Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, + b, \(A=\frac{x+2}{x-3}+\frac{2x-1}{x-1}-\frac{2x-1}{2x+1}\)DKXD : \(x\ne3;1;-\frac{1}{2}\)
\(=\frac{\left(x+2\right)\left(x-1\right)\left(2x+1\right)}{\left(x-3\right)\left(x-1\right)\left(2x+1\right)}+\frac{\left(4x^2-1\right)\left(x-3\right)}{\left(x-3\right)\left(x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)\left(x-1\right)\left(x-3\right)}{\left(2x+1\right)\left(x-1\right)\left(x-3\right)}\)
\(=\frac{2x^3+3x^2-3x-2+4x^3-12x^2-x+4-2x^3+9x^2-10x+3}{MTC}\)
\(=\frac{4x^3-14x+2x^3+5}{MTC}\)
Đề sai ko kiểm tra lại hộ nhé !!!
a + b , ĐKXĐ : \(x\ne2;-3\)
\(A=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+3\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-4-5}{\left(x-2\right)\left(x+3\right)}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-3}{x-2}\)
c, Thay x = 2 ta có : ... Vì ko thỏa mãn giá trị của phân thức x khác 2 nên ko có giá trị biểu thức
d, Ta có : \(\frac{x-3}{x-2}=\frac{x-2-1}{x-2}=-\frac{1}{x-2}\)
\(-x+2\inƯ\left(1\right)=\left\{\pm1\right\}\)
| -x + 2 | 1 | -1 |
| x | 1 | 3 |
a, \(B=\left(\frac{2x+1}{2x-1}+\frac{4}{1-4x^2}-\frac{2x-1}{2x+1}\right):\frac{x^2+2}{2x+1}\)
\(=\left(\frac{2x+1}{2x-1}+\frac{4}{\left(1-2x\right)\left(2x+1\right)}-\frac{2x-1}{2x+1}\right):\frac{x^2+2}{2x+1}\)
\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{4}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right):\frac{x^2+2}{2x+1}\)
\(=\left(\frac{4x^2+4x+1-4-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\right):\frac{x^2+2}{2x+1}\)
\(=\frac{8x-4}{\left(2x-1\right)\left(2x+1\right)}.\frac{2x+1}{x^2+2}=\frac{8x-4}{\left(2x-1\right)\left(x^2+2\right)}\)
b, Thay x = -1 ta được : \(\frac{9\left(-1\right)-4}{\left[2\left(-1\right)-1\right]\left[\left(-1\right)^2+2\right]}=-\frac{13}{-9}=\frac{13}{9}\)
\(P=\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\)
a) Điều kiện: \(x\ne3;x\ne-3\)
b) \(P=\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\)
\(P=\frac{3.\left(x-3\right)}{\left(x+3\right).\left(x-3\right)}+\frac{x+3}{\left(x-3\right).\left(x+3\right)}-\frac{-18}{\left(x-3\right).\left(x+3\right)}\)
\(P=\frac{3x-9+x+3+18}{\left(x+3\right).\left(x-3\right)}=\frac{4x+12}{\left(x-3\right).\left(x+3\right)}=\frac{4.\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}=\frac{4}{x-3}\)
c) \(\frac{4}{x-3}=4\Leftrightarrow4=\left(x-3\right).4\Leftrightarrow4x-12=4\Leftrightarrow4x=16\Leftrightarrow x=4\)