Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bằng phương pháp nào zậy bn????
547675675675678768768789980957457346242645657
Lời giải:
a) ĐKXĐ: $x\neq \pm 1$
\(\frac{x^4-4x^2+3}{x^4+6x^2-7}=\frac{x^2(x^2-1)-3(x^2-1)}{x^2(x^2-1)+7(x^2-1)}=\frac{(x^2-3)(x^2-1)}{(x^2-1)(x^2+7)}=\frac{x^2-3}{x^2+7}\)
b) ĐKXĐ: Với mọi $x\in\mathbb{R}$
\(\frac{x^4+x^3-x-1}{x^4+x^4+2x^2+x+1}=\frac{(x^4-x)+(x^3-1)}{(x^4+x^3+x^2)+(x^2+x+1)}=\frac{x(x^3-1)+(x^3-1)}{x^2(x^2+x+1)+(x^2+x+1)}\)
\(=\frac{(x^3-1)(x+1)}{(x^2+1)(x^2+x+1)}=\frac{(x-1)(x^2+x+1)(x+1)}{(x^2+1)(x^2+x+1)}=\frac{x^2-1}{x^2+1}\)
c) ĐK: $x\neq 1;-2$
\(\frac{x^3+3x^2-4}{x^3-3x+2}=\frac{x^2(x-1)+4(x^2-1)}{x^2(x-1)+x(x-1)-2(x-1)}=\frac{(x-1)(x^2+4x+4)}{(x-1)(x^2+x-2)}\)
\(=\frac{(x-1)(x+2)^2}{(x-1)(x-1)(x+2)}=\frac{x+2}{x-1}\)
d) ĐK: $x^2+3x-1\neq 0$
\(\frac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}=\frac{(x^2+3x)^2-1}{(x^2+3x)^2-2x^2-6x+1}\)
\(=\frac{(x^2+3x-1)(x^2+3x+1)}{(x^2+3x)^2-2(x^2+3x)+1}=\frac{(x^2+3x-1)(x^2+3x+1)}{(x^2+3x-1)^2}=\frac{x^2+3x+1}{x^2+3x-1}\)
a)\(3x^2-8x+4\)
\(=3x^2-2x-6x+4\)
\(=x\left(3x-2\right)-2\left(3x-2\right)\)
\(=\left(x-2\right)\left(3x-2\right)\)
b)\(4x^4+81\)
\(=4x^4+36x^2+81-36x^2\)
\(=\left(2x^2+9\right)^2-36x^2\)
\(=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)
c)\(x^8+98x^4+1\)
\(=\left(x^8+2x^4+1\right)+96x^4\)
\(=\left(x^4+1\right)^2+16x^2\left(x^4+1\right)+64x^4-16x^2\left(x^4+1\right)+32x^4\)
\(=\left(x^4+8x^2+1\right)^2-16x^2\left(x^4-2x^2+1\right)\)
\(=\left(x^4+8x^2+1\right)^2-16x^2\left(x^4-2x^2+1\right)\)
\(=\left(x^4+8x^2+1\right)^2-\left(4x^3-4x\right)^2\)
\(=\left(x^4+4x^3+8x^2-4x+1\right)\left(x^4-4x^3+8x^2+4x+1\right)\)
d)\(x^4+6x^3+7x^2-6x+1\)
\(=x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)
\(=x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)
\(=\left(x^2+3x-1\right)\left(x^2+3x-1\right)\)\(=\left(x^2+3x-1\right)^2\)
chắc bn nảy hỏi lun cả bài tâp về nhà quá, làm km 1 câu
a) = a+a+a + a +a +1 -a -a -a = a(a+a+1) +(a+a+1) - a(a+a+1)= (a+a+1)(a-a+1)
tự bn thêm mũ 4;3;2 vào được là bn làm dc cac câu sau
a) \(x^3-4x^2-8x+8\)
\(=x^3+8-4x^2-8x\)
\(=\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x+4-4x\right)\)
\(=\left(x+2\right)\left(x^2-6x+4\right)\)
\(=\left(x+2\right)\left(x^2-6x+9-5\right)\)
\(=\left(x+2\right)\left[\left(x-3\right)^2-5\right]\)
\(=\left(x+2\right)\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\)
b) \(1+6x-6x^2-x\)
\(=1-x+6x\left(1-x\right)\)
\(=\left(1-x\right)\left(6x+1\right)\)