tìm x biết : a) 25x(x-3000)-x+3000=0
b) x3 - 24x =0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
19 tháng 12 2021
a: \(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
12 tháng 12 2020
uses crt; var x:integer; begin clrscr; x:=0; while x*x*x<=3000 do begin write(x:4); x:=x+1; end; readln; end.
TM
Tìm x:
a) x3 +3x2 - 10x = 0
b) x3 - 5x2 - 14x =0
c) x3 + 5x2- 24x =0
Giải giúp mình với ạ !
Mình cảm ơn !
4
NF
24 tháng 10 2017
x3+3x2-10x=0
=>x(3+3.2-10)=0
=>x=0
x3-5x2-14x=0
=>x(3-5.2-14)=0
=>x=0
x3+5x2-24x=0
=>x(3+5.2-24)=0
=>x=0
24 tháng 10 2017
Câu a)
\(x^3+3x^2-10=0\Rightarrow x\left(x^2+3x-10\right)=0\Rightarrow x\left(x^2-2x+5x-10\right)=0\Rightarrow x\left(x\left(x-2\right)+5\left(x-2\right)\right)=0\Rightarrow x\left(x+5\right)\left(x-2\right)=0\)
\(\Rightarrow x=0;x=5;x=2\)
PA
17 tháng 7 2017
a) x^2 + 2x - 35 = 0
<=> (x - 5)(x + 7) = 0
<=> x = 5 hoặc x = - 7
b) 4x^2 - 12x - 27 = 0
<=> (2x - 9)(2x + 3) = 0
<=> x = 4,5 hoặc x = - 1,5
c) 9x^2 + 24x + 7 = 0
<=> (3x + 1)(3x + 7) = 0
<=> x = - 1/3 hoặc x = - 7/3
d) x^2 + y^2 - 4x + 6y + 13 = 0
<=> (x - 2)^2 + (y + 3)^2 = 0
<=> x = 2 và y = - 3
e) 25x^2 - 10x - 24 = 0
<=> (5x - 6)(5x + 4) = 0
<=> x = 1,2 hoặc x = - 0,8
LN
3 tháng 2 2022
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
LL
1
27 tháng 10 2021
b: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\\x=-1\end{matrix}\right.\)
c: \(\Leftrightarrow\left(x-1\right)\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\\x=-5\end{matrix}\right.\)
PT
1
OD
3
TD
7 tháng 1 2017
a. \(x^4-10x^3+25x^2-36=0\)
=> \(x^3\left(x-3\right)-7x^2\left(x-3\right)+4x\left(x-3\right)+12\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(x^3-7x^2+4x+12\right)=0\)
=>\(\left(x-3\right)\left[x^2\left(x-2\right)-5x\left(x-2\right)-6\left(x-2\right)\right]=0\)=> \(\left(x-3\right)\left(x-2\right)\left(x^2-5x-6\right)=0\)
=> \(\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(x-6\right)=0\)
=>\(\left[\begin{matrix}x=3\\x=2\\x=-1\\x=6\end{matrix}\right.\)
b) \(x^4\) - \(^{9x^2}\) - 24x - 16 = 0
=> \(x^3\left(x-4\right)+4x^2\left(x-4\right)+7x\left(x-4\right)+4\left(x-4\right)=0\)=>\(\left(x-4\right)\left(x^3+4x^2+7x+4\right)=0\)
=> \(\left(x-4\right)\left[x^2\left(x+1\right)+3x\left(x+1\right)+4\left(x+1\right)\right]=0\)=>\(\left(x-4\right)\left(x+1\right)\left(x^2+3x+4\right)=0\)
=> \(\left(x-4\right)\left(x+1\right)=0\) (vì x^2 + 3x + 4> 0)
=>\(\left[\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
TD
7 tháng 1 2017
a,pt\(\Leftrightarrow\left(x^4-10x^3+25x\right)-36=0\)\(\Leftrightarrow\left(x^2-5x\right)^2-36=0\)
\(\Leftrightarrow\left(x^2-5x-6\right)\left(x^2-5x+6\right)=0\)\(\Leftrightarrow\left[\begin{matrix}x^2-5x-6=0\\x^2-5x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}\left(x+1\right)\left(x-6\right)=0\\\left(x-2\right)\left(x-3\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=-1,x=6\\x=2,x=3\end{matrix}\right.\)
vậy pt có 4 nghiệm x=(-1,6,2,3)
b) x(x2-24)=0
=> x=0 hoặc x2-24=0<=>x=0 hoặc x2=24(loại)=>x=0