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Nguyễn Huệ Lam ơi cái câu b bn làm sai r cái đoạn đặt ntu chung là 2 x đầu tiên ấy bn
a)
\(\frac{9-\left(x+5\right)^2}{x^2+4x+4}=\frac{3^2-\left(x+5\right)^2}{x^2+2.x.2+2^2}=\frac{\left(3+x+5\right)\left(3-x-5\right)}{\left(x+2\right)^2}\)
\(=\frac{\left(x+8\right)\left(x-2\right)}{\left(x+2\right)^2}\)
b)
\(\frac{32x-8x^2+2x^3}{x^3+64}=\frac{2x\left(x^2-8x+16\right)}{x^3+4^3}=\frac{2x\left(x^2-2.x.4+4^2\right)}{\left(x+4\right)\left(x^2-4x+16\right)}\)
\(=\frac{2x\left(x-4\right)^2}{\left(x+4\right)\left(x^2-4x+16\right)}\)
a) (2x^2 +2xy - xy -y^2 ) / (2x^2 - 2xy - xy +y^2)
= 2x(x+y) - y(x+y) / 2x(x-y) - y(x-y)
= (2x-y)(x+y) / (2x-y)(x-y)
= x+y/x-y
Rút gọn cái sau:
\(\frac{32x+4x^2+2x^3}{x^3+64}\)
\(=\frac{2x\left(x^2+2x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}\)
Đề có vẻ sai sai ?
a) A=\(\frac{x+1}{6x^3-6x^2}-\frac{x-2}{8x^3-8x}=\frac{x+1}{6x^2\left(x-1\right)}-\frac{x-2}{8x\left(x-1\right)\left(x+1\right)}=\frac{4\left(x+1\right)^2-3x\left(x-2\right)}{24x^2\left(x-1\right)\left(x+1\right)}=\frac{4x^2+8x+4-3x^2+6x}{24x^2\left(x-1\right)\left(x+1\right)}=\frac{x^2+14x+10}{24x^2\left(x-1\right)\left(x+1\right)}\)
\(A =\frac{32x - 8x^{2} + 2x^{3}}{x^{3}+ 64}\)\(= \frac{2x(16 - 4x + x^{2})}{(x + 4)(x^{2} - 4x + 16)}= \frac{2x(x^{2} - 4x + 16)}{(x + 4)(x^{2} - 4x + 16)}= \frac{2x}{x + 4}\)
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{2x}{x^2-3x}+\frac{2x}{x^2-4x+3}+\frac{x}{x-1}\)
\(=\frac{2x}{x\left(x-3\right)}+\frac{2x}{x^2-3x-x+3}+\frac{x}{x-1}\)
\(=\frac{2}{x-3}+\frac{2x}{x\left(x-3\right)-\left(x-3\right)}+\frac{x}{x-1}\)
\(=\frac{2\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}+\frac{2x}{\left(x-3\right)\left(x-1\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x-1\right)}\)
\(=\frac{2x-2+2x+x^2-3x}{\left(x-3\right)\left(x-1\right)}\)
\(=\frac{x^2+x-2}{\left(x-3\right)\left(x-1\right)}=\frac{x^2-x+2x-2}{\left(x-3\right)\left(x-1\right)}=\frac{x\left(x-1\right)+2\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}=\frac{\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(x-1\right)}=\frac{x+2}{x-3}\)
Ta có \(\frac{32x-8x^2+2x^3}{x^3+64}=\frac{x\left(32-8x+2x^2\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{2x\left(x^2-4x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{2x}{x+4}\)