nhân hả bạn

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

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

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

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

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

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

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

ab(b - a) - bc(b - c) - ac(c - a)

= ab² - a²b - b²c +  bc² + ac(a - c)

= b²(a - c) - b(a² - c²) + ac(a - c)

= b²(a - c) - b(a - c)(a + c) + ac(a - c)

= (b² - ab - bc + ac)(a - c)

= [b(b - a) - c(b - a)](a - c)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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