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Bài 1:
\(a,27x^3+27x^2+9x+1\)
\(=\left(3x\right)^3+3.\left(3x\right)^2.1+3.3x.1^2+1^3\)
\(=\left(3x+1\right)^3\)
\(b,x^3+3\sqrt{2}x^2y+6xy^2+2\sqrt{2}y^3\)
\(=x^3+3.x^2.\sqrt{2}y+3.x.\left(\sqrt{2}y\right)^2+\left(\sqrt{2}y\right)^3\)
\(=\left(x+\sqrt{2}y\right)^3\)
Bài 2:
\(a,x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
\(b,\left(x+1\right)^3-x\left(x-2\right)^2+x-1=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3-4x^2+4x+x-1=0\)
\(\Leftrightarrow-x^2+8x=0\)
\(\Leftrightarrow-x\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
1)
a) = (3x+1)3
b) (x+\(\sqrt{2}\) )3
2)
a)\(x^3+9x^2+27x+27=0\\ \left(x+3\right)^3=0\\ =>x=-3\)
b) Bài cuối bạn tự làm nhé! Mình mắc học bài
# Chúc bạn học tốt !
a) x3-9x2+27x-27=0
<=>(x-3)3=0
<=>x-3=0
<=>x=3
b) x3-25x=0
<=>x.(x2-25)=0
<=>x.(x-5)(x+5)=0
<=>x=0 hoặc x-5=0 hoặc x+5=0
<=>x=0 hoặc x=5 hoặc x=-5
c)9x2-1=0
<=>(3x-1)(3x+1)=0
<=>3x-1=0 hoặc 3x+1=0
<=>x=1/3 hoặc x=-1/3
a, x^3 - 9x^2 + 27x - 27 = 0
=> ( x - 3)^3 = 0
=> x - 3 = 0
=> x = 3
b, x^3 - 25x = 0
=> x(x^2 - 25) = 0
=> x(x-5)(x + 5) = 0
=> x =0 hoặc x - 5 = 0 hoặc x + 5 = 0
=> x= 0 hoặc x =5 hoặc x = -5
c, 9x^2 - 1 = 0
=> (3x)^2 - 1^2 = 0
=> ( 3x- 1)(3x+ 1) = 0
=> 3x - 1 = 0 hoặc 3x + 1 = 0
=> x = 1/3 hoặc x = -1/3
Bài 1:
\(B=\dfrac{1}{9}x^2-2x+9\)
\(=\left(\dfrac{1}{3}x\right)^2-2\cdot\dfrac{1}{3}x\cdot3+3^2=\left(\dfrac{1}{2}x-3\right)^2\)
\(C=x^3-9x^2+27x-27=\left(x-3\right)^3\)
\(D=27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
\(E=\left(x-2y\right)^3\)
1) \(27+27x+9x^2+x^3\)
\(=3^3+3.3^2.x+3.3.x^2+x^3\)
\(=\left(3+x\right)^3\)
2) \(8-27x^3\)
\(=2^3-\left(3x\right)^3\)
\(=\left(2-3x\right)\left[2^2+2.3x+\left(3x\right)^2\right]\)
\(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
\(1,27+27+9x^2+x^3\)
\(=\left(3+x\right)^3\)
\(2,8-27x^3\)
\(=\left(2-3x\right).\left(4+6x+9x^2\right)\)
\(3,x^{64}+x^{32}+1\)
\(=\left(x^{64}+2x^{32}+1\right)-x^{32}\)
\(=\left(x^{32}+1\right)^2-x^{32}\)
\(=\left(x^{32}+1-x^{16}\right).\left(x^{32}+1+x^{16}\right)\)
Công nhận câu 3 hơi căng~
\(8-27x^3\)
\(=2^3-\left(3x\right)^3\)
\(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
a) \(8-27x^3=\left(2-x\right)\left(4+6x+9x^2\right)\)
b) \(27+27x+9x^2+x^3=\left(3+x\right)^3\)
c) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
\(a,\left(2x^3-27x^2+115x-150\right)\left(x-5\right)\)
\(=x\left(2x^3-27x^2+115-150\right)-5\left(2x^3-27x^2+115-150\right)\)
\(=2x^4-27x^3+115x-150x-10x^3+135x^2-575+750\)
\(=2x^4-37x^3+135x^2-35x+175\)
a) \(x^3-7x+6=x^3+3x^2-x^2-3x-2x^2-6x+2x+6\)
=\(x^2\left(x+3\right)-x\left(x+3\right)-2x\left(x+3\right)+2\left(x+3\right)\)
=\(\left(x+3\right)\left(x^2-x-2x+2\right)\)
=\(\left(x+3\right)\left(x-2\right)\left(x-1\right)\)
=\(\left\{\begin{matrix}x+3=0=>x=-3\\x-2=0=x=2\\x-1=0=>x=1\end{matrix}\right.\)
\(b...x^3-19x+30=0\)
\(=>x^3+5x^2-2x^2-10x-3x^2-15x+6x+30=0\)
=>\(x^2\left(x+5\right)-2x\left(x+5\right)-3x\left(x+5\right)+6\left(x+5\right)=0\)
=>\(\left(x+5\right)\left(x^2-2x-3x+6\right)=0\)
=>\(\left(x+5\right)\left(x-3\right)\left(x-2\right)=0\)
=>\(\left\{\begin{matrix}x-3=0=>x=3\\x-2=0=>x=2\\x+5=0=>x=-5\end{matrix}\right.\)
Vậy x=-5;2;3
a, \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
b, \(1-9x+27x^2-27x^3=-\left(3x-1\right)^3\)
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mày viết lại cái đề bài hộ tao cái
lm heets cmnr
Diệp Ẩn cút , troll vãi lìn
a: \(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+9x\left(x+3\right)=0\)
=>(x+3)(x^2+6x+9)=0
=>x=-3
b: \(\Leftrightarrow x^2-2x-x^3+6x^2-4=0\)
=>-x^3+6x^2-2x-4=0
hay \(x\in\left\{5.5;1.14;-0.64\right\}\)
c: =>(3x-1)^3=8
=>3x-1=2
=>3x=3
=>x=1