Rút gọn phương trình
P = \(\left(\frac{2x}{2x^2-5x+3}-\frac{5}{2x-3}\right):\left(3+\frac{2}{1-x}\right)\)
K
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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 :>>
L
20 tháng 1 2019
a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)
<=> \(6x^2-5x+3-2x+9x-6x^2=0\)
<=> \(2x+3=0\)
<=> \(x=\frac{-3}{2}\)
b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)
<=> \(10x-40-6-4x=20x+4-4x\)
<=> \(6x-46-16x-4=0\)
<=> \(-10x-50=0\)
<=> \(-10\left(x+5\right)=0\)
<=> \(x+5=0\)
<=> \(x=-5\)
c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)
<=> \(8x+9x-15=36x-18-14\)
<=> \(8x+9x-36x=+15-18-14\)
<=> \(-19x=-14\)
<=> \(x=\frac{14}{19}\)
d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
<=> \(12x+10-10x-3=8x+4x+2\)
<=> \(2x-7=12x+2\)
<=> \(2x-12x=7+2\)
<=> \(-10x=9\)
<=> \(x=\frac{-9}{10}\)
e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)
<=> \(x^2-6x-12-\left(x-4^2\right)=0\)
<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)
<=> \(x^2-6x-12-x^2+8x-16=0\)
<=> \(2x-28=0\)
<=> \(2\left(x-14\right)=0\)
<=> x-14=0
<=> x=14
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) \(Q=\left(\frac{x+1}{2x-1}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left(\frac{x+1}{2x-1}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right).\frac{4x^2-4}{5}\)
\(=\left(\frac{2\left(x+1\right)\left(x-1\right)\left(x+1\right)}{2\left(2x-1\right)\left(x-1\right)\left(x+1\right)}+\frac{2.3\left(2x-1\right)}{2\left(x-1\right)\left(x+1\right)\left(2x-1\right)}-\frac{\left(x+3\right)\left(2x-1\right)\left(x-1\right)}{2\left(x+1\right)\left(2x-1\right)\left(x-1\right)}\right).\frac{4x^2-4}{5}\)
\(=\frac{2\left(x+1\right)\left(x^2-1\right)+12x-6-\left(2x^2+5x-3\right)\left(x-1\right)}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)
\(=\frac{2\left(x^3+x^2-x-1\right)+12x-6-2x^3-5x^2+3x+2x^2+5x-3}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)
\(=\frac{2x^3+2x^2-2x-2+20x-2x^3-3x^2-9}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)
\(=\frac{-x^2+18x-11}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{-x^2+18x-11}{\left(2x-1\right)}.\frac{2}{5}\)
\(=\frac{-2x^2+36x-22}{5\left(2x-1\right)}\)
TT
17 tháng 8 2020
a) \(4x^2\left(5x^3-2x+3\right)\)
\(=20x^5-8x^3+12x^2\)
b) \(3y^2\left(4y^3+\frac{2}{3}y^2-\frac{1}{3}\right)\)
\(=12y^5+2y^4-y^2\)
c) \(\left(5x^2-4x\right)\left(x-2\right)\)
\(=5x^3-14x^2+8x\)
d) \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=6x^2+22x-55-6x^2-23x-21\)
\(=-x-76\)
A
17 tháng 8 2020
1, \(4x^2\left(5x^3-2x+3\right)=20x^5-8x^3+12x^2\)
2, \(3y^2\left(4y^3+\frac{2}{3}y^2-\frac{1}{3}\right)=12y^5+2y^4-y^2\)
3, \(\left(5x^2-4x\right)\left(x-2\right)=5x^3-10x^2-4x^2+8x=5x^3-14x^2+8x\)
4, \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)=6x^2+33x-10x-55-\left(6x^2+14x+9x+21\right)\)
\(=6x^2+23x-55-6x^2-23x-21=-76\)
VM
14 tháng 2 2020
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
3 tháng 3 2020
a, \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
\(=>\frac{1-x+x+1}{x+1}+2=\frac{1}{x+1}+2\)
\(=>\frac{2}{x+1}=\frac{1}{x+1}\)
\(=>2x+2=x+1\)
\(=>2x-x=1-2=-1\)
\(=>x=-1\)
vậy nghiệm của phương trình trên là {-1}
3 tháng 3 2020
À quên ĐKXĐ của câu a là \(x\ne-1\)
Nên \(x\in\varnothing\)nhé :v
Ta có: \(P=\left(\frac{2x}{2x^2-5x+3}-\frac{5}{2x-3}\right):\left(3+\frac{2}{1-x}\right)\)
\(=\left(\frac{2x}{\left(x-1\right)\left(2x-3\right)}-\frac{5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}\right):\left(3-\frac{2}{x-1}\right)\)
\(=\frac{2x-5\left(x-1\right)}{\left(x-1\right)\left(2x-3\right)}:\frac{3\left(x-1\right)-2}{x-1}\)
\(=\frac{2x-5x+5}{\left(x-1\right)\left(2x-3\right)}\cdot\frac{x-1}{3x-3-2}\)
\(=\frac{-3x+5}{2x-3}\cdot\frac{1}{3x-5}\)
\(=\frac{-\left(3x-5\right)\cdot1}{\left(2x-3\right)\cdot\left(3x-5\right)}\)
\(=\frac{-1}{2x-3}\)
\(=\frac{1}{3-2x}\)
thank bạn