Tìm x, biết:
a) 5x2\(-\)15x=0
b) 2x3\(-\)8x=0
c) (x+1)\(-\)x\(-\)1=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
H9
HT.Phong (9A5)
CTVHS
16 tháng 11 2023
1. a) \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=14x^5+21x^7\)
b) \(\left(x^3-x^2+x-1\right):\left(x-1\right)=\dfrac{x^3-x^2+x-1}{x-1}\)
\(=\dfrac{x^2\left(x-1\right)+\left(x-1\right)}{x-1}=\dfrac{\left(x-1\right)\left(x^2+1\right)}{x-1}=x^2+1\)
16 tháng 11 2023
2: \(x^2-8x+7=0\)
=>\(x^2-x-7x+7=0\)
=>\(x\left(x-1\right)-7\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x-7\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)
1:
a: \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=21x^7+14x^5\)
b: \(\dfrac{x^3-x^2+x-1}{x-1}=\dfrac{x^2\left(x-1\right)+\left(x-1\right)}{\left(x-1\right)}\)
\(=x^2+1\)
8 tháng 9 2023
a) \(4x^2-16+\left(3x+12\right)\left(4-2x\right)\)
\(=\left(2x-4\right)\left(2x+4\right)-3\left(x+4\right)\left(2x-4\right)\)
\(=\left(2x-4\right)\left(2x+4-3x-12\right)\)
\(=-\left(2x-4\right)\left(x+8\right)\)
b) \(x^3+x^2y-15x-15y\)
\(=x^2\left(x+y\right)-15\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-15\right)\)
c) \(3\left(x+8\right)-x^2-8x\)
\(=3\left(x+8\right)-x\left(x+8\right)\)
\(=\left(x+8\right)\left(3-x\right)\)
d) \(x^3-3x^2+1-3x\)
\(=x^3+1-3x^2-3x\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
d) \(5x^2-5y^2-20x+20y\)
\(=5\left(x^2-y^2\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y-4\right)\)
30 tháng 9 2021
\(a,A=\left\{0;1;2;3;4\right\}\\ b,B=\left\{-16;-13;-10;-7;-4;-1;2;5;8\right\}\\ c,C=\left\{-9;-8;-7;...;7;8;9\right\}\\ d,x^2-3x+1=0\\ \Delta=9-4=5\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{5}}{2}\\x=\dfrac{3+\sqrt{5}}{2}\end{matrix}\right.\\ \Leftrightarrow D=\left\{\dfrac{3-\sqrt{5}}{2};\dfrac{3+\sqrt{5}}{2}\right\}\)
\(e,2x^3-5x^2+2x=0\\ \Leftrightarrow x\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow E=\left\{0;2\right\}\\ f,F=\left\{0;3;6;9;12;15;18\right\}\)
28 tháng 12 2021
\(a,x+5x^2=0\\ \Rightarrow a,x\left(1+5x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\\ b,\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=0\\ \Rightarrow x^2+6x+9+16-x^2=0\\ \Rightarrow6x+25=0\\ \Rightarrow6x=-25\\ \Rightarrow x=-\dfrac{25}{6}\)
\(c,5x\left(x-1\right)=x-1\\ \Rightarrow c,5x\left(x-1\right)-\left(x-1\right)\\ \Rightarrow\left(x-1\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ d,x^2-2x-3=0\\ \Rightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Rightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
19 tháng 10 2021
5x2-15x=0
5x(x-3) =0
TH1: 5x=0 TH2: x-3=0
=>x=0 => x=3
Vậy x thuộc {0;3}
18 tháng 6 2023
a: P(x)=6x^3-4x^2+4x-2
Q(x)=-5x^3-10x^2+6x+11
M(x)=x^3-14x^2+10x+9
b: \(C\left(x\right)=7x^4-4x^3-6x+9+3x^4-7x^3-5x^2-9x+12\)
=10x^4-11x^3-5x^2-15x+21
29 tháng 1 2022
\(\Leftrightarrow\left(x-3\right)\left(2x^2+2\right)+5x\left(3-x\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x^2-5x+2\right)=0\)
=>(x-3)(x-2)(2x-1)=0
=>x=3 hoặc x=2 hoặc x=1/2
29 tháng 1 2022
\(2\left(x-3\right)\left(x^2+1\right)+15x-5x^2=0\\ \Leftrightarrow\left(x-3\right)\left(2x^2+2\right)-5x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(2x^2-5x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left[\left(2x^2-4x\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(x-3\right)\left[2x\left(x-2\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(x-3\right)\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
a) \(5x^2-15x=0\Leftrightarrow5x\left(x-3\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
vậy \(x=0;x=3\)
b) \(2x^3-8x=0\Leftrightarrow2x\left(x^2-4\right)=0\Leftrightarrow2x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-2=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\) vậy \(x=0;x=2;x=-2\)
c) \(\left(x+1\right)-x-1=0\Leftrightarrow x+1-x-1=0\Leftrightarrow0=0\)(đúng với mọi giá trị của \(x\))
vậy phương trình có vô số nghiệm
câu a và b bn đều lm sai ; nói chính sát là câu a thiếu còn câu b sai nhưng o nghiêm trọng