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3 tháng 8 2019
\(x\left(2x-7\right)-4x+14=0\Leftrightarrow\left(x-2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{2}\end{matrix}\right.\)
\(x^2\left(x-1\right)-4\left(x-1\right)=\left(x^2-4\right)\left(x-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=1\end{matrix}\right.\)
\(x^4-x^3-x^2+x=x\left(x^3+1\right)-x^2\left(x+1\right)=x\left(x+1\right)\left(x^2-x+1-x^2\right)=x\left(x+1\right)\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
3 tháng 8 2019
a) \(x\left(2x-7\right)-4x+14-0\Leftrightarrow2x^2-11x+14=0\Leftrightarrow2x^2-4x-7x+14=0\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=2\end{matrix}\right.\)
b) \(x^2\left(x-1\right)-4x+4=0\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)
c) \(x+x^2-x^3-x^4=0\Leftrightarrow x\left(x^3+x^2-x-1\right)=0\Leftrightarrow x\left[x\left(x^2-1\right)+\left(x^2-1\right)\right]=0\Leftrightarrow x\left(x+1\right)\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
d) \(2x^3+3x^2+2x+3=0\Leftrightarrow x^2\left(2x+3\right)+2x+3=0\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\Leftrightarrow x=-1,5\left(x^2+1>0\forall x\right)\)
e) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\Leftrightarrow2x-5=0\Leftrightarrow x=2,5\)
g) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
3 tháng 8 2017
a ) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\Leftrightarrow-2\left(2x-5\right)=0\)
\(\Leftrightarrow2x-5=0\Leftrightarrow x=\dfrac{5}{2}.\)
Vậy .........
b) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+4x\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-4\end{matrix}\right.\)
Vậy .........
c ) \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x^2=-1\left(loại\right)\end{matrix}\right.\)
Vậy .........
MA
7 tháng 10 2016
b) \(3x\left(x+5\right)-2x-10=0\)
\(\Leftrightarrow3x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-5\end{cases}}\)
c) \(x^3-9x=0\)
\(\Leftrightarrow x\left(x^2-9\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
TH1: \(x=0\)
TH2: \(x-3=0\Rightarrow x=3\)
\(x+3=0\Rightarrow x=-3\)
Vậy:..
d) \(\left(5+2x\right)\left(2x-7\right)=4x^2-25\)
\(\Leftrightarrow\left(5+2x\right)\left(2x-7\right)=\left(2x-5\right)\left(2x+5\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(2x-7-2x+5\right)=0\)
\(\Leftrightarrow-2\left(2x+5\right)=0\)
\(\Leftrightarrow2x+5=0\)
\(\Leftrightarrow x=-\frac{5}{2}\)
e) \(x^2-11x+30=0\)
\(\Leftrightarrow x^2-5x-6x+30=0\)
\(\Leftrightarrow x\left(x-5\right)-6\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=5\end{cases}}\)
KN
12 tháng 7 2019
g) \(\left(2x-1\right)^2-\left(2x+4\right)^2=0\)
\(\Leftrightarrow\left(2x-1+2x+4\right)\left(2x-1-2x-4\right)=0\)
\(\Leftrightarrow-5\left(4x+3\right)=0\)
\(\Leftrightarrow4x+3=0\)
\(\Leftrightarrow4x=-3\)
\(\Leftrightarrow x=\frac{-3}{4}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)
KN
12 tháng 7 2019
h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)
\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)
\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)
\(\Leftrightarrow-9x+2x-3-10x=30\)
\(\Leftrightarrow-17x-3=30\)
\(\Leftrightarrow-17x=33\)
\(\Leftrightarrow x=\frac{-33}{17}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)
Y
24 tháng 8 2018
a) \(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+7\right)+2\left(x-2\right)\left(x+2\right)-5\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left[x^2+2x+7+2\left(x+2\right)-5\right]=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+4x+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x^2+4x+6=0\end{matrix}\right.\)
Ta có:
\(x^2+4x+6\)
\(=x^2+2.x.2+4+2\)
\(=\left(x+2\right)^2+2\)
Vì \(\left(x+2\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x+2\right)^2+2\ge2\) với mọi x
\(\Rightarrow x^2+4x+6\) vô nghiệm
\(\Rightarrow x-2=0\)
\(\Rightarrow x=2\)
b) \(3x\left(x-1\right)+\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(3x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
c) \(2\left(x+3\right)x^2-3x=0\)
\(\Rightarrow x\left[2\left(x+3\right)x-3\right]=0\)
\(\Rightarrow x\left(2x^2+6x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x^2+6x-3=0\end{matrix}\right.\)
Ta có:
\(2x^2+6x-3\)
\(=2\left(x^2+3x-\dfrac{3}{2}\right)\)
\(=2\left(x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{9}{4}-\dfrac{3}{2}\right)\)
\(=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{15}{2}\)
Vì \(2\left(x+\dfrac{3}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow2\left(x+\dfrac{3}{2}\right)^2-\dfrac{15}{2}\ge-\dfrac{15}{2}\) với mọi x
\(\Rightarrow2x^2+6x-3\) vô nghiệm
\(\Rightarrow x=0\)
24 tháng 9 2020
a) ( 5 - 2x )( 2x + 7 ) - 4x2 + 25 = 0
<=> ( 5 - 2x )( 2x + 7 ) + ( 5 - 2x )( 5 + 2x ) = 0
<=> ( 5 - 2x )( 2x + 7 + 5 + 2x ) = 0
<=> ( 5 - 2x )( 4x + 12 ) = 0
<=> \(\orbr{\begin{cases}5-2x=0\\4x+12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)
b) ( 5x2 + 3x - 2 )2 - ( 4x2 - x - 5 )2 = 0 ( như này chứ nhỉ ? )
<=> [ ( 5x2 + 3x - 2 ) - ( 4x2 - x - 5 ) ][ ( 5x2 + 3x - 2 ) + ( 4x2 - x - 5 ) ] = 0
<=> ( 5x2 + 3x - 2 - 4x2 + x + 5 )( 5x2 + 3x - 2 + 4x2 - x - 5 ) = 0
<=> ( x2 + 4x + 3 )( 9x2 + 2x - 7 ) = 0
<=> ( x2 + x + 3x + 3 )( 9x2 + 9x - 7x - 7 ) = 0
<=> [ x( x + 1 ) + 3( x + 1 ) ][ 9x( x + 1 ) - 7( x + 1 ) ] = 0
<=> ( x + 1 )( x + 3 )( x + 1 )( 9x - 7 ) = 0
<=> ( x + 1 )2( x + 3 )( 9x - 7 ) = 0
<=> x + 1 = 0 hoặc x + 3 = 0 hoặc 9x - 7 = 0
<=> x = -1 hoặc x = -3 hoặc x = 7/9
c) 15x4 - 8x3 - 14x2 - 8x + 15 = 0
<=> 15x4 + 22x3 - 30x3 + 15x2 + 15x2 - 44x2 - 30x + 22x + 15 = 0
<=> ( 15x4 + 22x3 + 15x2 ) - ( 30x3 + 44x2 + 30x ) + ( 15x2 + 22x + 15 ) = 0
<=> x2( 15x2 + 22x + 15 ) - 2x( 15x2 + 22x + 15 ) + ( 15x2 + 22x + 15 ) = 0
<=> ( 15x2 + 22x + 15 )( x2 - 2x + 1 ) = 0
<=> ( 15x2 + 22x + 15 )( x - 1 )2 = 0
<=> \(\orbr{\begin{cases}15x^2+22x+15=0\\\left(x-1\right)^2=0\end{cases}}\)
+) ( x - 1 )2 = 0 <=> x = 1
+) 15x2 + 22x + 15 = 15( x2 + 22/15x + 121/225 ) + 104/15 = 15( x + 11/25 )2 + 104/15 ≥ 104/15 > 0 ∀ x
Vậy phương trình có nghiệm duy nhất là x = 1
CT
1
22 tháng 10 2017
\(\text{a) }4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\\ \Leftrightarrow\left[\left(2x\right)^2-5^2\right]-\left(2x-5\right)\left(2x+7\right)=0\\ \Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\\ \Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\\ \Leftrightarrow-2\left(2x-5\right)=0\\ \Leftrightarrow2x-5=0\\ \Leftrightarrow2x=5\\ \Leftrightarrow x=\dfrac{5}{2}\\ \text{Vậy }x=\dfrac{5}{2}\\ \)
\(\text{b) }2x^3+3x^2+2x+3=0\\ \Leftrightarrow\left(2x^3+3x^2\right)+\left(2x+3\right)=0\\ \Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(KTM\right)\\2x=-3\left(TM\right)\end{matrix}\right.\\ \Leftrightarrow x=-\dfrac{3}{2}\\ \text{Vậy }x=-\dfrac{3}{2}\)
H
16 tháng 12 2018
\(a,x^3=x\)
\(\Rightarrow x^3-x=0\)
\(\Rightarrow x\left(x^2-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-1=0\end{cases}}\)
\(\Rightarrow x^2=\left(-1\right)^2=1\)
\(KL:x=0;x=1\)
T
16 tháng 12 2018
c) \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}\right)\left(\frac{2x^3+3x^2+2x+3}{x+\frac{3}{2}}\right)=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}\right)\left(2x^2+2\right)=0\) (bạn tự thực hiện phép chia đa thức giúp mình)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{3}{2}=0\\2x^2+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\2\left(x^2+1\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x^2+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\left(C\right)\\x^2=-1\left(L\right)\end{cases}}\)
Vậy đa thức có nghiệm duy nhất \(x=-\frac{3}{2}\)
a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left[\left(2x\right)^2-5^2\right]-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(2x-5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(-2\right)=0\)
\(\Leftrightarrow10-4x=0\)
\(\Leftrightarrow4x=10\)
\(\Leftrightarrow x=\dfrac{10}{4}=\dfrac{5}{2}=2,5\)
Vậy: \(x=2,5\)
b) \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow2x^3+2x+3x^2+3=0\)
\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\2x+3=0\end{matrix}\right.\)\(\Leftrightarrow2x=-3\)\(\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy: \(x=-\dfrac{3}{2}\)
_Chúc bạn học tốt_
a) 4x2-25-(2x-5)(2x+7)=0
<=> (4x2-25)-(2x-5)(2x+7)=0
<=> [(2x)2-52]-(2x-5)(2x+7)=0
<=> (2x-5).(2x+5)-(2x-5)(2x+7)=0
<=> (2x-5).[(2x+5)-(2x+7)]=0
<=> (2x-5).(2x+5-2x-7)=0
<=> (2x-5).(-2)=0
=> 2x-5=0
<=> 2x=5
<=> x=5/2
Vậy x=5/2
b) 2x3+3x2+2x+3=0
<=> (2x3+2x)+(3x2+3)=0
<=> 2x(x2+1)+3(x2+1)=0
<=> (x2+1).(2x+3)=0
x2+1=0 x2= -1(vô lí)
<=> <=>
2x+3=0 x= -3/2
Vậy x= -3/2