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9 tháng 3 2022
\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)\(\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\)
\(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\)
\(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\)
\(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\)
\(=\frac{-2\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}.\frac{2x+1}{-2}\)
\(=\frac{2x+1}{x-3}\)
b)\(\left|x+1\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+1=\frac{1}{2}\\x+1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(koTMđkxđ\right)\\x=-\frac{3}{2}\left(TMđkxđ\right)\end{cases}}}\)
thay \(x=-\frac{3}{2}\) vào P tâ đc: \(P=\frac{2x+1}{x-3}=\frac{2.\left(-\frac{3}{2}\right)+1}{-\frac{3}{2}-3}=\frac{4}{9}\)
c)ta có:\(P=\frac{x}{2}\Leftrightarrow\frac{2x+1}{x-3}=\frac{x}{2}\)
\(\Rightarrow2.\left(2x+1\right)=x.\left(x-3\right)\)
\(\Leftrightarrow4x+2=x^2-3x\)
\(\Leftrightarrow x^2-7x-2=0\)
\(\Leftrightarrow x^2-2.\frac{7}{2}+\frac{49}{4}-\frac{57}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2-\frac{57}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}-\frac{\sqrt{57}}{2}\right).\left(x-\frac{7}{2}+\frac{\sqrt{57}}{2}\right)\)
bạn tự giải nốt nhé!!
d)\(x\in Z;P\in Z\Leftrightarrow\frac{2x+1}{x-3}\in Z\Leftrightarrow\frac{2x-6+7}{x-3}=2+\frac{7}{x-3}\in Z\)
\(2\in Z\Rightarrow\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
bạn tự làm nốt nhé
9 tháng 3 2022
a, \(\left(\dfrac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\dfrac{2x-1-2x-1}{2x+1}\right)\)
\(=\dfrac{-2x+6}{\left(x+3\right)\left(x-3\right)}:\dfrac{-2}{2x+1}=\dfrac{-2\left(x-3\right)\left(2x+1\right)}{-2\left(x+3\right)\left(x-3\right)}=\dfrac{2x+1}{x+3}\)
b, \(\left|x+1\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}-1\\x=-\dfrac{1}{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktmđk\right)\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Thay x = -3/2 ta được \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)
22 tháng 2 2020
ĐKXĐ : \(x\ne\pm\frac{1}{2}\)
\(E=\left(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}-\frac{\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}\right):\left(\frac{\left(1+2x\right)\left(1+2x\right)}{\left(1-2x\right)\left(1+2x\right)}-\frac{\left(1-2x\right)\left(1-2x\right)}{\left(1+2x\right)\left(1-2x\right)}\right)\)
\(E=\left(\frac{16x^4+8x^3+4x^2+2x+16x^4-8x^3-4x^2+2x}{1-16x^4}\right):\left(\frac{1+2x+x^2-1+2x-x^2}{1-4x^2}\right)\)
\(E=\frac{32x^4+4x}{1-16x^4}:\frac{4x}{1-4x^2}\)
\(E=\frac{4x\left(8x^3+1\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{4x}\)
\(E=\frac{8x^3+1}{1+4x^2}\)
Study well
LT
22 tháng 2 2020
E=\(\left(\frac{4x^2+2x}{1-4x^2}-\frac{4x^2-2x}{1+4x^2}\right):\left(\frac{1+2x}{1-2x}-\frac{1-2x}{1+2x}\right)\)
E=\(\left(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)-\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}\right):\)\(\left(\frac{\left(1+2x\right)^2-\left(1-2x\right)^2}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{4x^2+16x^4+2x+8x^3-\left(4x^2-16x^4-2x+8x^3\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{\left(1+4x+4x^2\right)-\left(1-4x+4x^2\right)}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{4x^2+16x^4+2x+8x^3-4x^2+16x^4+2x-8x^3}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{1+4x+4x^2-1+4x-4x^2}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{16x^4+2x+16x^4+2x}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{8x}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{32x^4+8x}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{8x}\)
E=\(\frac{8x\left(4x^3+1\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{8x}\)
E=\(\frac{4x^3+1}{1+4x^2}\)
22 tháng 2 2020
E=\(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)-\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}:\frac{\left(1+2x\right)^2-\left(1-2x\right)^2}{1-4x^2}\)
E=\(\frac{4x^2+16x^4+2x+8x^3-4x^2+16x^2+2x-8x^3}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{1+4x+4x^2-1+4x-4x^2}\)
E=\(\frac{32x^4+4x}{8x\left(1+4x^2\right)}=\frac{8x^3+1}{2\left(1+4x^2\right)}\)
19 tháng 1 2020
\(A=\left(\frac{3x}{1-2x}-\frac{2x}{1+2x}\right):\frac{2x^2+5}{1-4x+4x^2}\)
\(A=\frac{3x+2x}{1-2x}:\frac{2x^2+5}{\left(1-2x\right)^2}\)
\(A=\frac{5x}{1-2x}\cdot\frac{\left(1-2x\right)^2}{2x^2+5}\)
\(A=\frac{5x\left(1-2x\right)\left(1-2x\right)}{\left(1-2x\right)\left(2x^2+5\right)}\)
\(A=\frac{5x\left(1-2x\right)}{2x^2+5}\)
\(A=\frac{5x-10x^2}{2x^2+5}\)
VM
19 tháng 1 2020
\(A=\left(\frac{3x}{1-2x}-\frac{2x}{1+2x}\right):\frac{2x^2+5}{1-4x+4x^2}\)
\(A=\left(\frac{3x.\left(1+2x\right)}{\left(1-2x\right).\left(1+2x\right)}-\frac{2x.\left(1-2x\right)}{\left(1-2x\right).\left(1+2x\right)}\right):\frac{2x^2+5}{1-4x+4x^2}\)
\(A=\left(\frac{3x+6x^2}{\left(1-2x\right).\left(1+2x\right)}-\frac{2x-4x^2}{\left(1-2x\right).\left(1+2x\right)}\right):\frac{2x^2+5}{1-4x+4x^2}\)
\(A=\left(\frac{3x+6x^2}{\left(1-2x\right).\left(1+2x\right)}+\frac{-\left(2x-4x^2\right)}{\left(1-2x\right).\left(1+2x\right)}\right):\frac{2x^2+5}{1-4x+4x^2}\)
\(A=\left(\frac{3x+6x^2-2x+4x^2}{\left(1-2x\right).\left(1+2x\right)}\right):\frac{2x^2+5}{1-4x+4x^2}\)
\(A=\frac{x+10x^2}{\left(1-2x\right).\left(1+2x\right)}:\frac{2x^2+5}{1-4x+4x^2}\)
\(A=\frac{x+10x^2}{\left(1-2x\right).\left(1+2x\right)}:\frac{2x^2+5}{\left(1-2x\right)^2}\)
\(A=\frac{x+10x^2}{\left(1-2x\right).\left(1+2x\right)}.\frac{\left(1-2x\right)^2}{2x^2+5}\)
\(A=\frac{\left(x+10x^2\right).\left(1-2x\right)^2}{\left(1-2x\right).\left(1+2x\right).\left(2x^2+5\right)}\)
\(A=\frac{\left(x+10x^2\right).\left(1-2x\right)}{\left(1+2x\right).\left(2x^2+5\right)}\)
\(A=\frac{x-2x^2+10x^2-20x^3}{2x^2+5+4x^3+10x}\)
\(A=\frac{x+8x^2-20x^3}{2x^2+5+4x^3+10x}\)
Chúc bạn học tốt!
6 tháng 1 2018
\(P=\frac{2x^5-x^4-2x+1}{4x^2-1}+\frac{8x^2-4x+2}{ }\)
\(P=\frac{x^4\left(2x-1\right)-\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)
\(P=\frac{\left(x^4-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{2}{2x+1}\)
\(P=\frac{x^4-1}{2x+1}+\frac{2}{2x+1}\)
\(P=\frac{x^4+1}{2x+1}\)
Vậy \(P=\frac{x^4+1}{2x+1}\)
\(\left(1-2x\right)^2+\left(2-4x\right)\left(1+2x\right)+\left(1+2x\right)^2\)
\(=\left(1-2x\right)^2+2\left(1-2x\right)\left(1+2x\right)+\left(1+2x\right)^2\)
\(=\left(1-2x+1+2x\right)^2=2^2=4\)
Trả lời:
\(\left(1-2x\right)^2+\left(2-4x\right)\left(1+2x\right)+\left(1+2x\right)^2\)
\(=1-4x+4x^2+2+4x-4x-8x^2+1+4x+4x^2\)
\(=4\)