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

sao nhiều bt vại , sao làm hết nổi !!!!! ~ _~

1) \(ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)\)

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

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

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

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

\(=\left(a+b\right)\left[a\left(b-c\right)+c\left(b-c\right)\right]\)

\(=\left(a+b\right)\left(b-c\right)\left(a+c\right)\)

abc - (ab + ac + bc) + (a + b + c) - 1

= abc - bc - ab + b - ac + c + a - 1

= bc(a - 1) - b(a - 1) - c(a - 1) + (a - 1)

= (a - 1)(bc - b - c + 1)

= (a - 1)(b - 1)(c - 1)

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

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

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

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

\(=b\left(a-b\right)\left(a+c\right)-c\left(a+c\right)\left(a-b\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(a+c\right)\)

Tự lm đ