Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c) a(b3 - c3) + b(c3 - a3) + c(a3 - b3)
= a(b3 - a3) + (b - a)(c3 - a3) + c(a3 - b3)
= (a3 - b3)(c - a) - (a - b)(c - a)(c2 + ca + a2)
= (a - b)(c - a)(ab + b2 - c2 - ca)
= (a - b)(c - a)(b - c)(a + b + c)
b) Ta có:
a(b2 - c2) + b(c2 - a2) + c(a2 - b2)
= a(b2 - a2) + (b - a)(c2 - a2) + c(a2 - b2)
= (a - b)(a + b)(c - a) - (a - b)(c - a)(c + a)
= (a - b)(b - c)(c - 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 4:
a) Ta có: \(a^4+a^2+1\)
\(=a^4+2a^2+1-a^2\)
\(=\left(a^2+1\right)^2-a^2\)
\(=\left(a^2-a+1\right)\left(a^2+a+1\right)\)
b) Ta có: \(a^4+a^2-2\)
\(=a^4+2a^2-a^2-2\)
\(=a^2\left(a^2+2\right)-\left(a^2+2\right)\)
\(=\left(a^2+2\right)\left(a^2-1\right)\)
\(=\left(a^2+2\right)\left(a-1\right)\left(a+1\right)\)
c) Ta có: \(x^4+4x^2-5\)
\(=x^4+5x^2-x^2-5\)
\(=x^2\left(x^2+5\right)-\left(x^2+5\right)\)
\(=\left(x^2+5\right)\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)
d) Ta có: \(x^3-19x-30\)
\(=x^3-25x+6x-30\)
\(=x\left(x^2-25\right)+6\left(x-5\right)\)
\(=x\left(x-5\right)\left(x+5\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+5x\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+5x+6\right)\)
\(=\left(x-5\right)\left(x^2+2x+3x+6\right)\)
\(=\left(x-5\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)
e) Ta có: \(x^3-7x-6\)
\(=x^3-4x-3x-6\)
\(=x\left(x^2-4\right)-3\left(x+2\right)\)
\(=x\left(x-2\right)\left(x+2\right)-3\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x\right)-3\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x-3\right)\)
\(=\left(x+2\right)\left(x^2-3x+x-3\right)\)
\(=\left(x+2\right)\left[x\left(x-3\right)+\left(x-3\right)\right]\)
\(=\left(x+2\right)\left(x-3\right)\left(x+1\right)\)
f) Ta có: \(x^3-5x^2-14x\)
\(=x\left(x^2-5x-14\right)\)
\(=x\left(x^2-7x+2x-14\right)\)
\(=x\left[x\left(x-7\right)+2\left(x-7\right)\right]\)
\(=x\left(x-7\right)\left(x+2\right)\)
Sửa lại ạ!
a) \(\left(3x-1\right)^2-16\)
\(=\left(3x-1\right)^2-4^2\)
\(=\left(3x-1-4\right)\left(3x-1+4\right)\)
\(=\left(3x-5\right)\left(3x+3\right)\)
b) \(\left(5x-4\right)^2-49x^2\)
\(=\left(5x-4\right)^2-\left(7x\right)^2\)
\(=\left(5x-4-7x\right)\left(5x-4+7x\right)\)
\(=\left(-4-2x\right)\left(-4+12x\right)\)
c) \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)
\(=\left(x+14\right)\left(3x-4\right)\)
d) \(\left(3x+1\right)^2-4\left(x-2\right)^2\)
\(=\left(3x+1\right)^2-\left[2\left(x-2\right)\right]^2\)
\(=\left(3x+1\right)^2-\left(2x-4\right)^2\)
\(=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\)
\(=\left(x+5\right)\left(5x-3\right)\)
e) \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)
\(=\left[3\left(2x+3\right)\right]^2-\left[2\left(x+1\right)\right]^2\)
\(=\left(6x+9\right)^2-\left(2x+2\right)^2\)
\(=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\)
\(=\left(4x+7\right)\left(8x+11\right)\)
P/s: Ko chắc!
Bài 4:
a) Ta có: \(x^3+6x^2+12x+8\)
\(=x^3+2x^2+4x^2+8x+4x+8\)
\(=x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+4x+4\right)\)
\(=\left(x+2\right)^3\)
b) Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-x^2-2x^2+2x+x-1\)
\(=x^2\left(x-1\right)-2x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+1\right)\)
\(=\left(x-1\right)^3\)
c) Ta có: \(1-9x+27x^2-27x^3\)
\(=1-3x-6x+18x^2+9x^2-27x^3\)
\(=\left(1-3x\right)-6x\left(1-3x\right)+9x^2\left(1-3x\right)\)
\(=\left(1-3x\right)\left(1-6x+9x^2\right)\)
\(=\left(1-3x\right)^3\)
d) Ta có: \(x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)
\(=x^3+3\cdot x^2\cdot\frac{1}{2}+3\cdot x\cdot\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3\)
\(=\left(x+\frac{1}{2}\right)^3\)
e) Ta có: \(27x^3-54x^2y+36xy^2-8y^3\)
\(=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot2y+3\cdot3x\cdot\left(2y\right)^2-\left(2y\right)^3\)
\(=\left(3x-2y\right)^3\)
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 ^_^ !!#@##