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a, \(m^3+27\)
\(\Leftrightarrow m^3+3^3\)
\(\Leftrightarrow\left(m+3\right)\left(m^2-m.3+3^2\right)\)
\(\Leftrightarrow\left(m+3\right)\left(m^2-3m+9\right)\)
b,\(\frac{1}{27}+a^3\)
\(\Leftrightarrow\frac{1}{27}\left(1+27a^3\right)\)
\(\Leftrightarrow\frac{1}{27}.\left(1+3a\right)\left(1-3a+9a^2\right)\)
c,\(\left(a+b\right)^3-c^3\)
\(\Leftrightarrow\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]\)
\(\Leftrightarrow\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)
d,\(x^9+1\)
\(\Leftrightarrow\left(x^3+1\right)\left(x^6-x^3+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)\left(x^6-x^3+1\right)\)
e,\(x^3+9x^2+27x+27\)
\(\Leftrightarrow x^3+3.x^2.3+3x.9+3^3\)
\(\Leftrightarrow x^3+3x^2.3+3x+3^2+3^3\)
\(\Leftrightarrow\left(x+3\right)^3\)
Bài làm
a, ( x + 3 )10 = 45
=> ( x + 3 )10 = ( 22)5
=> ( x + 3 )10 = 210
=> x + 3 = 2
=> x = -1
b, x15 = 2710
=> ( x3 )5 = ( 272 )5
=> x3 = 272
=> x3 = 729
=> x3 = 93
=> x = 9
Vậy x = 9
c, ( 4 - 5x )3 = 27
=> ( 4 - 5x )3 = 33
=> 4 - 5x = 3
=> 5x = 4 - 3
=> 5x = 1
=> x = \(\frac{1}{5}\)
Vậy x = \(\frac{1}{5}\)
d, ( 1 - x )3 = 82
=> ( 1 - x )3 = ( 23 )2
=> ( 1 - x )3 = 26
=> ( 1 - x )3 = ( 22 )3
=> ( 1 - x )3 = 43
=> 1 - x = 4
=> x = -3
Vậy x = -3
# Học tốt #
a. (x + 3)10 = 45
<=> (x + 3)10 = (22)5
<=> x + 3 = 2
<=> x = -1
b. x15 = 2710
<=> x15 = (33)10
<=> x15 = (32)15
<=> x = 9
c. (4 - 5x)3 = 27
<=> (4 - 5x)3 = 33
<=> 4 - 5x = 3
<=> x = \(\frac{1}{5}\)
d. (1 - x)3 = 82
<=> (1 - x)3 = (23)2
<=> (1 - x)3 = 43
<=> 1 - x = 4
<=> x = -3
\(x^3-9x^2+27x-27=-8\Leftrightarrow\left(x^3-27\right)-\left(9x^2-27x\right)=\left(x-3\right)\left(x^2+3x+9\right)-9x\left(x-3\right)=\left(x-3\right)\left(x^2-6x+9\right)=\left(x-3\right)^3=-8=\left(-2\right)^3\Rightarrow x=\left(-2\right)+3=1\)
\(64x^3+48x^2+12x+1=\left(64x^3+1\right)+\left(48x^2+12x\right)=\left(4x+1\right)\left(16x^2-4x+1\right)+12x\left(4x+1\right)=\left(4x+1\right)\left(16x^2+8x+1\right)=\left(4x+1\right)^3=27\Rightarrow4x=2\Leftrightarrow x=\frac{1}{2}\)
c) \(\left(2x-1\right)^3-4x^2.\left(2x-3\right)=5\)
\(\Leftrightarrow\left(8x^3-12x^2+6x-1\right)-\left(8x^3-12x^2\right)=5\)
\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\)
\(\Leftrightarrow6x-1=5\)
\(\Leftrightarrow6x=6\)
\(\Leftrightarrow x=1\)
d) \(\left(x+4\right)^3-x^2.\left(x+12\right)=16\)
\(\Leftrightarrow\left(x^3+12x^2+48x+64\right)-\left(x^3+12x^2\right)=16\)
\(\Leftrightarrow x^3+12x^2+48x+64-x^3-12x^2=16\)
\(\Leftrightarrow48x+64=16\)
\(\Leftrightarrow48x=-48\)
\(\Leftrightarrow x=-1\)
#vì câu a,b có người làm rồi nên mình chỉ làm c,d thôi nhé ! :)
Học Tốt !!
a) x3 - 9x2 + 27x - 27 = -8
<=> x3 - 3x2.3 + 3x.32 - 33 = -8
<=> (x - 3)3 = -23
<=> x - 3 = -2
<=> x = 1 (T/m)
Vậy x = 1.
b) 64x3 + 48x2 + 12x + 1 = 27
<=> (4x)3 + 3.(4x)2.1 + 3.4x.12 + 13 = 27
<=> (4x + 1)3 = 33
<=> 4x + 1 = 3
<=> 4x = 2
<=> x = \(\frac{1}{2}\)(T/m)
Vậy x = \(\frac{1}{2}\).
1.
$27x^2-1=(\sqrt{27}x)^2-1^2=(\sqrt{27}x-1)(\sqrt{27}x+1)$
2.
a)
$x^3-9x^2+27x-27=-8$
$\Leftrightarrow x^3-3.3x^2+3.3^2.x-3^3=-8$
$\Leftrightarrow (x-3)^3=-8=(-2)^3$
$\Rightarrow x-3=-2$
$\Leftrightarrow x=1$
b)
$64x^3+48x^2+12x+1=27$
$\Leftrightarrow (4x)^3+3.(4x)^2.1+3.4x.1^2+1^3=27$
$\Leftrightarrow (4x+1)^3=3^3$
$\Rightarrow 4x+1=3$
$\Leftrightarrow x=\frac{1}{2}$
a: \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)
=>2x+41=1
=>2x=-40
hay x=-20
b: \(\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2-12=2\)
=>3x-40=2
=>x=14
\(a.\left(x+1\right)\left(x^2-x+1\right)-x\left(x^2-5\right)=71\)
\(\Leftrightarrow x^3+1-x^3+5x=71\)
\(\Leftrightarrow5x=71-1\)
\(\Leftrightarrow5x=70\)
\(\Leftrightarrow x=70:5=14\)
\(b.\left(2x-3\right)^3-8x\left(x-1\right)^2+4x\left(4x+1\right)+27=0\)
\(\Leftrightarrow8x^3-12x^2+18x-27-8x\left(x^2-2x+1\right)+16x^2+4x+27=0\)
\(\Leftrightarrow8x^3-12x^2+18x-27-8x^3+16x^2-8x+16x^2+4x+27=0\)
\(\Leftrightarrow20x^2+14x=0\)
\(\Leftrightarrow x\left(20x+14\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\20x+14=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{7}{10}\end{cases}}}\)
a) ta có: (x+1)(x^2 -x+1) -x(x^2 -5)=71
<=>x^3 +1 -x^3 +5x=71
<=>5x=70
<=>x=14
b) ta có:(2x-3)^3 -8x(x-1)^2 +4x(4x+1)+27=0
<=>[ (2x-3)^3 +27)] - [ 8x(x-1)^2 -4x(4x+1)]=0
<=> (2x-3+3)[ (2x-3)^2 - (2x-3).3 +3^2] - 2x [ 4(x^2 -2x +1) -2(4x+1)]=0
<=>2x( 4.x^2 - 12x +9 - 6x +9 +9) - 2x( 4.x^2 -8x+4 -8x -2)=0
<=>2x(4.x^2 -18x +27) - 2x(4.x^2 -16x +2)=0
<=>2x(4.x^2 -18x+27 -4.x^2 +16x-2)=0
<=>2x(25-2x)=0
<=>x=0 hoặc 25-2x=0 <=> x=0 hoặc x=25/2
b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)=14\)
\(\Rightarrow x^3+3^3-x\left(x^2-1^2\right)=14\)
\(\Rightarrow x^3+27-x^3+x=14\)
\(\Rightarrow27+x=14\)
\(\Rightarrow x=14-27=-13\)
c) \(3\left(x-1\right)^2-3x\left(x-5\right)=1\)
\(\Rightarrow3.\left(x^2-2x+1\right)-3x^2+15x=1\)
\(\Rightarrow3x^2-6x+3-3x^2+15x=1\)
\(\Rightarrow9x+3=1\)
\(\Rightarrow9x=1-3=-2\)
\(\Rightarrow x=-\frac{2}{9}\)
(x+1)3-27
= (x+1)3-33
=(x+1-3)((x+1)2+x+1.3+32)
=(x-2)(x2+x+13)
-Học tốt-