Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) A = \(\frac{3x^2+3x-3}{x^2+x-2}-\frac{x+1}{x+2}+\frac{x-2}{x}\cdot\left(\frac{1}{1-x}-1\right)\)
A = \(\frac{3x^2+3x-3}{x^2+2x-x-2}-\frac{x+1}{x+2}+\frac{x-2}{x}\cdot\left(\frac{1-1+x}{1-x}\right)\)
A = \(\frac{3x^2+3x-3}{\left(x-1\right)\left(x+2\right)}-\frac{x+1}{x+2}+\frac{x-2}{x}\cdot\frac{x}{1-x}\)
A = \(\frac{3x^2+3x-3}{\left(x-1\right)\left(x+2\right)}-\frac{x+1}{x+2}-\frac{x-2}{x-1}\)
A = \(\frac{3x^2+3x-3}{\left(x-1\right)\left(x+2\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}-\frac{\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)
A = \(\frac{3x^2+3x-3-x^2+1-x^2+4}{\left(x-1\right)\left(x+2\right)}\)
A = \(\frac{x^2+3x+2}{\left(x-1\right)\left(x+2\right)}\)
A = \(\frac{x^2+2x+x+2}{\left(x-1\right)\left(x+2\right)}\)
A = \(\frac{\left(x+1\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)
A = \(\frac{x+1}{x-1}\) (Đk: \(x-1\ge0\) => x \(\ge\)1)
b) Ta có: A = \(\frac{x+1}{x-1}=\frac{\left(x-1\right)+2}{x-1}=1+\frac{2}{x-1}\)
Để A \(\in\)Z <=> 2 \(⋮\)x - 1
<=> x - 1 \(\in\)Ư(2) = {1; -1; 2; -2}
<=> x \(\in\){2; 0; 3; -1}
c) Ta có: A < 0
=> \(\frac{x+1}{x-1}< 0\)
=> \(\hept{\begin{cases}x+1< 0\\x-1>0\end{cases}}\) hoặc \(\hept{\begin{cases}x+1>0\\x-1< 0\end{cases}}\)
=> \(\hept{\begin{cases}x< -1\\x>1\end{cases}}\)(loại) hoặc \(\hept{\begin{cases}x>-1\\x< 1\end{cases}}\)
=> -1 < x < 1
Edogawa Conan
Thiếu dòng đầu \(ĐKXĐ:\hept{\begin{cases}x\ne1\\x\ne-2\\x\ne0\end{cases}}\)
\(a,x\ne2;x\ne-2;x\ne0\)
\(b,A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\frac{6}{x+2}\)
\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(=\frac{1}{2-x}\)
\(c,\)Để A > 0 thi \(\frac{1}{2-x}>0\Leftrightarrow2-x>0\Leftrightarrow x< 2\)
a) Ta có :A = \(\left(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}+\frac{1}{x-1}\right):\frac{x^2+x}{x^3+x}\)
ĐK: \(\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)
A = \(\left(\frac{\left(x-1\right)^2}{x^2+x+1}-\frac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\right):\frac{x\left(x+1\right)}{x\left(x^2+1\right)}\)
= \(\frac{\left(x-1\right)^3-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)
= \(\frac{x^3-3x^2+3x-1+3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)
= \(\frac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}=1.\frac{x^2+1}{x+1}=\frac{x^2+1}{x+1}\)
b) Để A > - 1 <=> \(\frac{x^2+1}{x+1}>-1\)
<=> \(\frac{x^2+1}{x+1}+1>0\)
<=> \(\frac{x^2+x+2}{x+1}>0\)
Vì x2 + x + 2 >0 \(\forall x\)
=> A > 0 <=> x + 1 > 0 <=> x > -1
Câu 3 :
\(a,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\) ĐKXđ : \(x\ne\pm1\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{5\left(x-1\right)}\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{10}{x+1}\)
\(B=\left(\frac{x}{3x-9}+\frac{2x-3}{3x-x^2}\right).\frac{3x^2-9x}{x^2-6x+9}.\)
ĐKXđ : \(x\ne0;x\ne3\)
\(B=\left(\frac{x}{3\left(x-3\right)}+\frac{2x-3}{x\left(3-x\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\left(\frac{x^2}{3x\left(x-3\right)}+\frac{9-6x}{3x\left(x-3\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\frac{x^2-6x+9}{3x\left(x-3\right)}.\frac{3x\left(x-3\right)}{x^2-6x+9}=1\)
\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)
\(\left(DK:x\ne0;x\ne-1;x\ne\frac{1}{2}\right)\)
\(=\frac{\left(x+2\right)\left(x+1\right)+6x-9x\left(x+1\right)}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{x^2+3x+2+6x-9x^2-9x}{3x}.\frac{1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{-8x^2+2}{6x}.\frac{1}{1-2x}+\frac{x^2-3x-1}{3x}=\frac{-2\left(4x^2-1\right)}{6x}.\frac{1}{1-2x}+\)\(\frac{x^2-3x-1}{3x}\)
\(\frac{\left(1-2x\right)\left(1+2x\right)}{3x\left(1-2x\right)}+\frac{x^2-3x-1}{3x}=\frac{x^2-3x-1+1+2x}{3x}=\)\(=\frac{x\left(x-1\right)}{3x}=\frac{x-1}{3}\)
a)\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\left(DK:x\ne0;x\ne-1;x\ne\frac{1}{2}\right)\)
\(=\frac{\left(x+2\right)\left(x+1\right)+6x-9x\left(x+1\right)}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{x^2+3x+2+6x-9x^2-9x}{3x}.\frac{1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{-8x^2+2}{6x}.\frac{1}{1-2x}+\frac{x^2-3x-1}{3x}=\frac{-2\left(4x^2-1\right)}{6x}.\frac{1}{1-2x}+\frac{x^2-3x-1}{3x}\)
\(\frac{\left(1-2x\right)\left(1+2x\right)}{3x\left(1-2x\right)}+\frac{x^2-3x-1}{3x}=\frac{x^2-3x-1+1+2x}{3x}=\frac{x\left(x-1\right)}{3x}=\frac{x-1}{3}\)
b) \(\left|x\right|=\frac{1}{3}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\left(x\ge0\right)\\x=-\frac{1}{3}\left(x< 0\right)\end{cases}}\)
Thay vào \(\frac{x-1}{3}\)tính được A.
c) \(A< 0\Rightarrow\frac{x-1}{3}< 0\Rightarrow x-1< 0\Rightarrow x< 1\)
Kết hợp cùng với điều kiện của ở phần rút gọn.
d) \(A\in Z\Rightarrow\frac{x-1}{3}\in Z\Rightarrow x=3k+1\)(\(k\in Z\))