇔ 3x = 6
⇔ x = 2 ( KTM)
+) Với : \(\dfrac{-1}{2}\) ≤ x < \(\dfrac{2}{5}\) , ta có :
( 1) ⇔ 2x + 1 + 5x - 2 = 3
⇔ 7x = 4
⇔ x = \(\dfrac{4}{7}\) ( KTM)
+) Với : x ≥ \(\dfrac{2}{5}\) , ta có :
( 1) ⇔ 2x + 1 - 5x + 2 = 3
⇔ -3x = 0
⇔ x = 0 ( KTM)
Vậy , phương trình đã cho vô nghiệm
TT
31 tháng 5 2018
a)\(\left|1+4x\right|-\left|7x-2\right|=0\)
\(\left|1+4x\right|=\left|7x-2\right|\\\Leftrightarrow\left[{}\begin{matrix}1+4x=7x-2\\1+4x=-\left(7x-2\right)\end{matrix}\right.\)
TH1:
\(1+4x=7x-2\\ \Leftrightarrow4x-7x=-2-1\\ \Leftrightarrow-3x=-3\\ \Leftrightarrow x=1\)
TH2:
\(1+4x=-\left(7x-2\right)\\ \Leftrightarrow1+4x=-7x+2\\\Leftrightarrow4x+7x=2-1\\ \Leftrightarrow11x=1\\ \Leftrightarrow x=\dfrac{1}{11} \)
Vậy tập nghiệm của phương trình: S={1;\(\dfrac{1}{11}\)}
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 \(S=\left\{1;-1;2;\frac{1}{2}\right\}\)
ND
21 tháng 2 2018
3)
\(x^3-7x+6=0\)
\(\Leftrightarrow x^3+3x^2-3x^2-9x+2x+6=0\)
\(\Leftrightarrow\left(x^3+3x^2\right)-\left(3x^2+9x\right)+\left(2x+6\right)=0\)
\(\Leftrightarrow x^2\left(x+3\right)-3x\left(x+3\right)+2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
GH
21 tháng 2 2018
4) \(\left(2x+1\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)=0\)
\(\Leftrightarrow3x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vậy ................
NN
1
GH
5 tháng 3 2018
a) \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left(2x^2-2x-3x+3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x-1\right)-3\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy .................
b) \(\left(x-3\right)^2=\left(2x+1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(2x+1-x+3\right)\left(2x+1+x-3\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ...............
c) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
P/s: tới đây bn tự giải tiếp nha
TL
1
NH
13 tháng 2 2023
a)
`x^2 +5x+6=0`
`<=> x^2 + 3x +2x+6=0`
`<=> x(x+3)+2(x+3)=0`
`<=> (x+3)(x+2)=0`
`<=> x+3=0 hoặcx+2=0`
`<=> x=-3 hoặc x=-2`
b)
`x^2 -7x+6=0`
`<=> x^2 -6x-x+6=0`
`<=> x(x-6)-(x-6)=0`
`<=> (x-6)(x-1)=0`
`<=> x-6=0 hoặc x-1=0 `
`<=> x=6 hoặc x=1`
c)
`x^2 +x -12=0`
`<=> x^2 +4x-3x-12=0`
`<=> x(x+4)-3(x+4)=0`
`<=> (x+4)(x-3)=0`
`<=> x+4=0 hoặc x-3=0`
`<=> x=-4 hoặc x=3`
d)
`x^2 -x-6=0`
`<=>x^2 -3x+2x-6=0`
`<=> x(x-3)+2(x-3)=0`
`<=> (x-3)(x+2)=0`
`<=> x-3=0 hoặc x+2=0`
`<=> x=3 hoặc x=-2`
e)
`2x^2 -3x-5=0`
`<=> 2x^2 -5x+2x-5=0`
`<=> x(2x-5)+(2x-5)=0`
`<=> (2x-5)(x+1)=0`
`<=> 2x-5=0 hoặc x+1=0`
`<=> x=5/2 hoặc x=-1`
CT
1
12 tháng 7 2015
\(1;x^2+7x+10=0\Rightarrow x^2+2x+5x+10=0\Rightarrow x\left(x+2\right)+5\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x+5\right)=0\)
=> x + 2 = 0 hoặc x + 5 = 0
=> x = -2 hoặc x = - 5
2, x^4 - 5x^2 + 4 = 0
x^4 - 4x^2 - x^2 + 4 = 0
x^2 ( x^2 - 4) - ( x^2 - 4) = 0
( x^2 - 1)( x^2 - 4) = 0
( x - 1 )( x + 1)( x - 2)( x + 2) = 0
=> x= 1 hoặc x= -1 hoặc x = 2 hoặc x = - 2
Đúng cho mi8nhf mình giải tiếp cho
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 3 2020
a. \(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)\left(x+1\right)\left(2x-9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\2x+5=0\\x+1=0\\2x-9=0\end{matrix}\right.\) \(\Rightarrow x=\)
b. \(\Leftrightarrow x^3+x+3x^2+3=0\)
\(\Leftrightarrow x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+1=0\left(vn\right)\end{matrix}\right.\)
c. \(\Leftrightarrow2x\left(3x-1\right)^2-\left(9x^2-1\right)=0\)
\(\Leftrightarrow\left(6x^2-2x\right)\left(3x-1\right)-\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(6x^2-5x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-1\right)\left(6x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-1=0\\6x+1=0\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 3 2020
d.
\(\Leftrightarrow x^3-3x^2+2x-3x^2+9x-6=0\)
\(\Leftrightarrow x\left(x^2-3x+2\right)-3\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\\x-2=0\end{matrix}\right.\)
e.
\(\Leftrightarrow x^3+2x^2+x+3x^2+6x+3=0\)
\(\Leftrightarrow x\left(x^2+2x+1\right)+3\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow x\left(5x^2-7x+5x-7\right)=0\\ \Leftrightarrow x\left[5x\left(x+1\right)+7\left(x+1\right)\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\5x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=\dfrac{7}{5}\end{matrix}\right.\)
\(\Leftrightarrow5x^3+5x^2-7x^2-7x=0\)
\(\Leftrightarrow5x^2\left(x+1\right)-7x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(5x^2-7x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\5x^2-7x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)