phân tích đa thức sau thanh nhân tử :
\(12x^2-23xy+10y^2\)
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a) Ở trên
b) \(\left(x^2-8x\right)^2+36\) = \(x^4-16x^2+64+36\)
= \(\left(x^4+20x^2+100\right)-36x^2\)
= \(\left(x^2+10\right)^2-\left(6x\right)^2\)
= \(\left(x^2-6x+10\right)\left(x^2+6x+10\right)\)
a)
12x2 - 23xy + 10y2
= (12x2 - 8xy ) - ( 15xy - 10y2 )
= 4x(3x - 2y) - 5y(3x - 2y)
= (4x - 5y)(3x - 2y)
b)
( x2 - 8)2 + 36 = x4 - 16x2 + 64 +36
= (x4 + 20x2 +100) - 36x2
= (x2 + 10)2 - (6x)2
= (x2 + 10 - 6x)(x2 + 10 + 6x)
\(2x^2-8y^2+5xy^2-10y^3\)
\(=2\left(x^2-4y^2\right)+5y^2\left(x-2y\right)\)
\(=2\left(x-2y\right)\left(x+2y\right)+5y^2\left(x-2y\right)\)
\(=\left(2x+4y\right)\left(x-2y\right)+5y^2\left(x-2y\right)\)
\(=\left(x-2y\right)\left(5y^2+2x+4y\right)\)
\(3x^3-12x^2y+12-75xz^2\)
\(=3\left[x^3-4x^2y+4-25xz^2\right]\)
b)Ta có:\(A=2018^2+2019^2+2019^2.2018^2\)
\(=\left(2018^2-2.2018.2019+2019^2\right)+2.2018.2019+\left(2018.2019\right)^2\)
\(=\left(2019.2018\right)^2+2.2018.2019+1^2=\left(2019.2018+1\right)^2\)là số chính phương (đpcm)
c)Ta có:Xét hiệu a^2+b^2+c^2+d^2-a(b+c+d),ta có:
\(a^2+b^2+c^2+d^2-a\left(b+c+d\right)=a^2+b^2+c^2+d^2-ab-ac-ad\)
\(=\left(\frac{1}{4}a^2-ab+b^2\right)+\left(\frac{1}{4}a^2-ac+c^2\right)+\left(\frac{1}{4}a^2-ad+d^2\right)+\frac{a^2}{4}\)
\(=\left(\frac{a}{2}-b\right)^2+\left(\frac{a}{2}-c\right)^2+\left(\frac{a}{2}-d\right)^2+\left(\frac{a}{2}\right)^2\ge0\forall a,b,c,d\left(đpcm\right)\)
\(\Rightarrow a^2+b^2+c^2\ge a\left(b+c+d\right)-d^2\)
Dấu bằng xảy ra \(\Leftrightarrow\hept{\begin{cases}b=c=d=\frac{a}{2}\\\frac{a}{2}=0\end{cases}\Leftrightarrow}a=b=c=d=0\)
1) \(3x^3-12x\)
\(=3x\left(x^2-4\right)\)
\(=3x\left(x+2\right)\left(x-2\right)\)
2)\(x^3+x^2-x-1\)
\(=\left(x^3+x^2\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-1\right)\)
\(=\left(x+1\right)^2\left(x-1\right)\)
3) \(x^2-7xy+10y^2\)
\(=\left(x^2-2xy\right)-\left(5xy-10y^2\right)\)
\(=x\left(x-2y\right)-5y\left(x-2y\right)\)
\(=\left(x-5y\right)\left(x-2y\right)\)
1) \(3x^3-12x\)
\(=3x\left(x^2-4\right)\)
\(=3x\left(x-2\right)\left(x+2\right)\)
2) \(x^3+x^2-x-1\)
\(=x^2\left(x+1\right)-\left(x+1\right)\)
\(=\left(x^2-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+1\right)\)
Bài làm
a) 6x3 + x2yy + 23xy2 + 12y3
= ( 2x + y ) ( 3x2 - xy + 12y2 )
# Học tốt #
Ta có: \(12x^2-23xy+10y^2\) = \(\left(12x^2-8xy\right)-\left(15xy-10y^2\right)\)
= \(4x\left(3x-2y\right)\) \(-5y\left(3x-2y\right)\)
= \(\left(3x-2y\right)\left(4x-5y\right)\)