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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
\(\Rightarrow\left[\begin{array}{nghiempt}5-2x=0\\x+3=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-3\end{array}\right.\)
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) \(\left(x+1\right)^2=x+1\)
\(\Rightarrow\left(x+1\right)^2-\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x+1-1\right)=0\)
\(\Rightarrow\left(x+1\right)x=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
KN
28 tháng 7 2015
1) (2x-1)(x+3)(2-x)=0
=>2x-1 =0 hoặc x+3=0 hoặc 2-x=0
=>x=1/2 hoặc x=-3 hoặc x=2
2)x^3 + x^2 + x + 1 = 0
=>.x^2(x+1)+(x+1)=0
=>(x^2+1)(x+1)=0
=>x^2+1=0 hoặc x+1=0
=> x =-1
3) 2x(x-3)+5(x-3) =0
=>(2x+5)(x-3)=0
=>2x+5=0 hoặc x-3=0
=>x=-5/2 hoặc x=3
4)x(2x-7)-(4x-14)=0
=> (x-2)(2x-7)=0
=> x-2 =0 hoặc 2x-7=0
=>x=2 hoặc x=7/2
5)2x^3+3x^2+2x+3=0
=>x^2(2x+3)+2x+3=0
=>(x^2+1)(2x+3)=0
=>x^2+1=0 hoặc 2x+3=0
=> x =-3/2
PA
30 tháng 11 2016
\(2x^2-7x+5=0\)
\(2x^2-2x-5x+5=0\)
\(2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x-5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)
\(x\left(2x-5\right)-4x+10=0\)
\(x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(x-2\right)=0\)
\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)
\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)
\(x^2-25-x^2+2x=15\)
\(2x=15+25\)
\(2x=40\)
\(x=\frac{40}{2}\)
\(x=20\)
\(x^2\left(2x-3\right)-12+8x=0\)
\(x^2\left(2x-3\right)+4\left(2x-3\right)=0\)
\(\left(2x-3\right)\left(x^2+4\right)=0\)
\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))
\(2x=3\)
\(x=\frac{3}{2}\)
\(x\left(x-1\right)+5x-5=0\)
\(x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(x+5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)
\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)
\(4x^2-12x+9-4x^2+4x=5\)
\(-8x=5-9\)
\(-8x=-4\)
\(x=\frac{4}{8}\)
\(x=\frac{1}{2}\)
\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)
\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)
\(\left(2x-5\right)\left(x+11\right)=0\)
\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)
5 tháng 9 2019
a) 3x(4x - 3) - 2x(5 - 6x) = 0
=> 6x2 - 9x - 10x + 12x2 = 0
=> 18x2 - 19x = 0
=> x(18x - 19) = 0
=> \(\orbr{\begin{cases}x=0\\18x-19=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{19}{18}\end{cases}}\)
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
=> 10x - 15 + 4x2 - 8x + 6x - 4x2 = 0
=> 8x - 15 = 0
=> 8x = 15
=> x = 15 : 8 = 15/8
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
=> 6x - 3x2 + 2x2 - 2x = 5x2 + 15x
=> 4x - x2 - 5x2 - 15x = 0
=> -6x2 - 11x = 0
=> -x(6x - 11) = 0
=> \(\orbr{\begin{cases}-x=0\\6x-11=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{11}{6}\end{cases}}\)
2T
5 tháng 9 2019
a) \(3x\left(4x-3\right)-2x\left(5-6x\right)=0\)
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow-19x=0\Leftrightarrow x=0\)
b) \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\Leftrightarrow x=\frac{15}{8}\)
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 .)