P=\((\)\(\frac{2x}{2x^2-5x+3}\)- \(\frac{5}{2x-3}\)\()\):\((3+\frac{2}{1-x})\)
a, rút gọn P
b, P=? khi |2x-1|=3
c, x=? để P>1
D, tìm x\(\varepsilon\)Z để P\(\varepsilon\)Z
K
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NN
3 tháng 8 2018
\(P=\left(\frac{2x}{2x^2-5x+2}-\frac{5}{2x-3}\right):\left(3+\frac{2}{1-x}\right) \)(dk x khac 3/2 ; x khac 1)
\(P=\left(\frac{2x}{\left(2x-3\right)\left(x-1\right)}-\frac{5\left(x-1\right)}{\left(2x+3\right)\left(x-1\right)}\right):\left(\frac{3\left(x-1\right)}{x-1}-\frac{2}{x-1}\right)\)
\(P=\frac{2x-5x+5}{\left(2x-3\right)\left(x-1\right)}:\frac{3x-3-2}{x-1}\)
\(P=\frac{-\left(3x-5\right)}{\left(2x-3\right)\left(x-1\right)}\cdot\frac{x-1}{3x-5}\)
\(P=\frac{-1}{2x-3}\)
b) TC: \(|2x-1|=3\)
TH1) \(|2x-1|=2x-1\)khi \(x\ge\frac{1}{2}\)
2x-1=3 suy ra x=2 ( thoa dk)
TH2) \(|2x-1|=-2x+1\)khi \(x< \frac{1}{2}\)
-2x+1=3 suy ra x=-1 ( thoa dk)
khi x= 2 thi P=-1
khi x= -1 thi P=1/5
c) de P thuoc Z thi \(-\frac{1}{2x-3}\)thuoc Z
suy ra \(\frac{1}{3-2x}\)thuoc Z
suy ra 3-2x thuoc \(Ư\left(1\right)\in\left\{\pm1\right\}\)
khi 3-2x=1 thi x= 1 (ko thoa dk x khac 1)
khi 3-2x=-1 thi x=2(thoa dk)
vay x=2 thi P thuoc Z
d) giai tg tu cau c
23 tháng 6 2017
a) Điều kiện : \(x\ne2;x\ne3\)
\(B=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2x+4}{x-3}\)
\(=\frac{2x-9-\left(x-3\right)\left(x+3\right)+2\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}=\frac{x+4}{x-3}\)
23 tháng 6 2017
b) Điều kiện \(x\in Z;x\ne2;x\ne3\)
Có \(B=\frac{x+4}{x-3}\in Z\), mà x+4 và x-3 nguyên do x nguyên, nên
\(x+4⋮x-3\Leftrightarrow7⋮x-3\), do đó \(x-3\inƯ\left(7\right)=\left\{1;7;-1;-7\right\}\Rightarrow x\in\left\{4;10;2;-4\right\}\)
mà do x khác 2 (điều kiện) nên ta kết luận \(x\in\left\{4;10;-4\right\}\)
KN
18 tháng 8 2021
a. ĐKXĐ : \(x\ne\frac{1}{2};\frac{5}{2};4;-\frac{3}{2};\frac{1\pm\sqrt{43}}{2}\)
\(A=\left(\frac{2x-3}{4x^2-12x+5}+\frac{3x-8}{13x-2x^2-20}-\frac{3}{2x-1}\right):\frac{21+2x-2x^2}{4x^2+4x-3}+\)
\(=\left(\frac{2x-3}{\left(2x-1\right)\left(2x-5\right)}-\frac{3x-8}{\left(2x-5\right)\left(x-4\right)}-\frac{3}{2x-1}\right).\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)
\(=\frac{\left(2x-3\right)\left(x-4\right)-\left(3x-8\right)\left(2x-1\right)-3\left(2x-5\right)\left(x-4\right)}{\left(2x-1\right)\left(2x-5\right)\left(x-4\right)}.\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)
\(=\frac{-10x^2+47x-56}{\left(2x-5\right)\left(x-4\right)}.\frac{2x+3}{-2x^2+2x+21}+1\) số to wa
a) \(ĐKXĐ:\hept{\begin{cases}x\ne\frac{3}{2}\\x\ne1\\x\ne\frac{5}{3}\end{cases}}\)
\(P=\left(\frac{2x}{2x^2-5x+3}-\frac{5}{2x-3}\right):\left(3+\frac{2}{1-x}\right)\)
\(\Leftrightarrow P=\frac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}:\frac{3-3x+2}{1-x}\)
\(\Leftrightarrow P=\frac{2x-5x+5}{\left(2x-3\right)\left(x-1\right)}:\frac{-3x+5}{1-x}\)
\(\Leftrightarrow P=\frac{-3x+5}{\left(2x-3\right)\left(x-1\right)}\cdot\frac{1-x}{-3x+5}\)
\(\Leftrightarrow P=\frac{-1}{2x-3}\)
b) Khi |2x-1| = 3
\(\Leftrightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\Leftrightarrow P=\frac{-1}{4-3}=-1\\x=-1\Leftrightarrow P=\frac{-1}{-2-3}=\frac{1}{5}\end{cases}}\)
Vậy khi \(\left|2x-1\right|=3\Leftrightarrow P\in\left\{-1;\frac{1}{5}\right\}\)
c) Để \(P>1\)
\(\Leftrightarrow\frac{-1}{2x-3}>1\)
\(\Leftrightarrow-1>2x-3\)
\(\Leftrightarrow2x< 2\)
\(\Leftrightarrow x< 1\)
Vậy để \(P>1\Leftrightarrow x< 1\)
d) Để \(P\inℤ\)
\(\Leftrightarrow-1⋮2x-3\)
\(\Leftrightarrow2x-3\inƯ\left(-1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{1;2\right\}\)
Vì \(x\ne1\)
\(\Leftrightarrow x\in\left\{2\right\}\)
Vậy để \(P\inℤ\Leftrightarrow x\in\left\{2\right\}\)