A>B

bn nhé

tk mk ik

a>b chúc bạn học tốt

^__^

a) \(A=x^2-2.10x+100+1\)

\(A=\left(x-10\right)^2+1>=1\)với mọi x

Dấu = xảy ra khi x-10 =0

                           =>x=10

Min A=1 khi x=10

b) Câu b bạn viết sai đề rồi B= -x^2 +4x -3  mới làm dc

a)A= \(\left(x^2-2.x.10+100\right)+1\)

=\(\left(x-10\right)^2+1>=1\)

Dấu "=" xảy ra <=> \(\left(x-10\right)^2=0\)<=> \(x-10=0\)<=>\(x=10\)

Vậy MinA = 1 khi x=10

\(M=\left(a+4\right)^2+2\left(a+4\right)\left(6-a\right)+\left(6-a\right)^2\)

\(\Rightarrow M=\left(a+4+6-a\right)^2\)

\(\Rightarrow M=\left(1982+4+6-1982\right)^2\)\(=10^2=100\)

mk ghi đáp án, ko phân tích đc thì IB mk

a) \(x^2+6xy+9y^2=\left(x+3y\right)^2\)

b) \(4a^4-4a^2b^2+b^4=\left(2a^2-b^2\right)^2\)

c)  \(x^6+y^2-2x^3y=\left(x^3-y\right)^2\)

d)  \(\left(x+y\right)^3-\left(x-y\right)^3=2y\left(3x^2+y^2\right)\)

e)  \(25x^4-10x^2y^2+y^4=\left(5x^2-y^2\right)^2\)

f) \(-a^2-2a-1=-\left(a+1\right)^2\)

g)  \(27b^3-8a^3=\left(3b-2a\right)\left(9b^2+6ab+4a^2\right)\)

h)  \(x^3+9x^2y+27xy^2+27y^3=\left(x+3y\right)^3\)

i) \(16x^2-9\left(x+y\right)^2=\left(x-3y\right)\left(7x+3y\right)\)

Đặt \(A=4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2\)

\(=4\left(x+y\right)\left(x+z\right)x\left(x+y+z\right)+y^2z^2=4\left(x^2+xz+xy+yz\right)\left(x^2+xy+xz\right)+y^2z^2\)

Đặt x²+xy+xz=t, ta có:

\(A=4\left(t+yz\right)t+y^2z^2=4t^2+4tyz+y^2z^2=\left(2t+yz\right)^2=\left(2x^2+2xy+2xz+yz\right)^2\ge0\)

a) \(\left(2x-3\right)\left(2x+3\right)+\left(3-x\right)^2\)

\(=\left(2x\right)^2-3^2+\left(9-6x+x^2\right)\)

\(=4x^2-9+9-6x+x^2\)

\(=5x^2-6x\)

\(=x\left(5x-6\right)\)

b) Đặt tính rồi tính bình thường

\(\Rightarrow\left(x^3-x-6\right)\div\left(x-2\right)=x^2+2x+3\) (phép chia hết)

a, 4x^2 +6x -6x -9 +9 -6x+x^2

= 5x^2 -6x 

= x (5x - 6)

b, = x^2 +2x +3 

a3+b3+c3=(a+b+c)(a2+b2+c2−ab−bc−ac)+3abc

                    =(a+b+c)[a2+b2+c2+2ab+2ac+2bc−3ac−3bc−3ab)+3abc

                    =(a=b+c)[(a+b+c)2−3(ab+bc+ac)]+3abc

*Nếu a+b+c⋮3⇒a3+b3+c3⋮3

*Nếu a3+b3+c3⋮3⇒(a+b+c)[(a+b+c)2−3(ab+bc+ca)]⋮3⇒a+b+c⋮3

làm như vậy nha, mk xin lỗi , ko bt cách viết số mũ nha, k nha

    Xét \(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

                                                  \(=\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b\right)-3abc\)

                                                   \(=\left(a+b+c\right).\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)\)

                                                   \(=\left(a+b+c\right).\left[a^2+2ab+b^2-ac-bc+c^2\right]-3ab\left(a+b+c\right)\)

                                                   \(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)   

                                                   \(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

- Nếu \(a+b+c⋮3\)\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)⋮3\)

Mà 3abc chia hết cho 3 \(\Rightarrow a^3+b^3+c^3⋮3\)

- Nếu \(a^3+b^3+c^3⋮3\)mà \(3abc⋮3\Rightarrow a^3+b^3+c^3-3abc⋮3\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)⋮3\Rightarrow a+b+c⋮3\)

Chúc bạn học tốt.