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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) \(Q=\left(\frac{x+1}{2x-1}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left(\frac{x+1}{2x-1}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right).\frac{4x^2-4}{5}\)
\(=\left(\frac{2\left(x+1\right)\left(x-1\right)\left(x+1\right)}{2\left(2x-1\right)\left(x-1\right)\left(x+1\right)}+\frac{2.3\left(2x-1\right)}{2\left(x-1\right)\left(x+1\right)\left(2x-1\right)}-\frac{\left(x+3\right)\left(2x-1\right)\left(x-1\right)}{2\left(x+1\right)\left(2x-1\right)\left(x-1\right)}\right).\frac{4x^2-4}{5}\)
\(=\frac{2\left(x+1\right)\left(x^2-1\right)+12x-6-\left(2x^2+5x-3\right)\left(x-1\right)}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)
\(=\frac{2\left(x^3+x^2-x-1\right)+12x-6-2x^3-5x^2+3x+2x^2+5x-3}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)
\(=\frac{2x^3+2x^2-2x-2+20x-2x^3-3x^2-9}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)
\(=\frac{-x^2+18x-11}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{-x^2+18x-11}{\left(2x-1\right)}.\frac{2}{5}\)
\(=\frac{-2x^2+36x-22}{5\left(2x-1\right)}\)
19 tháng 3 2020
=[x(x-2)/2(x2+4)-2x2/(4+x2)(2-x)][x(x-2)(x+1)/x3]
={[x(x-2)(2-x)-4x2 ]/2(2-x)(4+x2)} .[x(x-2)(x+1)/x3 ]
=[-x(x2+4)/2(2-x)(4+x2)].[x(x-2)(x+1)/x3 ]
=-x.x(x-2)(x+1)/2(2-x)x3
=(x+1)/2x
CH
Cô Hoàng Huyền
Admin
VIP
21 tháng 12 2017
ĐK: \(\hept{\begin{cases}x\ne1\\x\ne\frac{3}{2}\end{cases}}\)
\(\left(\frac{2x}{2x^2-5x+3}-\frac{5}{2x-3}\right)\div\left(3+\frac{2}{1-x}\right)\)
\(=\frac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}:\frac{3-3x+2}{1-x}\)
\(=\frac{5-3x}{\left(2x-3\right)\left(x-1\right)}.\frac{1-x}{5-3x}\)
\(=\frac{1}{3-2x}\)
21 tháng 12 2017
\(\left(\frac{2x}{2x^2-5x+3}-\frac{5}{2x-3}\right):\left(3+\frac{2}{1-x}\right)\)\(ĐKXĐ:x\ne1;x\ne\frac{3}{2}\)
\(=\)\(\left[\frac{2x}{\left(2x-3\right)\left(x-1\right)}-\frac{5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}\right]:\left[\frac{3x-3-2}{x-1}\right]\)
\(=\frac{2x-5x+5}{\left(2x-3\right)\left(x-1\right)}:\frac{3x-5}{x-1}\)
\(=\frac{-3x+5}{\left(2x-3\right)\left(x-1\right)}.\frac{x-1}{3x-5}\)
\(=\frac{-\left(3x-5\right)}{2x-3}.\frac{1}{3x-5}\)
\(=\frac{-1}{2x-3}\)
31 tháng 10 2020
Bài làm
Như đã nhắn là mình sẽ làm theo quan điểm của mình là 5/(x^2 - 1) nha
\(A=\left[\frac{3\left(x+2\right)}{2x^3+2x+2x^2+2}+\frac{2x^2-x-10}{2x^3-2-2x^2+2x}\right]:\left[\frac{5}{x^2-1}+\frac{3}{2x+2}-\frac{3}{2x-2}\right]\)
\(A=\left[\frac{3\left(x+2\right)}{2x^2\left(x+1\right)+2\left(x+1\right)}+\frac{2x^2+4x-5x-10}{\left(2x^3-2x^2\right)+\left(2x-2\right)}\right]:\left[\frac{5}{x^2-1}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)
\(A=\left[\frac{3\left(x+2\right)}{\left(2x^2+2\right)\left(x+1\right)}+\frac{2x\left(x+2\right)-5\left(x+2\right)}{2x^2\left(x-1\right)+2\left(x-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)
\(A=\left[\frac{3\left(x+2\right)}{\left(2x^2+2\right)\left(x+1\right)}+\frac{\left(2x-5\right)\left(x+2\right)}{\left(2x^2+2\right)\left(x-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)
\(A=\left[\frac{3\left(x+2\right)\left(x-1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}+\frac{\left(2x-5\right)\left(x+2\right)\left(x+1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3\left(x-1\right)}{2\left(x^2-1\right)}-\frac{3\left(x+1\right)}{2\left(x^2-1\right)}\right]\)
\(A=\left[\frac{3\left(x+2\right)\left(x-1\right)+\left(2x-5\right)\left(x+2\right)\left(x+1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{10}{2\left(x^2-1\right)}+\frac{3x-3}{2\left(x^2-1\right)}-\frac{3x+3}{2\left(x^2-1\right)}\right]\)
\(A=\left[\frac{\left(x+2\right)\left[3x-3+\left(2x-5\right)\left(x+1\right)\right]}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{10+3x-3-3x-3}{2\left(x^2-1\right)}\right]\)
\(A=\left[\frac{\left(x+2\right)\left(3x-3+2x^2+2x-5x-5\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\frac{4}{2\left(x^2-1\right)}\)
\(A=\frac{\left(x+2\right)\left(2x^2-8\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\cdot\frac{\left(x^2-1\right)}{2}\)
\(A=\frac{\left(x+2\right)2\left(x^2-4\right)}{2\left(2x^2+2\right)}\)
\(A=\frac{2\left(x+2\right)\left(x-2\right)\left(x+2\right)}{4\left(x^2+1\right)}\)
\(A=\frac{\left(x+2\right)^2\left(x-2\right)}{2\left(x^2+1\right)}\)
:>>> Chả biết đúng không nữa nhưng số to quá :>>
22 tháng 6 2018
A = \(\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right).\left(x+1\right)}-\frac{x+3}{2\left(x+2\right)}\right).\frac{4x^2-4}{5}\)
A = \(\left(\frac{\left(x+1\right)^2+3.2-\left(x+3\right).\left(x-1\right)}{2\left(x-1\right).\left(x+1\right)}\right).\frac{4x^2-4}{5}\)
A = \(\left(\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right).\left(x+1\right)}\right).\frac{4\left(x^2-1\right)}{5}\)
A = \(\frac{10}{2\left(x-1\right).\left(x+1\right)}.\frac{4\left(x-1\right).\left(x+1\right)}{5}\)
A = 4
HN
0
29 tháng 7 2016
\(3x^3-\frac{3}{2}x^2-x^3-\frac{1}{2}x+\frac{1}{2}x+2=2x^3-\frac{3}{2}x^2+2\)
\(2x^2-10x-3x-2x^2=26\)
-13x=26
x=-2
\(\frac{-56x-19}{60}\)