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a) x3 + 3x2 + 3x + 1 = 64
=> (x + 1)3 = 64
=> (x + 1)3 = 43
=> x + 1 = 4 => x = 3
b) x3 + 6x2 + 9x = 4x
=> x3 + 6x2 + 9x - 4x = 0
=> x3 + 6x2 + 5x = 0
=> x3 + 5x2 + x2 + 5x = 0
=> x2(x + 5) + x(x + 5) = 0
=> (x + 5)(x2 + x) = 0
=> (x + 5)x(x + 1) = 0
=> \(\hept{\begin{cases}x=-5\\x=0\\x=-1\end{cases}}\)
c) 4(x - 2)2 = (x + 2)2
=> 4(x2 - 4x + 4) = x2 + 4x + 4
=> 4x2 - 16x + 16 = x2 + 4x + 4
=> 4x2 - 16x + 16 - x2 - 4x - 4 = 0
=> 3x2 - 20x + 12 = 0
=> 3x2 - 18x - 2x + 12 = 0
=> 3x(x - 6) - 2(x - 6) = 0
=> (x - 6)(3x - 2) = 0
=> \(\orbr{\begin{cases}x=6\\x=\frac{2}{3}\end{cases}}\)
d) x4 - 16x2 = 0
=> x2(x2 - 16) = 0
=> \(\orbr{\begin{cases}x^2=0\\x^2=16\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
e) x4 - 4x3 + x2 - 4x = 0
=> x4 + x2 - 4x3 - 4x = 0
=> x2(x2 + 1) - 4x(x2 + 1) = 0
=> (x2 - 4x)(x2 + 1) = 0
=> x(x - 4)(x2 + 1) = 0
=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)(vì x2 + 1 \(\ge\)1 > 0 \(\forall\)x)
f) x3 + x = 0 => x(x2 + 1) = 0 => x = 0 (vì x2 + 1 \(\ge1>0\forall\)x)
Bài 2 :
a, Ta có : \(\left(x+4\right)\left(x-1\right)=0\)
=> \(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)
b, Ta có : \(\left(3x-2\right)\left(4x-7\right)=0\)
=> \(\left[{}\begin{matrix}3x-2=0\\4x-7=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}3x=2\\4x=7\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{7}{4}\end{matrix}\right.\)
c, Ta có : \(\left(x+5\right)\left(x^2+1\right)=0\)
=> \(\left[{}\begin{matrix}x+5=0\\x^2+1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-5\\x^2+1=0\left(VL\right)\end{matrix}\right.\)
d, Ta có : \(x\left(x-1\right)\left(x^2+4\right)=0\)
=> \(\left[{}\begin{matrix}x=0\\x-1=0\\x^2+4=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\\x=1\\x^2+4=0\left(VL\right)\end{matrix}\right.\)
e, Ta có : \(\left(3x+2\right)\left(x+\frac{1}{2}\right)=0\)
=> \(\left[{}\begin{matrix}3x+2=0\\x+\frac{1}{2}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-\frac{1}{2}\end{matrix}\right.\)
f, Ta có : \(\left(x+2\right)\left(x+3\right)\left(x^2+7\right)=0\)
=> \(\left[{}\begin{matrix}x+2=0\\x-3=0\\x^2+7=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-2\\x=3\\x^2+7=0\left(VL\right)\end{matrix}\right.\)
Bài 1 :
a, Ta có : \(1-\frac{x+3}{4}-\frac{x-2}{6}=0\)
=> \(\frac{12}{12}-\frac{3\left(x+3\right)}{12}-\frac{2\left(x-2\right)}{12}=0\)
=> \(12-3\left(x+3\right)-2\left(x-2\right)=0\)
=> \(12-3x-9-2x+4=0\)
=> \(-5x=-7\)
=> \(x=\frac{7}{5}\)
a) \(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2\right)^2=9\)
\(\Rightarrow\left(x+2\right)^2=3^2\)
\(\Rightarrow x+2=3\)
\(\Rightarrow x=3-2=1\)
a) ( x + 2 )2 = 9
=> ( x + 2 ) 2 = 9
=> ( x + 2 )2 = 32
=> x + 2 = + 3
=> \(\orbr{\begin{cases}x+2=-3\\x+2=3\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
Vậy x = -1; 5
b) ( x + 2 )2 - x2 + 4 = 0
=> ( x + 2 )2 - ( x2 - 4 ) = 0
=> ( x + 2 )2 - ( x + 2 ) ( x - 2 ) = 0
=> ( x + 2 ) ( x + 2 - x + 2 ) = 0
=> ( x + 2 ) . 4 = 0
=> x + 2 = 0
=> x = - 2
Vậy x = - 2
c) 5 ( 2x - 3 )2 - 5 ( x + 1 )2 - 15( x + 4 ) ( x - 4 ) = - 10
=> 5 ( 4x2 - 12x + 9 ) - 5 ( x2 + 2x + 1 ) - 15 ( x2 - 42 ) = - 10
=> 20x2 - 60x + 45 - 5x2 - 10x - 5 - 15x2 + 240 = -10
=> - 70x + 280 = - 10
=> - 70x = - 290
=> x = \(\frac{29}{7}\)
Vậy x = \(\frac{29}{7}\)
d) x ( x + 5 ) ( x - 5 ) - ( x + 2 ) ( x2 - 2x + 4 ) = 3
=> x ( x2 - 25 ) - ( x3 - 8 ) = 3
=> x3 - 25x - x3 + 8 = 3
=> - 25x + 8 = 3
=> - 25x = -5
=> x = \(\frac{1}{5}\)
Vậy x = \(\frac{1}{5}\)
làm nốt
d) (2x-1)(3x+2)(3-x)
=(6x2+x-2)(3-x)
=-6x3+17x2+5x-6
e) (x+3)(x2+3x-5)
=x3+6x2+4x-15
f) (xy-2)(x3-2x-6)
=x4y-2x3-2x2y-6xy+4x+12
g) (5x3-x2+2x-3)(4x2-x+2)
=20x5-9x4+19x3-16x2+7x-6
Bài 1:
a) (x-2)(x2+3x+4)
=x(5x+4)-2(5x+4)
= 5x2+4x-10x-8
=5x2-6x-8
a) \(\left(x+2\right)^2-\left(x+4\right)^2=0\)
\(\Rightarrow\left(x+2-x-4\right)\left(x+2+x+4\right)=0\)
\(\Rightarrow\left(-2\right)\left(2x+6\right)=0\)
\(\Rightarrow\left(-2\right).2.\left(x+3\right)=0\)
\(\Rightarrow x+3=0\) (vì \(-4\ne0\) )
\(\Rightarrow x=-3\)
Vậy \(x=-3\) (câu này mk có sửa đề ko biết có đúng ko 
!!!)
b) \(\left(x-3\right)^2-9=0\Rightarrow\left(x-3\right)^2=9\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^2=3^2\\\left(x-3\right)^2=\left(-3\right)^2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-3=3\\x-3=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=0\end{matrix}\right.\)
Vậy \(x=6\) hoặc \(x=0\)
c) \(x^2+6x+9=0\Rightarrow\left(x+3\right)^2=0\)
\(\Rightarrow x+3=0\Rightarrow x=-3\)
Vậy \(x=-3\)
d) \(-x^3+9x^2-27x+27=0\)
\(\Rightarrow-\left(x^3-9x^2+27x-27\right)=0\)
\(\Rightarrow-\left(x-3\right)^3=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
Ta có
x 4 + 4 x 3 + 4 x 2 = 0
⇔ x 2 x 2 + 4 x + 4 = 0
⇔ x 2 x + 2 2 = 0 ⇔ x 2 = 0 x + 2 2 = 0
⇔ x = 0 x + 2 = 0 ⇔ x = 0 x = - 2
Vậy x = 0; x = -2
Đáp án cần chọn là: C