Rút gọn phân thức
a)\(\frac{9-\left(x+5\right)^2}{x^2+4x+4}\)
b)\(\frac{32x-8x+2x^3}{x^3+64}\) . Giúp mk nha,mk tik cho .
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TH
29 tháng 6 2017
Nguyễn Huệ Lam ơi cái câu b bn làm sai r cái đoạn đặt ntu chung là 2 x đầu tiên ấy bn
29 tháng 6 2017
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)
\(\frac{32x-8x^2+2x^3}{x^3+64}=\frac{2x\left(x^2-8x+16\right)}{x^3+4^3}=\frac{2x\left(x^2-2.x.4+4^2\right)}{\left(x+4\right)\left(x^2-4x+16\right)}\)
\(=\frac{2x\left(x-4\right)^2}{\left(x+4\right)\left(x^2-4x+16\right)}\)
14 tháng 2 2020
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
PT
29 tháng 6 2017
a)= \(\frac{\left(2x+3\right)^2}{2x^2+3x-4x-6}\)
=\(\frac{\left(2x+3\right)^2}{x\left(2x+3\right)-2\left(2x+3\right)}\)
= \(\frac{\left(2x+3\right)^2}{\left(x-2\right)\left(2x+3\right)}\)
=\(\frac{2x+3}{x-2}\)
PT
29 tháng 6 2017
b) = \(\frac{3\left|x-4\right|}{3x^2-3x-1296}\)
= \(\frac{3\left|x-4\right|}{3\left(x^2-x-432\right)}\)
=\(\frac{\left|x-4\right|}{x^2-x-432}\)
3 tháng 9 2016
1) \(\frac{8xy\left(3x-1\right)^3}{12x^3\left(1-3x\right)}=-\frac{8xy\left(3x-1\right)^3}{12x^3\left(3x-1\right)}=-\frac{2y\left(3x-1\right)^2}{3x^2}\)
2) \(\frac{5x^3+5x}{x^4-1}=\frac{5x\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2-1\right)}=\frac{5x}{x^2-1}\)
3) \(\frac{9-\left(x+5\right)^2}{x^2+4x+4}=\frac{\left(3-x-5\right)\left(3+x+5\right)}{\left(x+2\right)^2}=\frac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}=-\frac{x+8}{x+2}\)
3) \(\frac{32x-8x^2+2x^3}{x^3+64}=\frac{2x\left(16-4x+x^2\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{2x}{x+4}\)
1 tháng 11 2019
a) \(P=\frac{2}{2x+3}+\frac{3}{2x+1}-\frac{6x+5}{\left(2x+3\right)\left(2x-3\right)}\)
\(=\frac{2\left(2x+1\right)\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}+\frac{3\left(2x+3\right)\left(2x-3\right)}{\left(2x+1\right)\left(2x+3\right)\left(2x-3\right)}-\frac{\left(6x+5\right)\left(2x+1\right)}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)
\(=\frac{\left(4x+2\right)\left(2x-3\right)+3\left(4x^2-9\right)-12x^2-16x-5}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)
\(=\frac{8x^2-8x-6+12x^2-27-12x^2-16x-5}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)
\(=\frac{8x^2-24x-38}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)
Check hộ mình xem nghi nghi sai sai
1 tháng 11 2019
b) \(Q=\left(\frac{x+1}{2x-1}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left(\frac{x+1}{2x-1}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right).\frac{4x^2-4}{5}\)
\(=\left(\frac{2\left(x+1\right)\left(x-1\right)\left(x+1\right)}{2\left(2x-1\right)\left(x-1\right)\left(x+1\right)}+\frac{2.3\left(2x-1\right)}{2\left(x-1\right)\left(x+1\right)\left(2x-1\right)}-\frac{\left(x+3\right)\left(2x-1\right)\left(x-1\right)}{2\left(x+1\right)\left(2x-1\right)\left(x-1\right)}\right).\frac{4x^2-4}{5}\)
\(=\frac{2\left(x+1\right)\left(x^2-1\right)+12x-6-\left(2x^2+5x-3\right)\left(x-1\right)}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)
\(=\frac{2\left(x^3+x^2-x-1\right)+12x-6-2x^3-5x^2+3x+2x^2+5x-3}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)
\(=\frac{2x^3+2x^2-2x-2+20x-2x^3-3x^2-9}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)
\(=\frac{-x^2+18x-11}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{-x^2+18x-11}{\left(2x-1\right)}.\frac{2}{5}\)
\(=\frac{-2x^2+36x-22}{5\left(2x-1\right)}\)
HA
1
12 tháng 7 2017
Ta có \(\frac{32x-8x^2+2x^3}{x^3+64}=\frac{x\left(32-8x+2x^2\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{2x\left(x^2-4x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{2x}{x+4}\)
khó qua mik mới hc lớp 7 thôi