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a) \(x^3-5x^2+8x-4=x^3-x^2-4x^2+4x+4x-4\)
\(=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)
b) \(x^3-3x+2=x^3+2x^2-2x^2-4x+x+2\)
\(=x^2\left(x+2\right)-2x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x+1\right)=\left(x+2\right)\left(x-1\right)^2\)
c) \(x^3-5x^2+3x+9=x^3+x^2-6x^2-6x+9x+9\)
\(=x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-6x+9\right)=\left(x+1\right)\left(x-3\right)^2\)
d) \(x^3+8x^2+17x+10=x^3+2x^2+6x^2+12x+5x+10\)
\(=x^2\left(x+2\right)+6x\left(x+2\right)+5\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+6x+5\right)=\left(x+2\right)\left(x+5\right)\left(x+1\right)\)
e) \(x^3+3x^2+6x+4=x^3+x^2+2x^2+2x+4x+4\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+4\right)\)
f) \(x^3+3x^2+3x+2=x^3+2x^2+x^2+2x+x+2\)
\(=x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+x+1\right)\)
Câu 2 nha
\(a,x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(c,x^2-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
Bài 1 :
a, \(\left(a-2\right)^2-b^2=\left(a-2-b\right)\left(a-2+b\right)\)
b, \(2a^3-54b^3=2\left(a^3-27b^3\right)=2\left(a-3b\right)\left(a^2+3ab+9b\right)\)
Bài 2 : tự kết luận nhé, ngại mà lười :(
a, \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
\(\Leftrightarrow\frac{4x-3}{5}-\frac{5x-4}{3}=\frac{6x-2}{7}+3\)
\(\Leftrightarrow\frac{12x-9-25x+20}{15}=\frac{6x-2+21}{7}\)
\(\Leftrightarrow\frac{-13x-29}{15}=\frac{6x+19}{7}\Rightarrow-91x-203=90x+285\)
\(\Leftrightarrow181x=-488\Leftrightarrow x=-\frac{488}{181}\)
b, \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)
\(\Leftrightarrow\frac{4x+8+9\left(2x-1\right)}{12}-\frac{10x-6}{12}=\frac{12x+5}{12}\)
\(\Rightarrow4x+8+18x-9-10x+6=12x+5\)
\(\Leftrightarrow12x+5=12x+5\Leftrightarrow0x=0\)
Vậy phương trình có vô số nghiệm
c, \(\left|2x-3\right|=4\)
Với \(x\ge\frac{3}{2}\)pt có dạng : \(2x-3=4\Leftrightarrow x=\frac{7}{2}\)
Với \(x< \frac{3}{2}\)pt có dạng : \(2x-3=-4\Leftrightarrow x=-\frac{1}{2}\)
d, \(\left|3x-1\right|-x=2\Leftrightarrow\left|3x-1\right|=x+2\)
Với \(x\ge\frac{1}{3}\)pt có dạng : \(3x-1=x+2\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Với \(x< \frac{1}{3}\)pt có dạng : \(3x-1=-x-2\Leftrightarrow4x=-1\Leftrightarrow x=-\frac{1}{4}\)
a) x^3−3x^2−4x+12
=(x^3-3x^2)-(4x-12)
=x^2(x-3)-4(x-3)
=(x-3)(x^2-4)=(x-3)(x-2)(x+2)
b) x^4-5x^2+4=x^4-x^2-4x^2+4
=(x^4-x^2) - ( 4x^2-4)
=x^2(x^2-1) - 4(x^2-1)
=(x^2-1)(x^2-4)
=(x-1)(x+1)(x-2)(x+2)
c) (x+y+z)^3-x^3-y^3-z^3
=x^3+y^3+z^3+3x^2yz+3xy^2z+3xyz^2-x^3-y^3-z^3
=3x^2yz+3xy^2z+3xyz^2
3xyz(x+y+z)
\(C=x^3+5x^2+8x+4\)
\(=x^3+x^2+4x^2+4x+4x+4\)
\(=x^2\left(x+1\right)+4x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x^2+4x+4\right)\left(x+1\right)\)
\(=\left(x+2\right)^2.\left(x+1\right)\)
\(D=x^3-x^2-4\)
\(=x^3-2x^2+x^2-2x+2x-4\)
\(=x^2\left(x-2\right)+x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x^2+x+2\right)\left(x-2\right)\)
Chúc bạn học tốt.
