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TK
6
26 tháng 8 2020
a) 5x( x - 1 ) = x - 1
<=> 5x2 - 5x = x - 1
<=> 5x2 - 5x - x + 1 = 0
<=> 5x2 - 6x + 1 = 0
<=> 5x2 - 5x - x + 1 = 0
<=> 5x( x - 1 ) - 1( x - 1 ) = 0
<=> ( x - 1 )( 5x - 1 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
b) 2( x + 5 ) - x2 - 5x = 0
<=> 2x + 10 - x2 - 5x = 0
<=> -x2 - 3x + 10 = 0
<=> -x2 - 5x + 2x + 10 = 0
<=> -x( x + 5 ) + 2( x + 5 ) = 0
<=> ( x + 5 )( 2 - x ) = 0
<=> \(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
c) x2 - 2x - 3 = 0
<=> x2 + x - 3x - 3 = 0
<=> x( x + 1 ) - 3( x + 1 ) = 0
<=> ( x + 1 )( x - 3 ) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
d) 2x2 + 5x - 3 = 0
<=> 2x2 - x + 6x - 3 = 0
,<=> x( 2x - 1 ) + 3( 2x - 1 ) = 0
<=> ( 2x - 1 )( x + 3 ) = 0
<=> \(\orbr{\begin{cases}2x-1=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)
PN
26 tháng 8 2020
a) 5x ( x - 1 ) = x - 1 <=> 5x2 - 5x - x + 1 = 0
<=> 5x2 - 6x + 1 = 0 <=> 5x2 - x - ( 5x - 1 ) = 0
<=> x ( 5x - 1 ) - ( 5x - 1 ) = 0 <=> ( x - 1 )( 5x - 1 ) = 0
<=> x = 1 hoặc x = 1/5
b) 2 ( x + 5 ) - x2 - 5x = 0 <=> 2 ( x + 5 ) - x ( x + 5 ) = 0
<=> ( 2 - x ) ( x + 5 ) = 0 <=> x = 2 hoặc x = -5
c) x2 - 2x - 3 = 0 <=> x2 + x - 3x - 3 = 0
<=> x ( x + 1 ) - 3 ( x + 1 ) = 0 <=> ( x - 3 ) ( x + 1 ) = 0
<=> x = 3 hoặc x = -1
d) 2x2 + 5x - 3 = 0
Ta có : delta = 52 - 4.2.3 = 25 - 24 = 1
Khi đó : x = -1 hoặc x = 3/2
A
1 tháng 9 2020
Bài 1 :
a, \(\left(x-3\right)^2-4=0\Leftrightarrow\left(x-3\right)^2=4\Leftrightarrow\left(x-3\right)^2=\left(\pm2\right)^2\)
TH1 : \(x-3=2\Leftrightarrow x=5\)
TH2 : \(x-3=-2\Leftrightarrow x=1\)
b, \(x^2-2x=24\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)
TH1 : \(x-6=0\Leftrightarrow x=6\)
TH2 : \(x+4=0\Leftrightarrow x=-4\)
c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-4\right)=0\)
\(\Leftrightarrow2x+30=0\Leftrightarrow x=-15\)
d, tương tự
5 tháng 12 2017
1)⇔x2+1x-3x+3=0
⇔x(x+1)-3(x+1)=0
⇔(x+1)(x-3)=0
⇔x+1=0 hoặc x-3=0
⇔x=-1 hoặc x=3
5 tháng 12 2017
4)⇔x(1+5x)=0
⇔x=0 hoặc 1+5x=0
⇔x=0 hoặc 5x=-1
⇔x=0 hoặc x=-0.2
NK
4
N
24 tháng 8 2017
a, \(2\left(x+5\right)-x^2-5x=0\)
\(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}\)
HH
25 tháng 6 2018
a/ \(\left(x+2\right)^2-9=0\)
<=> \(\left(x+2-3\right)\left(x+2+3\right)=0\)
<=> \(\left(x-1\right)\left(x+5\right)=0\)
<=> \(\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
b/ \(x^2-2x+1=25\)
<=> \(\left(x-1\right)^2=25\)
<=> \(\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}}\)
<=> \(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
MA
1
B
30 tháng 10 2020
a) \(5x\left(x+4\right)-x\left(5x+1\right)=0\)
\(\Leftrightarrow x\left[5\left(x+4\right)-5x-1\right]=0\)
\(\Leftrightarrow x\left(5x+20-5x-1\right)=0\Leftrightarrow x=0\)
b) \(3x\left(5-x\right)+4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(4-3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{4}{3}\end{cases}}\)
c) \(x\left(x-3\right)+4x-12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
d) \(x^2-36=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
e) \(x^2+3x+1=2\)
\(\Leftrightarrow x^2+3x+1-2=0\)
\(\Leftrightarrow x^2+3x-1=0\)
\(\Leftrightarrow x^2+3x+\frac{3}{2}-\frac{5}{2}=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2-\frac{5}{2}=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}+\frac{\sqrt{5}}{\sqrt{2}}\right)\left(x+\frac{3}{2}-\frac{\sqrt{5}}{\sqrt{2}}\right)=0\)
Còn lại ........... Tự lm nất nha
BA
26 tháng 7 2015
Câu a) (2X-1)2-(X+3)2=0
(2X-1-X-3)(2X-1+X+3)=0
(X-4)(3X+2)=0
-> X-4=0 HOẶC 3X+2=0
xét:X-4=0
X=0+4=4
Xét:3X+2=0
3X=0-2=-2
X= -2:3= -2/3
=>X=4 HOẶC -2/3
CÂU b) 2(X+5)-X2-5X=0
2(X+5)-(x2+5X)=0
2(X+5)-X(x+5)=0
(X+5)(2-x)=0
-> X+5=0 HOẶC 2-X=0
XÉT: X+5=0 XÉT: 2-X=0
X=0-5= -5 X=2-0=2
=>X= -5 HOẶC 2
K
2
B
8 tháng 12 2016
Sửa: a)\(3x^2-12=0\)
\(\Rightarrow3x^2=12\)
\(\Rightarrow x^2=\frac{12}{3}=4\)
\(\Rightarrow x=\sqrt{4}=2\)
Vậy: x=2
b)\(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+5=0\\2-x=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=-5\\x=2\end{array}\right.\)
Vậy: \(x=-5;2\)
c)\(\Rightarrow2x^2-2x+5x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-1=0\\2x+5=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=1\\x=-\frac{5}{2}\end{array}\right.\)
Vậy: \(x=1;-\frac{5}{2}\)
18 tháng 12 2016
a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
\(\Leftrightarrow-13x=26\Leftrightarrow x=-2\)
b) \(5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=\frac{1}{5}\end{array}\right.\)
c) \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\x=2\end{array}\right.\)
d) \(\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=8\\x=-\frac{2}{3}\end{array}\right.\)
e) \(3x^3-48x=0\)
\(\Leftrightarrow3x\left(x^2-16\right)=0\)
\(\Leftrightarrow3x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=4\\x=-4\end{array}\right.\)
f) \(x^3+x^2-4x=4\)
\(\Leftrightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\\x=-2\end{array}\right.\)
Pt <=> (2-x)(x+5)=0
<=> x=2 hoặc x=-5
\(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)