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KK
8 tháng 12 2018
a.(2x - 5)(3x + 4) - x(6x - 5) = 4
⇔ 6x2 +8x -15x-20-6x2+5x=4
⇔-2x=24
⇔ x=-12
vậy x=12
b.(x - 2)2 + x(x - 2) = 0
⇔(x-2)(x-2+x)=0
⇔(x-2) (2x-2)=0
⇔ (x-2)2(x-2)=0
⇔(x-2)2.2=0
⇔(x-2)2=0
⇔x-2=0
⇔x=2
vậy x=2
c.(x3 + 4x2 - x - 4) : (x + 4) = 0
⇔[(x3+4x2)-(x+4)] :(x+4)=0
⇔ [x2(x+4)-(x+4)] :(x+4)=0
⇔ (x+4)(x2-1):(x+4)=0
⇔(x-1)(x+1)=0
⇔ \(\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
vậy \(\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
6 tháng 6 2016
a, \(x^2-25-\left(x+5\right)=0\)
\(\Rightarrow x^2-5^2-\left(x+5\right)=0\)
\(\Rightarrow\left(x-5\right)\times\left(x+5\right)-\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\times\left(x-5-1\right)=0\)
\(\Rightarrow\left(x+5\right)\times\left(x-6\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+5=0\\x-6=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=0-5=\left(-5\right)\\x=0+6=6\end{cases}}\)
b, \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Rightarrow\left(2x-1\right)^2-\left(\left(2x\right)^2-1^2\right)=0\)
\(\Rightarrow\left(2x-1\right)^2-\left(2x-1\right)\times\left(2x+1\right)=0\)
\(\Rightarrow\left(2x-1\right)\times\left(2x-1-\left(2x+1\right)\right)=0\)
\(\Rightarrow\left(2x-1\right)\times\left(2x-1-2x-1\right)=0\)
\(\Rightarrow\left(2x-1\right)\times\left(-2\right)=0\)\(\Rightarrow\left(-4x\right)+2=0\)
\(\Rightarrow\left(-4x\right)=0-2=-2\)
\(\Rightarrow x=\frac{-2}{-4}=\frac{1}{2}\)
c, \(x^2\times\left(x^2+4\right)-x^2-4=0\)
\(\Rightarrow x^2\times\left(x^2+4\right)-\left(x^2+4\right)=0\)
\(\Rightarrow\left(x^2-1\right)\times\left(x^2+4\right)=0\)
\(\Rightarrow\hept{\begin{cases}x^2-1=0\\x^2+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x^2=1\\x^2=\left(-4\right)\end{cases}}\)
\(\Rightarrow x=1\)
A
1 tháng 10 2020
a, \(x^2-12x-2x+24=0\Leftrightarrow x^2-14x+24=0\Leftrightarrow\left(x-12\right)\left(x-2\right)=0\)
TH1 : x = 12 ; TH2 : x = 2
b, \(x^2-5x-24=0\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
TH1 : x = 8 ; TH2 : x = -3
c, \(4x^2-12x-7=0\Leftrightarrow\left(2x+1\right)\left(2x-7\right)=0\)
TH1 : x = -1/2 ; TH2 : x = 7/2
d, \(x^3+6x^2+12x+8=0\Leftrightarrow\left(x+2\right)^3=0\Leftrightarrow x=-2\)
Tương tự HĐT thôi :)
1 tháng 10 2020
a) x2 - 12x - 2x + 24 = 0
<=> x( x - 12 ) - 2( x - 12 ) = 0
<=> ( x - 12 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}x-12=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=12\\x=2\end{cases}}\)
b) x2 - 5x - 24 = 0
<=> x2 + 3x - 8x - 24 = 0
<=> x( x + 3 ) - 8( x + 3 ) = 0
<=> ( x + 3 )( x - 8 ) = 0
<=> \(\orbr{\begin{cases}x+3=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)
c) 4x2 - 12x - 7 = 0
<=> 4x2 + 2x - 14x - 7 = 0
<=> 2x( 2x + 1 ) - 7( 2x + 1 ) = 0
<=> ( 2x + 1 )( 2x - 7 ) = 0
<=> \(\orbr{\begin{cases}2x+1=0\\2x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
d) x3 + 6x2 + 12x + 8 = 0
<=> ( x + 2 )3 = 0
<=> x + 2 = 0
<=> x = -2
e) ( x + 2 )2 - x2 + 4 = 0
<=> x2 + 4x + 4 - x2 + 4 = 0
<=> 4x + 8 = 0
<=> 4x = -8
<=> x = -2
f) 2( x + 5 ) = x2 + 5x
<=> x2 + 5x - 2x - 10 = 0
<=> x( x + 5 ) - 2( x + 5 ) = 0
<=> ( x + 5 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}x+5=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
m) 16( 2x - 3 )2 - 25( x - 5 )2 = 0
<=> 42( 2x - 3 )2 - 52( x - 5 )2 = 0
<=> [ 4( 2x - 3 ) ]2 - [ 5( x - 5 ) ]2 = 0
<=> ( 8x - 12 )2 - ( 5x - 25 )2 = 0
<=> [ 8x - 12 - ( 5x - 25 ) ][ 8x - 12 + ( 5x - 25 ) ] = 0
<=> ( 8x - 12 - 5x + 25 )( 8x - 12 + 5x - 25 ) = 0
<=> ( 3x + 13 )( 13x - 37 ) = 0
<=> \(\orbr{\begin{cases}3x+13=0\\13x-37=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{3}\\x=\frac{37}{13}\end{cases}}\)
n) x2 - 6x + 4 = 0
<=> ( x2 - 6x + 9 ) - 5 = 0
<=> ( x - 3 )2 - ( √5 )2 = 0
<=> ( x - 3 - √5 )( x - 3 + √5 ) = 0
<=> \(\orbr{\begin{cases}x-3-\sqrt{5}=0\\x-3+\sqrt{5}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3+\sqrt{5}\\x=3-\sqrt{5}\end{cases}}\)
NC
1 tháng 10 2020
a) \(x^2-12x-2x+24=0\)
\(\Leftrightarrow x\left(x-12\right)-2\left(x-12\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-12\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=12\\x=2\end{cases}}\)
b) \(x^2-5x-24=0\)
\(\Leftrightarrow\left(x^2+3x\right)-\left(8x+24\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)
c) \(4x^2-12x-7=0\)
\(\Leftrightarrow\left(4x^2-14x\right)+\left(2x-7\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}\)
d) \(x^3+6x^2+12x+8=0\)
\(\Leftrightarrow\left(x+2\right)^3=0\)
\(\Rightarrow x=-2\)
NC
1 tháng 10 2020
e) \(\left(x+2\right)^2-x^2+4=0\)
\(\Leftrightarrow4x+8=0\)
\(\Rightarrow x=-2\)
f) \(2\left(x+5\right)=x^2+5x\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
m) \(16\left(2x-3\right)^2-25\left(x-5\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}8x-12=5x-25\\8x-12=25-5x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=-13\\13x=37\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{13}{3}\\x=\frac{37}{13}\end{cases}}\)
n) \(x^2-6x+4=0\)
\(\Leftrightarrow\left(x-3\right)^2-5=0\)
\(\Leftrightarrow\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3+\sqrt{5}\\x=3-\sqrt{5}\end{cases}}\)
NT
1
KB
22 tháng 10 2018
a ) \(x^2-25-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5\right)-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5-1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
b ) \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left[\left(2x-1\right)-\left(2x+1\right)\right]=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
c ) \(x^2\left(x^2+4\right)-x^2-4=0\)
\(\Leftrightarrow x^2\left(x^2+4\right)-\left(x^2+4\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x^2+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x^2=-4\left(VL\right)\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
LD
18 tháng 9 2018
Bài 1:
a) \(9\left(4x+3\right)^2=16\left(3x-5\right)^2\)
\(114x^2+216x+81=114x^2-480x+400\)
\(144x^2+216x=144x^2-480x+400-81\)
\(114x^2+216=114x^2-480x+319\)
\(696x=319\)
\(\Rightarrow x=\frac{11}{24}\)
b) \(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)
\(\left(x-1\right)^2\left(x^2+2\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0\)
\(\Rightarrow x=1\)
c) \(x^5+x^4+x^3+x^2+x+1=0\)
\(\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)=0\)
\(\Rightarrow x=-1\)
Bài 2:
a) \(5x^3-7x^2-15x+21=0\)
\(\left(5x-7\right)\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)=0\)
\(\Rightarrow x=\frac{7}{5}\)
b) \(\left(x-3\right)^2=4x^2-20x+25\)
\(x^2-6x+9-25=4x^2-20x+25\)
\(x^2-6x+9=4x^2-20x+25-25\)
\(x^2-6x-16=4x^2-20x\)
\(x^2+14x-16=4x^2-4x^2\)
\(-3x^2+14x-16=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{8}{3}\end{cases}}\)
c) \(\left(x-1\right)^2-5=\left(x+2\right)\left(x-2\right)-x\left(x-1\right)\)
\(x^2-2x=x-4\)
\(x^2-2x=x-4+4\)
\(x^2-2x=x-x\)
\(x^2-3x=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
d) \(\left(2x-3\right)^3-\left(2x+3\right)\left(4x^2-1\right)=-24\)
\(-48x^2+56x-24=-24\)
\(-48x^2+56x=-24+24\)
\(-48x^2+56=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{7}{6}\end{cases}}\)
mình ko chắc
6 tháng 10 2020
a) \(4x^3-9x=0\)
\(\Leftrightarrow x\left(4x^2-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\4x^2=9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\frac{3}{2}\end{cases}}\)
b) \(3x\left(x-2\right)-5x+10=0\)
\(\Leftrightarrow\left(3x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}\)
c) \(4x\left(x+3\right)-x^2+9=0\)
\(\Leftrightarrow4x\left(x+3\right)-\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(3x+3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)
d) \(\left(2x+5\right)\left(x-4\right)=\left(x-4\right)\left(5-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow3x\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
6 tháng 10 2020
e) \(16x^2-25=\left(4x-5\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(4x-5\right)\left(4x+5\right)-\left(4x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(4x-5\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=-2\end{cases}}\)
f) \(\left(x+\frac{1}{5}\right)^2=\frac{64}{9}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{8}{3}\\x+\frac{1}{5}=-\frac{8}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{37}{15}\\x=-\frac{43}{15}\end{cases}}\)
g) \(9\left(x+2\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}3x+6=x+3\\3x+6=-x-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-3\\4x=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{9}{4}\end{cases}}\)
a, x^2(x^2+4)-x^2-4=0
=) x^2(x^2+4)-(x^2+4)=0
=) (x^2+4)(x^2-1)=0
=)(x^2+4)(x-1)(x+1)=0
=) x^2+4=0 =)x^2=4 =)x=2
x-1=0 =)x=1
x+1=0 =)x=-1
b,x^2-25-x-5=0
=)(x+5)(x-5)-(x+5)=0
=)(x+5)(x-6)=0
=)x+5=0 =) x=-5
x-6=0 =) x=6