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)

bạn phân tích thành:(2x+y)(3x^2-xy+12y^2)

ơ,sao đăng đáp án mà nó éo hiện nhỉ,cái trang này đểu r

bạn cứ đánh lên hộ mk đi

sai đề hay sao vậy bạn

đề đúng mak

\(6x^3+x^2y+23xy^2+12y^3=\left(6x^3-2x^2y+24xy^2\right)+\left(3x^2y-xy^2+12y^3\right)=2x\left(3x^2-xy+12y^2\right)+y\left(3x^2-xy+12y^2\right)=\left(2x+y\right)\left(3x^2-xy+12y^2\right)\)

      \(2x^2-8y^2+5xy^2-10y^3\)

\(=2\left(x^2-4y^2\right)+5y^2\left(x-2y\right)\)

\(=2\left(x-2y\right)\left(x+2y\right)+5y^2\left(x-2y\right)\)

\(=\left(2x+4y\right)\left(x-2y\right)+5y^2\left(x-2y\right)\)

\(=\left(x-2y\right)\left(5y^2+2x+4y\right)\)

      \(3x^3-12x^2y+12-75xz^2\)

\(=3\left[x^3-4x^2y+4-25xz^2\right]\)

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 \(\Leftrightarrow\hept{\begin{cases}b=c=d=\frac{a}{2}\\\frac{a}{2}=0\end{cases}\Leftrightarrow}a=b=c=d=0\)

a) 4x² - 12x + 23xy - 35y² + 15y

= 4x² + 23xy - 35y² - (12x - 15y) 

= 4x² - 5xy + 28xy - 35y² - 3(4x - 5y)

= x(4x - 5y) + 7y(4x - 5y) - 3(4x - 5y) 

= (4x - 5y)(x + 7y - 3) 

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) \(x^3+x^2-x-1\)

\(=x^2\left(x+1\right)-\left(x+1\right)\)

\(=\left(x^2-1\right)\left(x+1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x+1\right)\)

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)      \(x^2-7xy+10y^2\)

\(=\left(x^2-2xy\right)-\left(5xy-10y^2\right)\)

\(=x\left(x-2y\right)-5y\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x-5y\right)\)

Bài làm

a) 6x³ + x²yy + 23xy² + 12y³

= ( 2x + y ) ( 3x² - xy + 12y² )

# Học tốt #

(2x+y)(3x²-xy+12y²)