\(M=\left(\frac{x}{x-3}-\frac{x+3}{3x^2-6x-9}+\frac{1}{3x+3}\right)\frac{x^2-2x-3}{x^2+x+2}\)
a, rút gọn
b,chứng tỏ rằng giá trị của M luôn nhỏ hơn 1
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7 tháng 11 2018
\(ĐKXĐ:\hept{\begin{cases}x-3\ne0\\3x^2-6x-9\ne0\\3x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne3\\3\left(x^2-2x-3\right)\ne0\\3\left(x+1\right)\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)
\(M=\left(\frac{x}{x-3}-\frac{x+3}{3x^2-6x-9}+\frac{1}{3x+3}\right).\frac{x^2-2x-3}{x^2+x+2}\)
\(=\left[\frac{x\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}-\frac{x+3}{3\left(x^2-2x-3\right)}+\frac{1}{3\left(x+1\right)}\right].\frac{x^2-2x-3}{x^2+x+2}\)
\(=\left[\frac{3x\left(x+1\right)}{3\left(x+1\right)\left(x-3\right)}-\frac{x+3}{3\left(x+1\right)\left(x-3\right)}+\frac{x-3}{3\left(x+1\right)\left(x-3\right)}\right].\frac{x^2-2x-3}{x^2+x+2}\)
\(=\frac{3x\left(x+1\right)-x-3+x-3}{3\left(x+1\right)\left(x-3\right)}.\frac{x^2-2x-3}{x^2+x+2}\)
\(=\frac{3x^2+3x-6}{3\left(x+1\right)\left(x-3\right)}.\frac{x^2-2x-3}{x^2+x+2}\)
\(=\frac{x^2+x-2}{\left(x+1\right)\left(x-3\right)}.\frac{\left(x+1\right)\left(x-3\right)}{x^2+x+2}=\frac{x^2+x-2}{x^2+x+2}\)
7 tháng 11 2018
\(=\frac{x^2+x-2}{x^2+x+2}=1-\frac{4}{x^2+x+2}\)
b,\(\text{Với }x\ne-1\text{ và }x\ne3\text{ ta có:}\)
\(\text{Để }M=1-\frac{4}{x^2+x+2}< 1\)
\(\Leftrightarrow-\frac{4}{x^2+x+2}< 0\)
\(\Leftrightarrow\frac{4}{x^2+x+2}>0\)
\(\Leftrightarrow4>0\left(\text{hiển nhiên}\right)\)
Vậy ... đpcm
24 tháng 11 2019
Ta có:
a) M = \(\left(\frac{6x}{x^2-9}-\frac{1}{x+3}+\frac{5}{3-x}\right):\frac{4}{x^2-3x}\)
M = \(\left(\frac{6x}{\left(x-3\right)\left(x+3\right)}-\frac{x-3}{\left(x+3\right)\left(x-3\right)}-\frac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\cdot\frac{x^2-3x}{4}\)
M = \(\left(\frac{6x-x+3-5x-15}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x\left(x-3\right)}{4}\)
M = \(\frac{-12.x\left(x-3\right)}{\left(x-3\right)\left(x+3\right).4}\)
M = \(-\frac{3x}{x+3}\)
b) Với x = 2 => M = \(-\frac{3.2}{3+2}=-\frac{6}{5}\)
DL
5 tháng 10 2019
a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)
b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)
\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)
H
30 tháng 1 2019
Câu 3 :
\(a,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\) ĐKXđ : \(x\ne\pm1\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{5\left(x-1\right)}\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{10}{x+1}\)
H
30 tháng 1 2019
\(B=\left(\frac{x}{3x-9}+\frac{2x-3}{3x-x^2}\right).\frac{3x^2-9x}{x^2-6x+9}.\)
ĐKXđ : \(x\ne0;x\ne3\)
\(B=\left(\frac{x}{3\left(x-3\right)}+\frac{2x-3}{x\left(3-x\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\left(\frac{x^2}{3x\left(x-3\right)}+\frac{9-6x}{3x\left(x-3\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\frac{x^2-6x+9}{3x\left(x-3\right)}.\frac{3x\left(x-3\right)}{x^2-6x+9}=1\)
M = \(\left(\frac{x}{x-3}-\frac{x+3}{3x^2-6x-9}+\frac{1}{3x+3}\right)\)\(\frac{x^2-2x-3}{x^2+x+2}\)
= \(\left(\frac{x\left(3x+3\right)}{3\left(x-3\right)\left(x+1\right)}-\frac{x+3}{3\left(x-3\right)\left(x+1\right)}+\frac{x-3}{3\left(x+1\right)\left(x-3\right)}\right)\)\(\frac{\left(x+1\right)\left(x-3\right)}{x^2+x+2}\)
= \(\frac{3\left(x^2+x-2\right)}{3\left(x-3\right)\left(x+1\right)}\)* \(\frac{\left(x+1\right)\left(x-3\right)}{x^2+x+2}\) = \(\frac{x^2+x-2}{x^2+x+2}\)
Ta thấy x2 + x - 2 < x2 + x + 2
nên M < 1