4x(x-2004)- x + 2004 =0
K
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29 tháng 8 2019
\(PT\Leftrightarrow\left(4x-1\right).\left(x-2004\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=0\\x-2004=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{4}\\x=2004\end{matrix}\right.\)
Vậy : ...
K
2
12 tháng 6 2018
Đơn giản như đang dỡn :V
a )
\(5\left(x+3\right)-2x\left(3+x\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(5-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\5-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy..........................
b )
\(4x\left(x-2004\right)-x+2004=0\)
\(\Leftrightarrow4x\left(x-2004\right)-\left(x-2004\right)=0\)
\(\Leftrightarrow\left(x-2004\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2004=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2004\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy.....................
c )
\(\left(x+1\right)^2=x+1\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy.............
V
12 tháng 6 2018
Tìm x:
5(x+3)-2x(3+x)=0
<=>(x+3)(5-2x)=0<=>\(\left\{{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
(x+1)^2=x+1
<=> (x+1).(x+1-1)=0
<=>x(x+1)=0
<=>\(\left\{{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
(bạn ơi , mk ko biết làm câu : 4x(x-2004)-x+2004=0 đâu . Tại vì mk mới học lớp 6 nâng cao nên ko biết làm bài lớp 7 đâu .)
HP
4
19 tháng 9 2016
a/ \(5\left(x+3\right)-2x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(5-2x\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=\frac{5}{2}\end{array}\right.\)
b/ \(4x\left(x-2004\right)-x+2004=0\)
\(\Leftrightarrow4x\left(x-2004\right)-\left(x-2004\right)=0\)
\(\Leftrightarrow\left(x-2007\right)\left(4x-1\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=2007\\x=\frac{1}{4}\end{array}\right.\)
c/ \(\left(x+1\right)^2=x+1\Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-1\end{array}\right.\)
19 tháng 9 2016
A) 5(x+3)-2x(3+x)=0
=> 5(x+3)-2x(x+3)=0
=> (5-2x)(x+3)=0
\(\Rightarrow\left[\begin{array}{nghiempt}5-2x=0\\x+3=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-3\end{array}\right.\)
SL
2
LC
24 tháng 9 2016
Tìm x
a) 5(x+3)-2x(3+x)=0
\(\Leftrightarrow\left(x+3\right)\left(5-2x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=\frac{5}{2}\end{array}\right.\)
b) 4x(x-2004)-x+2004
\(\Leftrightarrow4x\left(x-2004\right)-\left(x-2004\right)=0\)
\(\Leftrightarrow\left(x-2007\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2007\\x=\frac{1}{4}\end{array}\right.\)
c) (x+1)2=x+1
\(\Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-1\end{array}\right.\)
24 tháng 9 2016
c) \(\left(x+1\right)^2=x+1\)
\(\Rightarrow\left(x+1\right)^2-\left(x+1\right)=0\)
\(\Rightarrow x+1.\left(x+1-1\right)=0\)
\(\Rightarrow\left(x+1\right).x=0\)
\(\Rightarrow x+1=0\) hoặc \(x=0\)
+) \(x+1=0\Rightarrow x=-1\)
Vậy x = 0 hoặc x = -1
28 tháng 6 2017
a.\(5\left(x+3\right)-2x\left(3+x\right)=0\)
\(\Leftrightarrow\left(3+x\right)\left(5-2x\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\5-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{5}{2}\end{cases}}}\)
c.\(\left(x+1\right)^2=x+1\Leftrightarrow\left(x+1\right)x=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
28 tháng 6 2017
a)\(5\left(x+3\right)-2x\left(x+3\right)=0\)
\(\left(5-2x\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5-2x=0\\x+3=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)
b)\(4x\left(x+2004\right)-x+2004=0\)
\(4x^2+8016x-x+2004 =0\)
\(4x^2+8015x+2004=0\)
Xem lại đề
NT
1
NN
1
5 tháng 1 2019
\(\dfrac{x^2-2x+2004}{x^2}=\dfrac{\dfrac{2003}{2004}x^2+\dfrac{1}{2004}x^2-2x+2004}{x^2}=\dfrac{2003}{2004}+\dfrac{\dfrac{1}{2004}\left(x^2-2.2004+2004^2\right)}{x^2}=\dfrac{2003}{2004}+\dfrac{\dfrac{1}{2004}\left(x-2004\right)^2}{x^2}\ge\dfrac{2003}{2004}\)
\("="\Leftrightarrow x=2004\)
TT
23 tháng 9 2017
Bài 1:
a.\(y.\left(x-z\right)+7\left(z-x\right)\)
\(=y\left(x-z\right)-7\left(x-z\right)\)
\(=\left(y-7\right)\left(x-z\right)\)
b,\(27x^2\left(y-1\right)-9x^3\left(1-y\right)\)
\(=27x^2\left(y-1\right)+9x^3\left(y-1\right)\)
\(=\left(27x^2+9x^3\right)\left(y-1\right)\)
Bài 2
a.\(5\left(x+3\right)-2x\left(3+x\right)=0\)
\(\left(5-2x\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5-2x=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2,5\\x=-3\end{matrix}\right.\)
b.\(4x\left(x-2004\right)-x+2004=0\)
\(4x\left(x-2004\right)-\left(x-2004\right)=0\)
\(\left(4x-1\right)\left(x-2004\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=0\\x-2004=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0,25\\x=2004\end{matrix}\right.\)
c.\(\left(x+1\right)^2=x+1\)
\(\left(x+1\right)^2-x-1=0\)
\(\left(x+1\right)^2-\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+1-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+1-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
HN
23 tháng 9 2017
bài 1
a) y(x-z)+7(z-x)= y(x-z)-7(x-z)= (x-z)(y-7)
b) 27x2.(y-1)-9x3.(1-y)= 27x2.(y-1)+9x3.(y-1)= (y-1)(27x2-9x3)
bài 2
a) 5(x+3)+2x(x+3)=0
=(x+3)(5+2x)=0
\(\Leftrightarrow\)x+3=0 hoặc 5+2x=0
=>x=-3 hoặc x=\(\dfrac{-5}{2}\)
b)=4x(x-2014)-(x-2014)=0
= (x-2014)(4x-1)=0
\(\Leftrightarrow\)x-2014=0 hoặc 4x-1=0
=>x=2014 hoặc x= \(\dfrac{1}{4}\)
câu c) thấy kì kì, k biết làm
AJ
3
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 5 2019
Giả sử tồn tại \(x\) sao cho \(x^2+x+1=0\) (ví dụ trên trường số phức)
\(\Leftrightarrow x+1+\frac{1}{x}=0\Rightarrow x+\frac{1}{x}=-1\)
Ta có \(\left(x+\frac{1}{x}\right)^2=\left(-1\right)^2\Leftrightarrow x^2+\frac{1}{x^2}=-1\)
\(\left(x+\frac{1}{x}\right)^3=\left(-1\right)^3\Leftrightarrow x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=-1\Leftrightarrow x^3+\frac{1}{x^3}=2\)
Đặt \(a_n=x^n+\frac{1}{x^n}\Rightarrow a_1=-1;a_2=-1;a_3=2\)
\(a_1.a_n=\left(x+\frac{1}{x}\right)\left(x^n+\frac{1}{x^n}\right)=x^{n+1}+\frac{1}{x^{n+1}}+x^{n-1}+\frac{1}{x^{n-1}}=a_{n+1}+a_{n-1}\)
\(\Rightarrow a_{n+1}=a_1.a_n-a_{n-1}=-a_n-a_{n-1}=-\left(a_n+a_{n-1}\right)\)
Thay \(n=3;4;5;6...\) vào ta được:
\(a_4=-\left(a_3+a_2\right)=-1;a_5=-\left(a_4+a_3\right)=-1;a_6=-\left(a_5+a_4\right)=2\)
Nhìn vào quy luật ta thấy: \(a_k=-1\) nếu \(k⋮̸3\)
Và \(a_k=2\) nếu \(k⋮3\)
Do \(2004⋮3\Rightarrow a_{2004}=2\) hay \(x^{2004}+\frac{1}{x^{2004}}=2\)
\(4x\left(x-2004\right)-x+2004=0\)
\(\Leftrightarrow4x\left(x-2004\right)-\left(x-2004\right)=0\)
\(\Leftrightarrow\left(4x-1\right).\left(x-2004\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=0\\x-2004=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=1\\x=2004\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{4}\\x=2004\end{matrix}\right.\)
Vậy : \(x\in\left\{\frac{1}{4},2004\right\}\)