Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ai có thể giảng cho mình dạng toán tìm số tự nhiên thỏa mãn đièu kiện chia hết ko
hãy nêu ra cách giải cụ thể cho câu sau 3a-11 chia hết cho a+2 tìm a
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+bc\right)+\left(a+b+c\right)ac-abc\)
\(=\left(ab+b^2+bc\right)\left(a+c\right)+\left(a+c\right)ac+abc-abc\)
\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Bài 1:
a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left(3-2x+5\right)\)
\(=\left(2x+1\right)\left(8-2x\right)\)
\(=2\left(4-x\right)\left(2x+1\right)\)
b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\left(3x-2\right)\left(3x-6\right)\)
\(=3\left(3x-2\right)\left(x-2\right)\)
Bài 2:
a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)
\(=\left(a-b\right)\left(2a-4b\right)\)
\(=2\left(a-b\right)\left(a-2b\right)\)
f: Ta có: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
Phân tích đa thức thành nhân tử
a) (1-2x)(1+2x)-x(x+2)(x-2)
\(=1-4x^2-x\left(x^2-4\right)\)
\(=1-4x^2-x^3+4x\)
\(=\left(1-x^3\right)+\left(4x-4x^2\right)\)
\(=\left(1-x\right)\left(1+x+x^2\right)+4x\left(1-x\right)\)
\(=\left(1-x\right)\left(1+x+x^2+4x\right)\)
\(=\left(1-x\right)\left(x^2+5x+1\right)\)
\(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
\(=a^4+6a^3b+12a^2b^2+8b^3a-8a^3b-12a^2b^2+6ab^3-b^4\)
\(=a^4+6a^3b+8b^3a-8a^3b-6ab^3-b^4\)
\(=\left(a^4-b^4\right)+\left(6a^3b-6ab^3\right)+\left(8b^3a-8a^3b\right)\)
\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3\right)+6ab\left(a^2-b^2\right)+8ab\left(b^2-a^2\right)\)
\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3\right)+6ab\left(a-b\right)\left(a+b\right)-8ab\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3+6a^2b+6ab^2-8a^2b-8ab^2\right)\)
\(=\left(a-b\right)\left(a^3-a^2b-ab^2+b^3\right)\)
\(=\left(a-b\right)\left[a^2\left(a-b\right)-b^2\left(a-b\right)\right]\)
\(=\left(a-b\right)^3\left(a+b\right)\)
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+bc\right)+\left(a+b+c\right)ac-abc\)
\(=\left(ab+b^2+bc\right)\left(a+c\right)+\left(a+c\right)ac+abc-abc\)
\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+bc\right)+\left(a+b+c\right)ac-abc\)
\(=\left(ab+b^2+bc\right)\left(a+c\right)+\left(a+c\right)ac+abc-abc\)
\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
a) \(5a^2-5ax-7a+7x\)
\(=5a\left(a-x\right)-7\left(a-x\right)\)
\(=\left(5a-7\right)\left(a-x\right)\)
c) \(x^2-\left(a+b\right).x+ab\)
\(=x^2-ax-bx+ab\)
\(=x\left(x-a\right)-b\left(x-a\right)\)
\(=\left(x-b\right)\left(x-a\right)\)
Xửa đề:
a/ \(a\left(a+2b\right)^3-b\left(b+2a\right)^3=\left(a-b\right)^3\left(a+b\right)\)
b/ \(\left(b-a^2\right)\left(c-b^2\right)\left(c^2-a\right)\)