\(\left(a+b+c\right)^5-a^5-b^5-c^5\)

\(=5\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(a^2+b^2+c^2+ab+bc+ca\right)\)

Đặt \(\left\{{}\begin{matrix}a+b-c=x\\b+c-a=y\\c+a-b=z\end{matrix}\right.\Leftrightarrow x+y+z=a+b+c\)

Do đó \(A=\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(\Leftrightarrow A=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)-x^3-y^3-z^3\\ \Leftrightarrow A=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(\Leftrightarrow A=3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(c+a-b+a+b-c\right)\\ \Leftrightarrow A=3\cdot2b\cdot2c\cdot2a=24abc\)

a) xy+3x-7y-21

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

=(x-7)(y+3)

b)2xy-15-6x-5y

=2x(y-3)-5(-3+y)

=(2x-5)(y-3)

c)2x^2y+2xy^2-2x-2y

=2x(xy-1)+2y(xy-1)

=(2x+2y)(xy-1)

x(x+3)-5x(x-5)-5(x+3)

=(x-5)(x+3)-5x(x-5)

=(x-5)(x+3-5x)

Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn

a) xy+3x-7y-21=(xy+3x)-(7y+21)= x(y+3)-7(y+3)=(y+3)(x-7)

b)2xy-15-6x+5y=(2xy-6x)+(5y-15)=2x(y-3)+5(y-3)=(y-3)(2x+5)

c)2x^2y+2xy^2-2x-2y=2xy(x+y)-2(x+y)=2(x+y)(xy-1)

d) x(x+3)-5x(x-5)-5(x+3)=[x(x+3)-5(x+3)]-5x(x-5)=(x+3)(x-5)-5x(x-5)=(x-5)(x+3-5x)=(x-5)(3-4x)

a. \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x+11=t.\)Thay vào ta được :
\(\left(t+1\right)\left(t-1\right)-24\)

\(=t^2-1-24=t^2-25=\left(t+5\right)\left(t-5\right)\)

Thay \(t=x^2+7x+11\)Ta được :
\(\left(x^2+7x+11+5\right)\left(x^2+7x+11-5\right)\)

\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)

a) - Đặt \(A=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

    + Ta có: \(A=\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right).\left(x+4\right)\right]-24\)

      \(\Leftrightarrow A=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)

    - Đặt \(a=x^2+7x+10\)

    + Ta lại có: \(A=a.\left(a+2\right)-24\)

               \(\Leftrightarrow A=a^2+2a-24\)

               \(\Leftrightarrow A=\left(a^2-4a\right)+\left(6a-24\right)\)

               \(\Leftrightarrow A=a.\left(a-4\right)+6.\left(a-4\right)\)

               \(\Leftrightarrow A=\left(a-4\right).\left(a+6\right)\)

    - Thay \(a=x^2+7x+10\)vào phương trình \(A\), ta có:

                     \(A=\left(x^2+7x+10-4\right).\left(x^2+7x+10+6\right)\)

              \(\Leftrightarrow A=\left(x^2+7x+6\right).\left(x^2+7x+16\right)\)

              \(\Leftrightarrow A=\left[\left(x^2+x\right)+\left(6x+6\right)\right].\left(x^2+7x+16\right)\)

              \(\Leftrightarrow A=\left[x.\left(x+1\right)+6.\left(x+1\right)\right].\left(x^2+7x+16\right)\)

              \(\Leftrightarrow A=\left(x+1\right).\left(x+6\right).\left(x^2+7x+16\right)\)

^_^ Chúc bạn hok tốt ^_^ !!#@##

a) Câu hỏi của a - Toán lớp 8 - Học toán với OnlineMath

b) Câu hỏi của c - Toán lớp 8 - Học toán với OnlineMath

a) ( 3 x   -   2 y ) 3 .        b) ( x   -   1 ) ( x   +   3 ) 2 .