tìm x:
5(x+3)-2x(3+x)=0
4x(x-2004)-x+2004=0
(x+1)^2=x+1
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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
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
18 tháng 8 2018
a ) \(5\left(x+3\right)-6x-2x^2=0\)
\(\Leftrightarrow5\left(x+3\right)-2x\left(3+x\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5-2x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy ...
b ) \(\left(x-2004\right)=8016x-4x^2\)
\(\Leftrightarrow x-2004=-4x\left(x-2004\right)\)
\(\Leftrightarrow x-2004+4x\left(x-2004\right)=0\)
\(\Leftrightarrow\left(x-2004\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2004=0\\4x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2004\\4x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2004\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy ...
c ) \(\left(x+1\right)^2=x+1\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+1-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
Vậy ...
Y
18 tháng 8 2018
a) \(5\left(x+3\right)-6x-2x^2=0\)
\(\Rightarrow5\left(x+3\right)-2x\left(3+x\right)=0\)
\(\Rightarrow\left(x+3\right)\left(5-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\5-2x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
b) \(\left(x-2004\right)=8016x-4x^2\)
\(\Rightarrow\left(x-2004\right)=4x\left(2004-x\right)\)
\(\Rightarrow\left(x-2004\right)-4x\left(2004-x\right)=0\)
\(\Rightarrow\left(x-2004\right)+4x\left(x-2004\right)=0\)
\(\Rightarrow\left(x-2004\right)\left(1+4x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2004=0\\1+4x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2004\\x=-\dfrac{1}{4}\end{matrix}\right.\)
c) \(\left(x+1\right)^2=x+1\)
\(\Rightarrow\left(x+1\right)^2-\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x+1-1\right)=0\)
\(\Rightarrow\left(x+1\right)x=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
27 tháng 5 2022
Câu 2:
a: Ta có: \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{378}{395}\\z=2004\end{matrix}\right.\)
b: \(\left|x-\dfrac{1}{2}\right|+\left|y+\dfrac{3}{2}\right|+\left|x-y-z-\dfrac{1}{2}\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+\dfrac{3}{2}=0\\x-y-z-\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{3}{2}\\z=\dfrac{3}{2}\end{matrix}\right.\)
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.............
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 .)