PK
1
K
Khách
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26 tháng 9 2016
a ) \(9x^2-49=9\)
\(\Leftrightarrow9x^2=58\)
\(\Leftrightarrow x^2=29\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=29\\x=-29\end{array}\right.\)
Vậy ......................
b ) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)
\(\Leftrightarrow\left(x^3+3^3\right)-x.\left(x^2-1^2\right)-27=0\)
\(\Leftrightarrow x^3+27-x^3+x-27=0\)
\(\Leftrightarrow x=0\)
c ) \(\left(x-1\right)\left(x+2\right)-x-2=0\)
\(\Leftrightarrow x^2+2x-x-2-x-2=0\)
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-2\end{array}\right.\)
Vây .....................
LL
1
15 tháng 12 2020
a, \(x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow x=0;4\)
b, \(x^3+x^2-9x-9=0\Leftrightarrow x^2\left(x+1\right)-9\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-9\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=-1;\pm3\)
c, \(x^2-3x-10=0\Leftrightarrow x^2+2x-5x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\Leftrightarrow x=5;-2\)
PT
3
21 tháng 8 2017
a. \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\Leftrightarrow x^2-4x+4-x^2+9-6=0\Leftrightarrow-4x+7=0\Leftrightarrow x=\dfrac{7}{4}\)
b. \(9x^2-4-\left(3x-2\right)\left(4x-5\right)=0\Leftrightarrow9x^2-4-12x^2+23x-10=0\Leftrightarrow-3x^2+23x-14=0\Leftrightarrow\left(x-7\right)\left(-3x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-7=0\\-3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)
c. \(x^2\left(x+3\right)-x^2-3x=0\Leftrightarrow x^3+2x^2-3x=0\Leftrightarrow x\left(x^2+2x-3\right)=0\Leftrightarrow x\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-3\end{matrix}\right.\)
21 tháng 8 2017
a) \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Rightarrow\left(x^2-4x+4\right)-\left(x^2-9\right)=6\)
\(\Rightarrow x^2-4x+4-x^2+9=6\)
\(\Rightarrow-4x+13=6\)
\(\Rightarrow-4x=-7\)
\(\Rightarrow x=1,75\)
TH
1
S
26 tháng 11 2015
a/ => 3x(x2 - 4) = 0
=> 3x = 0 => x = 0
hoặc x2 - 4 = 0 => x2 = 4 => x = 2 hoặc x = -2
Vậy x = 0 ; x = 2 ; x = -2
b/ => (x - 3)(x - 3 - 3 + x2) = 0
=> (x - 3) (x2 + x - 6) = 0
=> (x - 3) (x2 - 2x + 3x - 6) = 0
=> (x - 3) [x(x - 2) + 3(x - 2)] = 0
=> (x - 3)(x - 2)(x + 3) = 0
=> x - 3 = 0 => x = 3
hoặc x - 2 = 0 => x = 2
hoặc x + 3 = 0 => x = -3
Vậy x = 3 ; x = 2 ; x =-3
TN
23 tháng 7 2017
\(a,x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
\(b,\left(x-2\right)^3+6\left(x+1\right)^2-x+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x+12=0\)\(\Leftrightarrow x^3+23x+10=0\) (1)
Đặt \(t=\dfrac{x}{\dfrac{2\sqrt{69}}{3}}\Leftrightarrow x=\dfrac{2\sqrt{69}}{3}t\)
Khi đó: (1) \(\Leftrightarrow4t^3+3t=-0,2355375386\)
Đặt a= \(\sqrt[3]{-0,2355375386+\sqrt{-0,2355375386^2+1}}\)
Và \(\alpha=\dfrac{1}{2}\left(a-\dfrac{1}{a}\right)\) , ta được:
\(4\alpha^3+3\alpha=-0,2355375386\) , vậy \(t=\alpha\) là nghiệm của pt
Vậy t= \(\dfrac{1}{2}\left(\sqrt[3]{-0,2355375386}+\sqrt{-0,2355375386^2+1}\right)\) \(\left(\sqrt[3]{-0,2355375386-\sqrt{-0,2355375386^2+1}}\right)\)\(=-0,07788262891\)
\(\Rightarrow x=\dfrac{2\sqrt{69}}{3}.t=-0,4312944692\)
\(c,x^3+6x^2+12x+8=0\)
\(\Leftrightarrow\left(x+2\right)^3=0\)
\(\Leftrightarrow x+2=0\Rightarrow x=-2\)
\(d,x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
\(e,8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Rightarrow2x-1=0\Rightarrow x=\dfrac{1}{2}\)
\(f,x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Rightarrow x+3=0\Rightarrow x=-3\)
NN
1
14 tháng 10 2020
9x2 - 4 - ( 3x - 2 )( x + 5 ) = 0
<=> ( 3x - 2 )( 3x + 2 ) - ( 3x - 2 )( x + 5 ) = 0
<=> ( 3x - 2 )( 3x + 2 - x - 5 ) = 0
<=> ( 3x - 2 )( 2x - 3 ) = 0
<=> \(\orbr{\begin{cases}3x-2=0\\2x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{3}{2}\end{cases}}\)
x3 + 64 + ( x + 4 )( 2x - 3 ) = 0
<=> ( x + 4 )( x2 - 4x + 16 ) + ( x + 4 )( 2x - 3 ) = 0
<=> ( x + 4 )( x2 - 4x + 16 + 2x - 3 ) = 0
<=> ( x + 4 )( x2 - 2x + 13 ) = 0
<=> \(\orbr{\begin{cases}x+4=0\\x^2-2x+13=0\end{cases}}\Leftrightarrow x=-4\)( vì x2 - 2x + 13 = ( x2 - 2x + 1 ) + 12 = ( x - 1 )2 + 12 ≥ 12 > 0 ∀ x )
( x - 3 )( x2 + 4x + 9 ) + 2( x2 - 9 ) - 10( x - 3 ) = 0
<=> ( x - 3 )( x2 + 4x + 9 ) + 2( x - 3 )( x + 3 ) - 10( x - 3 ) = 0
<=> ( x - 3 )( x2 + 4x + 9 + 2x + 6 - 10 ) = 0
<=> ( x - 3 )( x2 + 6x + 5 ) = 0
<=> ( x - 3 )( x + 1 )( x + 5 ) = 0
<=> x = 3 hoặc x = -1 hoặc x = -5
<=> ( x - 3 )(
6 tháng 9 2016
a.\(x^3-6x^2+12x-8=0\Rightarrow\)\(\left(x-2\right)^3=0\Rightarrow x=2\)
b.\(x^3+9x^2+27x+27=0\Rightarrow\left(x+3\right)^3=0\)\(\Rightarrow x=-3\)
29 tháng 7 2017
c. \(8x^3-12x^2+6x-1=0\)
\(\Rightarrow\left(2x-1\right)^3=0\)
\(\Rightarrow x=\frac{1}{2}\)
KN
29 tháng 5 2019
a)\(2x+143=557\)
\(\Leftrightarrow2x=557-143\)
\(\Leftrightarrow2x=414\)
\(\Leftrightarrow x=414\div2\)
\(\Leftrightarrow x=207\)
Vậy x = 207
a/ => x2 + 3x - 10 = 0
=> x2 - 2x + 5x - 10 = 0
=> x(x - 2) + 5(x - 2) = 0
=> (x - 2)(x + 5) = 0
=> x - 2 = 0 => x = 2
hoặc x + 5 = 0 => x = -5
Vậy x = 2; x = -5
b/ => 3(x3 + 3x2 + 3x + 1) = 0
=> x3 + 3x2 + 3x + 1 = 0
=> (x + 1)3 = 0
=> x + 1 = 0 => x = -1
Vậy x = -1