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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) \(\left(2x+5\right)\left(x-4\right)=\left(x-4\right)\left(5-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow3x\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
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) 4x3 - 9x = 0
<=> x( 4x2 - 9 ) = 0
<=> x( 2x - 3 )( 2x + 3 ) = 0
<=> x = 0 hoặc 2x - 3 = 0 hoặc 2x + 3 = 0
<=> x = 0 hoặc x = ±3/2
b) 3x( x - 2 ) - 5x + 10 = 0
<=> 3x( x - 2 ) - 5( x - 2 ) = 0
<=> ( x - 2 )( 3x - 5 ) = 0
<=> x - 2 = 0 hoặc 3x - 5 = 0
<=> x = 2 hoặc x = 5/3
c) 4x( x + 3 ) - x2 + 9 = 0
<=> 4x( x + 3 ) - ( x2 - 9 ) = 0
<=> 4x( x + 3 ) - ( x - 3 )( x + 3 ) = 0
<=> ( x + 3 )[ 4x - ( x - 3 ) ] = 0
<=> ( x + 3 )( 4x - x + 3 ) = 0
<=> ( x + 3 )( 3x + 3 ) = 0
<=> x + 3 = 0 hoặc 3x + 3 = 0
<=> x = -3 hoặc x= -1
d) ( 2x + 5 )( x - 4 ) = ( x - 4 )( 5 - x )
<=> ( 2x + 5 )( x - 4 ) - ( x - 4 )( 5 - x ) = 0
<=> ( x - 4 )[ ( 2x + 5 ) - ( 5 - x ) ] = 0
<=> ( x - 4 )( 2x + 5 - 5 + x ) = 0
<=> ( x - 4 ).3x = 0
<=> x - 4 = 0 hoặc 3x = 0
<=> x = 4 hoặc x = 0
e) 16x2 - 25 = ( 4x - 5 )( 2x + 1 )
<=> ( 4x - 5 )( 4x + 5 ) - ( 4x - 5 )( 2x + 1 ) = 0
<=> ( 4x - 5 )[ ( 4x + 5 ) - ( 2x + 1 ) ] = 0
<=> ( 4x - 5 )( 4x + 5 - 2x - 1 ) = 0
<=> ( 4x - 5 )( 2x + 4 ) = 0
<=> 4x - 5 = 0 hoặc 2x + 4 = 0
<=> x = 5/4 hoặc x = -2
f) ( x + 1/5 )2 = 64/9
<=> ( x + 1/5 )2 = ( ±8/3 )2
<=> x + 1/5 = 8/3 hoặc x + 1/5 = -8/3
<=> x = 37/15 hoặc x = -43/15
g) 9( x + 2 )2 = ( x + 3 )2
<=> 32( x + 2 )2 - ( x + 3 )2 = 0
<=> [ 3( x + 2 ) ]2 - ( x + 3 )2 = 0
<=> ( 3x + 6 )2 - ( x + 3 )2 = 0
<=> [ ( 3x + 6 ) - ( x + 3 ) ][ ( 3x + 6 ) + ( x + 3 ) ] = 0
<=> ( 3x + 6 - x - 3 )( 3x + 6 + x + 3 ) = 0
<=> ( 2x + 3 )( 4x + 9 ) = 0
<=> 2x + 3 = 0 hoặc 4x + 9 = 0
<=> x = -3/2 hoặc x = -9/4
a) \(x^3+64=0\)
\(x^3=0-64\)
\(x^3=-64\)
\(x^3=-4^3\)
\(\Rightarrow x=-4\)
b)Tương tự
a) x^4 - 2x^2 + 1 = 0
=> ( x^2 - 1 )^2 = 0
=> x^2 - 1 = 0
=> x^2 = 1
=> x = 1 hoặc x = -1
a) x4-2x2+1=0
(thang Tran giải rồi nhé)
b) x4-2x2-8=0
<=> x^4 - 2x^2 +1 -9 =0
<=> (x^2 -1)^2 -9 =0
\(\Leftrightarrow\orbr{\begin{cases}x^2-1=-3\\x^2-1=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-2\left(VN\right)\\x=+_-\sqrt{2}\end{cases}}}\)
Vậy x=+- căn 2
c) x4-4x2-60=0
\(\Leftrightarrow x^4-4x^2+4-64=0\)
\(\Leftrightarrow\left(x^2-2\right)-64=0\)
\(\Leftrightarrow\left(x^2+62\right)\left(x^2-66\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+62=0\\x^2-66=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-62\left(VN\right)\\x^2=+_-\sqrt{66}\end{cases}}}\)
Vậy x=+- căn 66
d) x6-16x2+64=0
b, ( x + 2)^2 - x^2 + 4 = 0
( x + 2)^2 - ( x^2 - 2^2) = 0
( x + 2)^2 - ( x - 2)( x +2) = 0
( x+ 2)( x + 2 - x + 2) = 0
4( x + 2) = 0
=> x + 2 = 0
=> x = - 2
a, ( 2x - 1)^2 -9 = 0
( 2x - 1)^2 - 3^2 = 0
( 2x- 1 - 3)( 2x - 1 + 3 ) = 0
( 2x - 4)( 2x + 2) = 0
=> 2x - 4 = 0 hoặc 2x+ 2 = 0
=> x = 2 hoặc x = -2
a) 4x2 - 9=0
(2x)2 - 32 = 0
=》(2x - 3)(2x+3) =0
=》 2x - 3 = 0 hoặc 2x +3 = 0
=》x = 1,5 hoặc x = - 1,5
b) (x + 1)2 - 16 = 0
=》( x + 1)2 - 42 = 0
=》( x - 3 )( x + 5 ) =0
=》 x - 3 = 0 hoặc x + 5 = 0
=》 x = 3 hoặc x = -5
c) ( x + 1)2 - (2x + 3)2 = 0
=》 ( x + 1 - 2x - 3)(x+1 +2x +3 ) =0
=》 ( -x - 2 )( 3x + 4 ) = 0
=》 -x -2 =0 hoặc 3x + 4 = 0
=》 x = -2 hoặc x = -4/3
d) 4(3x +2)2 - 9( x + 1 )2 =0
=》 [ 2(3x +2) ]2 - [3 (x + 1)] 2 = 0
=> ( 6x +4 )2 - ( 3x + 3)2 = 0
=》 ( 6x +4 -3x -3 )( 6x + 4 + 3x + 3 )=0
=》 (3x +1)(9x + 7 ) =0
=》 3x + 1 =0 hoặc 9x + 7 =0
=》 x = -1/3 hoặc x = -7/9
b)(x+3)\(^2\)-x\(^2\)=9
(x+3-x)(x+3+x)=9
3(2x+3)=9
2x+3=3
2x=0
x=0
c)\(x^2+4x+4+x^2-6x+9-2x^2+2=9\)
-2x+15=9
-2x=-6
x=3
a) x(x-2) - 5x + 10 = 0
x(x - 2) - 5(x - 2)=0
(x - 2)(x - 5)=0
=> x=2 ; x=5
a) x2 - 9 = 0
=> x2 = 0 + 9
=> x2 = 9
=> x2 = 32
=> x = 3
b) 64 - x2 = 0
=> - x2 = 0 - 64
=> - x2 = - 64
=> x2 = 64
=> x2 = 82
=> x = 8
X phải =8 và -8
3 và -3 chứ bạn??