a) Ta có: A = ax + bx + cx + ay + by + cy + az + bz + cz

                  = x.(a+b+c) + y.(a+b+c) + z.(a+b+c)

                  = (a+b+c).(x+y+z) (1)

Lại có: a + b + c = -3 (2)

            x + y + z = -6 (3)

Từ (1) ; (2) ; (3) => A = -3.(-6) = 18

           Vậy A = 18

b) B = ax - bx - cx - ay + by + cy - az + bz +cz

       = x.(a-b-c) - y.(a-b-c) - z.(a-b-c)

       = (a-b-c).(x-y-z)

Lại có: a - b - c = 0 ; x - y - z = 2016

=> B = 0.2016 = 0

Vậy B = 0

a) \(ab+ac=a.\left(b+c\right)\)

b) \(ab-ac+ad=a.\left(b-c+d\right)\)

c) \(ax-bx-cx-dx=x.\left(a-b-c-d\right)\)

d) \(a.\left(b+c\right)-d.\left(b+c\right)=ab+ac-db-dc=b.\left(a-d\right)+c.\left(a-d\right)=\left(a-d\right).\left(b+c\right)\)

e) \(ac-ad+bc-bd=a.\left(c-d\right)+b.\left(c-d\right)=\left(c-d\right).\left(a+b\right)\)

f) \(ax+by+bx+ay=a.\left(x+y\right)+b.\left(y+x\right)=\left(x+y\right).\left(a+b\right)\)

CHÚC BN HỌC TỐT!!!!!

\(\left\{{}\begin{matrix}ax+by=c\\bx+cy=a\\cx+ay=b\end{matrix}\right.\)

Cộng đại số => \(ax+by+bx+cy+cx+ay=a+b+c\)

<=>\(\left(a+b+c\right)x+\left(a+b+c\right)y=a+b+c\)

<=>\(\left(a+b+c\right)\left(x+y\right)=a+b+c\)

<=>\(\left(a+b+c\right)\left(x+y\right)-\left(a+b+c\right)=0\)

<=>\(\left(a+b+c\right)\left(x+y-1\right)=0\)

+TH1:\(\left(a+b+c\right)=0\)

=>\(a+b=-c\)

=>\(\left(a+b\right)^3=-c^3\)

=>\(a^3+b^3+3a^2b+3ab^2=-c^3\)

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

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

Mà a+b=-c => -3ab(a+b)=-3ab(-c)=3abc

=>\(a^3+b^3+c^3=3abc\)

+TH2:x+y=1

<=>y=1-x

=>\(\left\{{}\begin{matrix}ax+b\left(1-x\right)=c\\bx+c\left(1-x\right)=a\\cx+a\left(1-x\right)=b\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}ax+b-bx=c\\bx+c-cx=a\\cx+a-ax=b\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(a-b\right)x=c-b\\\left(b-c\right)x=a-c\\\left(c-a\right)x=b-a\end{matrix}\right.\)

Nếu \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)

=>a=b=c 

\(\Rightarrow a^3+b^3+c^3=3a^3\\ 3abc=3a^3\\ \Rightarrow a^3+b^3+c^3=3abc\)

Nếu \(\left\{{}\begin{matrix}a-b\ne0\\b-c\ne0\\c-a\ne0\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=\dfrac{c-b}{a-b}\left(1\right)\\x=\dfrac{a-c}{b-c}\left(2\right)\\x=\dfrac{b-a}{c-a}\end{matrix}\right.\)

Ta có : (1)=(2)=x  suy ra \(\dfrac{c-b}{a-b}=\dfrac{a-c}{b-c}\Rightarrow\dfrac{b-c}{b-a}=\dfrac{a-c}{b-c}\Rightarrow\left(b-c\right)\left(b-c\right)=\left(a-c\right)\left(b-a\right)^{ }\Rightarrow b^2-2bc+c^2=a^2+ab-bc+ca\)

\(\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Rightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=0\\ \\ \\ \Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

=>a=b=c(đưa về trường hợp như trên)

Phương Ann Nhã Doanh đề bài khó wá Mashiro Shiina Đinh Đức Hùng

Nguyễn Huy Tú Lightning Farron Akai Haruma

ab + ac = a(b + c)

ab - ac + ad = a(b - c + d)

ax - bx - cx + dx

=x(a - b - c + d)

chuẩn men

ab+ac=a(b+c)

ab-ac+ad=a(b-c+d)

ax-bx-cx+dx=x(a-b-c+d)

\(x^2+2xy+y^2-xz-yz\)

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

\(=\left(x+y\right)\left(x+y-z\right)\)

mk chỉnh lại đề

\(x^2-2xy+y^2-z^2+2zt+t^2\)

\(=\left(x-y\right)^2-\left(z-t\right)^2\)

\(=\left(x-y-z+t\right)\left(x-y+z-t\right)\)

mk chỉnh lại đề:

\(ax^2+cx^2-ay+ay^2-cy+cy^2\)

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

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

\(ax^2+ay^2-bx^2-by^2+b-a\)

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

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

\(ac^2-ad-bc^2+cd+bd-c^3\)

\(=a\left(c^2-d\right)-b\left(c^2-d\right)-c\left(c^2-d\right)\)

\(=\left(c^2-d\right)\left(a-b-c\right)\)

trả lời giùm mình với

có điều kiện x, y > 0 ko bạn?