\(A=\frac{x^2-3}{\left(x+1\right)\left(x-3\right)}-\frac{2\left(x-3\right)}{x+1}-\frac{x+3}{x-3}\)
Rút gọn biểu thức A
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
17 tháng 10 2018
\(a)\)\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt{x-3}}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}-3}{\sqrt{x}-3}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\frac{3\sqrt{x}+3}{\sqrt{x}+3}.\frac{\sqrt{x}-3}{\sqrt{x+1}}\)
\(R=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
\(b)\) Ta có : \(R< -1\)
\(\Leftrightarrow\)\(\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}< -1\)
\(\Leftrightarrow\)\(\frac{\sqrt{x}-3}{\sqrt{x}+3}< \frac{-1}{3}\)
\(\Leftrightarrow\)\(3\sqrt{x}-9< -\sqrt{x}-3\)
\(\Leftrightarrow\)\(4\sqrt{x}< 6\)
\(\Leftrightarrow\)\(\sqrt{x}< \frac{3}{2}\)
\(\Leftrightarrow\)\(x< \frac{9}{4}\)
Chúc bạn học tốt ~
A
25 tháng 12 2018
\(\left(\frac{1}{x}+1-\frac{3}{x^3+1}-\frac{3}{x^2-x+1}\right)\cdot\frac{3x^2-3x+3}{\left(x+1\right).\left(x+2\right)}-\frac{2x-2}{x^2+2x}\)
\(=\left(\frac{x+1}{x}-\frac{3}{\left(x+1\right).\left(x^2-x+1\right)}+\frac{3.\left(x+1\right)}{\left(x+1\right).\left(x^2-x+1\right)}\right)\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\left[\frac{\left(x+1\right)^2.\left(x^2-x+1\right)-3x+3x^2+3x}{x.\left(x+1\right).\left(x^2-x+1\right)}\right]\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\left[\frac{x^4+x^3+x+1+3x^2}{x.\left(x+1\right).\left(x^2-x+1\right)}\right]\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\frac{3x^4+3x^3+3x+3+9x^2}{x.\left(x+1\right)^2.\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}=\frac{3x^4+3x^3+3x+3+9x^2}{x.\left(x+1\right)^2.\left(x+2\right)}-\frac{2x^3+2x^2-2x-2}{x.\left(x+1\right)^2.\left(x+2\right)}\)
\(=\frac{3x^4+x^3+7x^2+5x+5}{x.\left(x+1\right)^2.\left(x+2\right)}\)
27 tháng 2 2020
\(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right):\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\left(x\ne-1;x\ne0;x\ne-2\right)\)
\(=\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right):\frac{3x^3-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\left(\frac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3x+3}{\left(x+1\right)\left(x^2-x+1\right)}\right)\)\(:\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2-x+1-3+3x+3}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+1\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\frac{\left(x+1\right)\left(x+2\right)}{3\left(x^2-x+1\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x+2\right)^2\left(x+1\right)}{3\left(x^2-x+1\right)^2}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
8 tháng 3 2020
B=\(\frac{3\left(2x^8+5x^6+6x^4+5x^2+2\right)}{x\left(x^2+1\right)\left(2x^4+x^2+2\right)}\)
\(A=\frac{x^2-3}{\left(x+1\right).\left(x-3\right)}-\frac{2.\left(x-3\right)^2}{\left(x+1\right).\left(x-3\right)}-\frac{\left(x+1\right).\left(x+3\right)}{\left(x-3\right).\left(x+1\right)}\)
\(A=\frac{x^2-3-2x^2+12x-18-x^2-4x+3}{\left(x+1\right).\left(x-3\right)}=\frac{8x-18}{\left(x+1\right).\left(x-3\right)}=\frac{2.\left(4x-9\right)}{\left(x+1\right).\left(x-3\right)}\)