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\(B=\left(\dfrac{1}{2}x-1\right)^3=\left(-1-1\right)^3=\left(-2\right)^3=-8\)
\(C=\left(x+1\right)^3-1000\)
\(=100^3-1000=999000\)
\(D=27x^3+54x^2+36x+8-4\)
\(=\left(3x+2\right)^3-4=\left(-6+2\right)^3-4\)
\(=-64-4=-68\)
Giải:
1) \(a^3-3a^2+3a-1\)
\(=a^3-3a^2.1+3a.1^2-1^3\)
\(=\left(a-1\right)^3\)
Vậy ...
2) \(x^3+6x^2+12x+8\)
\(=x^3+3.x^2.2+3.x.2^2+2^3\)
\(=\left(x+2\right)^3\)
Vậy ...
3) \(8x^3-12x^2+6x-1\)
\(=\left(2x\right)^3-3.\left(2x\right)^2.1+3.2x.1^2-1^3\)
\(=\left(2x-1\right)^3\)
Vậy ...
4) \(x^3-6x^2y+12xy^2-8y^3\)
\(=x^3-3.x^2.2y+3.x.\left(2y\right)^2-\left(2y\right)^3\)
\(=\left(x-2y\right)^3\)
Vậy ...
a) \(\left(2x+1\right)^3\)
\(=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1+1\)
\(=8x^3+12x^2+6x+1\)
b) \(\left(x-3\right)^3\)
\(=x^3-3.x^2.3+3.x.3^2-3^3\)
\(=x^3-9x^2+27x-27\)
Bài 2:
a: \(x^3+15x^2+75x+125=\left(x+5\right)^3\)
b: \(1-15y+75y^2-125y^3=\left(1-5y\right)^3\)
c: \(8x^3+4x^2y+\dfrac{3}{2}xy^2+8y^3=\left(2x+2y\right)^3\)
1.x2-y2+2x+1=(x2+2x+1)-y2=(x+1)2-y2=(x+1-y)(x+1+y)
2.(x2+9)2-36x2=(x2+9)2-(6x)2=(x2+9-6x)(x2+9+6x)=(x-3)2(x+3)2
3.\(8x^3+\dfrac{1}{27}=\left(2x\right)^3+\left(\dfrac{1}{3}\right)^3\\ =\left(2x+\dfrac{1}{3}\right)\text{[}\left(2x\right)^2-2x.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\text{]}\\ =\left(2x+\dfrac{1}{3}\right)\left(4x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)\)4.x3-8y3=x3-(2y)3=(x-2y)(x2+2xy+4y2)
Giải:
1) \(x^3-3x^2+3x-2=0\)
\(\Leftrightarrow x^3-3x^2+3x-1-1=0\)
\(\Leftrightarrow\left(x-1\right)^3-1=0\)
\(\Leftrightarrow\left(x-1-1\right)\left[\left(x-1\right)^2+x-1+1\right]=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy ...
2) \(8x^3+12x^2+6x+\dfrac{7}{8}=0\)
\(\Leftrightarrow8x^3+12x^2+6x+1-\dfrac{1}{8}=0\)
\(\Leftrightarrow\left(2x+1\right)^3-\left(\dfrac{1}{2}\right)^3=0\)
\(\Leftrightarrow\left(2x+1-\dfrac{1}{2}\right)\left[\left(2x+1\right)^2+\left(2x+1\right)\dfrac{1}{2}+\dfrac{1}{4}\right]=0\)
\(\Leftrightarrow2x+\dfrac{1}{2}=0\)
\(\Leftrightarrow x=-\dfrac{1}{4}\)
Vậy ...
3) \(x^3-9x^2+27x-19=0\)
\(\Leftrightarrow x^3-9x^2+27x-27+8=0\)
\(\Leftrightarrow\left(x-3\right)^3+2^3=0\)
\(\Leftrightarrow\left(x-3+2\right)\left[\left(x-3\right)^2-2\left(x-3\right)+4\right]=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy ...
Chúc chị học tốt trong thời gian tới nha! ^^
a: \(F=a^3-b^3-3ab\)
\(=\left(a-b\right)^3+3ab\left(a-b\right)-3ab\)
\(=1^3+3ab-3ab=1\)
b: \(G=a^3-b^3+3ab\)
\(=\left(a-b\right)^3+3ba\left(a-b\right)+3ab\)
=-1
c: \(H=\left(x-1\right)^3\)
(x+1)3=8
=>x+1=2
=>x=1
\(H=\left(1-1\right)^3=0\)
1) \(B=\dfrac{1}{8}x^3-\dfrac{3}{4}x^2+\dfrac{3}{2}x-1=\left(\dfrac{1}{2}x-1\right)^3\)
thay x =-2 vào B, ta được:
\(B=\left(\dfrac{1}{2}\cdot\left(-2\right)-1\right)^3=\left(-2\right)^3=-8\)
2) \(C=x^3+3x^2+3x-999=\left(x+1\right)^3-1000\)
thay x =99 vào B, ta được:
\(C=\left(99+1\right)^3-1000=999000\)
3) \(D=27x^3+54x^2+36x+4=\left(3x+2\right)^3-4\)
thay x =-2 vào D, ta được:
\(D=\left(3\left(-2\right)+2\right)^3-4=-68\)
1: \(\left(x-y\right)^3+\left(x+y\right)^3-6xy^2\)
\(=x^3-3x^2y+3xy^2-y^3+x^3+3x^2y+3xy^3+y^3-6xy^2\)
\(=2x^3\)
2: \(8x^3-36x^2y+54xy^2-28y^3\)
\(=8x^3-36x^2y+54xy^2-27y^3-y^3\)
\(=\left(2x-3y\right)^3-y^3\)
3: \(x^3-3x^2+3x-2\)
\(=x^3-3x^2+3x-1-1\)
\(=\left(x-1\right)^3-1\)