B1:
a)
\(A=x^3-9x^2+27x-27\\ =x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3\\ =\left(x-3\right)^3\)
Thay x=13 vào A ta có:
\(A=\left(13-3\right)^3=10^3=1000\)
b)
\(A=x^3+6x^2+12x+12\\ =\left(x^3+6x^2+12x+8\right)+4\\ =\left(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3\right)+4\\ =\left(x+2\right)^3+4\)
Thay x = 8 vào A ta có:
\(A=\left(8+2\right)^3+4=10^3+4=1000+4=1004\)
Bài 2:
a: \(x^3+9x^2+27x+27=0\)
=>\(x^3+3\cdot x^2\cdot3+3\cdot x\cdot3^2+3^3=0\)
=>\(\left(x+3\right)^3=0\)
=>x+3=0
=>x=-3
b: \(\left(x+1\right)^3-x\left(x+1\right)\left(x-1\right)=3x^2\)
=>\(x^3+3x^2+3x+1-x\left(x^2-1\right)-3x^2=0\)
=>\(x^3+3x+1-x^3+x=0\)
=>4x+1=0
=>4x=-1
=>\(x=-\dfrac{1}{4}\)
Bài 1:
a: \(A=x^3-9x^2+27x-27\)
\(=x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=\left(x-3\right)^3\)
Thay x=13 vào A, ta được:
\(A=\left(13-3\right)^3=10^3=1000\)
b: \(B=x^3+6x^2+12x+8=x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3\)
\(=\left(x+2\right)^3\)
Thay x=8 vào B, ta được:
\(B=\left(8+2\right)^3=10^3=1000\)
a)
\(x^3+9x^2+27x+27=0\\\Leftrightarrow x^3+3\cdot x^2\cdot3+3\cdot x\cdot3^2+3^3=0\\\Leftrightarrow \left(x+3\right)^3=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\)
b)
\(\left(x+1\right)^3-x\left(x+1\right)\left(x-1\right)=3x^2\\\Leftrightarrow \left(x^3+3x^2+3x+1\right)-x\left(x^2-1\right)=3x^2\\ \Leftrightarrow x^3+3x^2+3x+1-x^3+x-3x^2=0\\ \Leftrightarrow4x+1=0\\ \Leftrightarrow4x=-1\\ \Leftrightarrow x=-\dfrac{1}{4}\)
