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\(x^3-9x^2+27x-27=-8\Leftrightarrow\left(x^3-27\right)-\left(9x^2-27x\right)=\left(x-3\right)\left(x^2+3x+9\right)-9x\left(x-3\right)=\left(x-3\right)\left(x^2-6x+9\right)=\left(x-3\right)^3=-8=\left(-2\right)^3\Rightarrow x=\left(-2\right)+3=1\)
\(64x^3+48x^2+12x+1=\left(64x^3+1\right)+\left(48x^2+12x\right)=\left(4x+1\right)\left(16x^2-4x+1\right)+12x\left(4x+1\right)=\left(4x+1\right)\left(16x^2+8x+1\right)=\left(4x+1\right)^3=27\Rightarrow4x=2\Leftrightarrow x=\frac{1}{2}\)
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 !!
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)
$64x^3+48x^2+12x+1=27$
$\Leftrightarrow (4x)^3+3.(4x)^2.1+3.4x.1^2+1^3=27$
$\Leftrightarrow (4x+1)^3=3^3$
$\Rightarrow 4x+1=3$
$\Leftrightarrow x=\frac{1}{2}$
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
a) Ta có: \(x^2+2x+1\)
\(=x^2+2\cdot x\cdot1+1^2\)
\(=\left(x+1\right)^2\)
b) Ta có: \(1-2y+y^2\)
\(=y^2-2\cdot y\cdot1+1^2\)
\(=\left(y-1\right)^2\)
c) Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-x^2-2x^2+2x+x-1\)
\(=x^2\left(x-1\right)-2x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+1\right)\)
\(=\left(x-1\right)^3\)
d) Ta có: \(27+27x+9x^2+x^3\)
\(=x^3+3x^2+6x^2+18x+9x+27\)
\(=x^2\left(x+3\right)+6x\left(x+3\right)+9\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+6x+9\right)\)
\(=\left(x+3\right)^3\)
e) Ta có: \(8-125x^3\)
\(=2^3-\left(5x\right)^3\)
\(=\left(2-5x\right)\left(4+10x+25x^2\right)\)
f) Ta có: \(64x^3+\frac{1}{8}\)
\(=\left(4x\right)^3+\left(\frac{1}{2}\right)^3\)
\(=\left(4x+\frac{1}{2}\right)\left(16x^2-2x+\frac{1}{4}\right)\)
g) Ta có: \(1-x^2y^4\)
\(=1^2-\left(xy^2\right)^2\)
\(=\left(1-xy^2\right)\left(1+xy^2\right)\)
a) \(x^2+2x+1=x^2+2x.1+1^2=\left(x+1\right)^2\)
b) \(1-2y+y^2=1^2-2y.1+y^2=\left(1-y\right)^2\)
c) \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
d) \(27+27x+9x^2+x^3=3^3+3.3^2x+3.3x^2+x^3=\left(3+x\right)^3\)
e) \(8-125x^3=2^3-\left(5x\right)^3=\left(2-5x\right)\left[2^2+2.5x+\left(5x\right)^2\right]=\left(2-5x\right)\left(4+10x+25x^2\right)\)
f) \(64x^3+\frac{1}{8}=\left(4x\right)^3+\left(\frac{1}{2}\right)^3=\left(4x+\frac{1}{2}\right)\left[\left(4x\right)^2-4x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]=\left(4x+\frac{1}{2}\right)\left(16x^2-2x+\frac{1}{4}\right)\)
Ko chắc ạ!
\(a,x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
\(b,\left(x-2\right)^3+6\left(x+1\right)^2-x+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x+12=0\)\(\Leftrightarrow x^3+23x+10=0\) (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= \(\dfrac{1}{2}\left(\sqrt[3]{-0,2355375386}+\sqrt{-0,2355375386^2+1}\right)\) \(\left(\sqrt[3]{-0,2355375386-\sqrt{-0,2355375386^2+1}}\right)\)\(=-0,07788262891\)
\(\Rightarrow x=\dfrac{2\sqrt{69}}{3}.t=-0,4312944692\)
\(c,x^3+6x^2+12x+8=0\)
\(\Leftrightarrow\left(x+2\right)^3=0\)
\(\Leftrightarrow x+2=0\Rightarrow x=-2\)
\(d,x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
\(e,8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Rightarrow2x-1=0\Rightarrow x=\dfrac{1}{2}\)
\(f,x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Rightarrow x+3=0\Rightarrow x=-3\)
CẢM ƠN NHÌU
1. 8 - 12x + 6x2 - x3
= 23 - 3.22.x + 3.x2.2 - x3
=(2-x)3
2. 125x3 - 75x2 +15x - 1
=(5x)3 - 3.(5x)2.1 + 3.5x.12 - 13
=(5x - 1)3
3, 4 (sai đề)
5. x3 + 2x2 - 6x - 27
=(x3 - 27) + (2x2 - 6x)
=(x3 - 33) + (2x2 - 6x)
=(x -3)(x2 + 3x + 9) + 2x(x-3)
=(x-3)(x2 + 3x +9 +2x)
=(x-3)(x2 + 5x +9)
6. 12x3 + 4x2- 27x -9
=(12x3 + 4x2) - (27x + 9)
=4x2(3x + 1) - 9(3x +1)
=(3x -1)(4x2 -9)
=(3x-1)(2x-3)(2x+3)
Bài 1:
\(B=\dfrac{1}{9}x^2-2x+9\)
\(=\left(\dfrac{1}{3}x\right)^2-2\cdot\dfrac{1}{3}x\cdot3+3^2=\left(\dfrac{1}{2}x-3\right)^2\)
\(C=x^3-9x^2+27x-27=\left(x-3\right)^3\)
\(D=27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
\(E=\left(x-2y\right)^3\)
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 = \(\frac{1}{2}\).