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LP
1
ML
16 tháng 8 2015
a)x2-20-x=0
<=>(x2-5x)+(4x-20)=0
<=>x(x-5)+4(x-5)=0
<=>(x-5)(x+4)=0
<=>x-5=0 hoặc x+4=0
<=>x=5 hoặc x=-4
b)(2x+3)2-(4x2-9)=0
<=>(2x+3)(2x+3)-(2x-3)(2x+3)=0
<=>(2x+3)(2x+3-2x+3)=0
<=>(2x+3).6=0
<=>2x+3=0
<=>2x=-3
<=>x=-1,5
c)(2x2+5x+3):(x+1)=4x-5
<=>2x2+5x+3=(4x-5)(x+1)
<=>2x2+5x+3=4x2-x-5
<=>4x2-x-5-2x2-5x-3=0
<=>2x2-6x-8=0
<=>x2-3x-4=0
<=>(x2-4x)+(x-4)=0
<=>x(x-4)+(x-4)=0
<=>(x-4)(x+1)=0
<=>x+1=0 hoặc x-4=0
<=>x=-1 hoặc x=4
NN
8 tháng 12 2019
Ko viết lại đề
Câu 1: chia ra làm 3 trường hợp
Câu 2:
\(\left(x+2-x+2\right)\left(x+2\right)=0\)
\(4\left(x+2\right)=0\)
\(\Rightarrow x+2=0\)
\(x=-2\)
5 tháng 11 2017
\(a,x^4-4x^3+x^2-4x=0\)
\(\Rightarrow\left(x^4-4x^3\right)+\left(x^2-4x\right)=0\)
\(\Rightarrow x^3\left(x-4\right)+x\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x^2+x\right)=0\)
\(\Rightarrow x\left(x-4\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-1\end{matrix}\right.\)
\(b,x^3-5x^2+4x-20=0\)
\(\Rightarrow\left(x^3-5x^2\right)+\left(4x-20\right)=0\)
\(\Rightarrow x^2\left(x-5\right)+4\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x^2+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x^2+4=0\end{matrix}\right.\)
\(\Rightarrow x=5\)
AN
5 tháng 11 2017
a) \(x^4-4x^3+x^2-4x=0\)
\(\Leftrightarrow\left(x^4-4x^3\right)+\left(x^2-4x\right)=0\)
\(\Leftrightarrow x^3\left(x-4\right)+x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^3+x\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x^2+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x^2=-1\left(loai\right)\end{matrix}\right.\)
Vậy x=0; x=4
b) \(x^3-5x^2+4x-20=0\)
\(\Leftrightarrow\left(x^3-5x^2\right)+\left(4x-20\right)=0\)
\(\Leftrightarrow x^2\left(x-5\right)+4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x^2=-4\left(loai\right)\end{matrix}\right.\)
Vậy x=5
1 tháng 10 2016
a) \(x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-4=0\\x-5=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=5\end{array}\right.\)
b) \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+6=0\\x-7=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-6\\x=7\end{array}\right.\)
d) \(x^2-9x+8=0\)
\(\Leftrightarrow x^2-x-8x+8=0\)
\(\Leftrightarrow x\left(x-1\right)-8\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-8=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=8\end{array}\right.\)
g) \(3x^2-5x+2=0\)
\(\Leftrightarrow3x^2-3x-2x+2=0\)
\(\Leftrightarrow3x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=\frac{2}{3}\end{array}\right.\)
15 tháng 10 2017
Bài 4 :
\(\left(5x-20\right)+\left(3x^2-12x\right)=0\)
\(\Leftrightarrow5\left(x-4\right)+3x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(5+3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5+3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(x=4\) hoặc \(x=-\dfrac{5}{3}\)
Bài 5 :
\(\left(1-x\right)-3x^2+3x=0\)
\(\Leftrightarrow\left(1-x\right)-\left(3x^2-3x\right)=0\)
\(\Leftrightarrow\left(1-x\right)-3x\left(x-1\right)=0\)
\(\Leftrightarrow\left(1-x\right)+3x\left(1-x\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(1+3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\1+3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=-\dfrac{1}{3}\)
Bài 6 :
\(\left(4x+20\right)-\left(x+5\right)^2=0\)
\(\Leftrightarrow4\left(x+5\right)-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)\left(4-x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\-x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=-1\end{matrix}\right.\)
Vậy \(x=-5\) hoặc \(x=-1\)
TN
1
16 tháng 7 2016
b) \(x^3-5x^2+4x-20=0\)
\(=\left(x^3-5x^2\right)+\left(4x-20\right)=0\)
\(=x^2\left(x-5\right)+4\left(x-5\right)=0\)
\(=\left(x^2+4\right)\left(x-5\right)=0\)
\(x^2\ge0\)
\(\Rightarrow x^2+4\ge4>0\)
\(\Rightarrow x-5=0\)
\(\Rightarrow x=5\)
NN
0
TN
0
x2 - 4x + 5x - 20 = 0
(x2 - 4x)+(5x-20) = 0
x(x-4) + 5(x-4) = 0
(x-4).(x+5) = 0
x-4 = 0 hay x+5 = 0
x=4 hay x=-5