Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P=\left(\frac{2x}{2x^2-5x+2}-\frac{5}{2x-3}\right):\left(3+\frac{2}{1-x}\right) \)(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
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 \(P\inℤ\Leftrightarrow x\in\left\{2\right\}\)
Rút gọn \(A=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\left(ĐKXĐ:x\ne2;x\ne3\right)\)
\(\Rightarrow A=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}\)
\(=\frac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\frac{x^2+3x-4x-12}{\left(x+3\right)\left(x-2\right)}=\frac{x.\left(x+3\right)-4.\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x-4}{x-2}\)
b) Để A > 0 <=> x-4/x-2 > 0
<=> x-4>0 <=>x>4
c) Ta có: x-4/x-2 = x-2-2/x-2 = 1-2/x-2
Để A nguyên dương <=> 2 chia hết cho x-2
<=> x-2 thuộc Ư(2) = {-2;2;-1;1}
giải như bài lớp 6 bình thương (loại những giá trị giống ĐKXĐ)
cảm ơn nạ rất rất rất....nhìu. Sư phụ hãy nhận của đồ đệ 1 lạy