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VT
2
20 tháng 6 2019
2-x2(x2+x+1)=-x4-x3-x2+m
2-x4-x3-x2=-x4-x3-x2+m (=) m=2
vậy ..
chúc bn hc tốt
20 tháng 6 2019
\(2-x^2\left(x^2+x+1\right)=-x^4-x^3-x^2+m\)
\(\Leftrightarrow-x^4-x^3-x^2-m=-x^4-x^3-x^2+2\)
\(\Leftrightarrow-x^4-x^3-x^2-m+x^4=-x^4-x^3-x^2+2+x^4\)
\(\Leftrightarrow-x^3-x^2-m=-x^3-x^2+2\)
\(\Leftrightarrow-x^3-x^2-m+x^3=-x^3-x^2+2+x^3\)
\(\Leftrightarrow-x^3-m=-x^2+2\)
\(\Leftrightarrow-x^2-m+x^2=-x^2+2+x^2\)
\(\Leftrightarrow-m=2\)
\(\Leftrightarrow\frac{-m}{-1}=\frac{-2}{-1}\)
\(\Leftrightarrow x=2\)
Vậy: x = 2
VT
0
VV
24 tháng 12 2018
a) Điều kiện xác định :
x ≠ 3; x ≠ -3; x ≠ 0
M = \(\dfrac{x}{x^2-9}\) - \(\dfrac{1}{x+3}\): ( \(\dfrac{x}{x\left(x-3\right)}\) - \(\dfrac{x-3}{x\left(x-3\right)}\) )
M = \(\dfrac{x}{x^2-9}\) - \(\dfrac{1}{x+3}\) : ( \(\dfrac{x-x+3}{x\left(x-3\right)}\) )
M = \(\dfrac{x}{x^2-9}\) - \(\dfrac{1}{x+3}\) : \(\dfrac{3}{x\left(x-3\right)}\)
M = \(\dfrac{x}{x^2-9}\) - \(\dfrac{x\left(x-3\right)}{3\left(x+3\right)}\) = \(\dfrac{x}{\left(x-3\right)\left(x+3\right)}\) - \(\dfrac{x\left(x-3\right)}{3\left(x+3\right)}\)
M = \(\dfrac{3x}{3\left(x-3\right)\left(x+3\right)}\) - \(\dfrac{x\left(x-3\right)^2}{3\left(x-3\right)\left(x+3\right)}\)
M = \(\dfrac{3x-x\left(x-3\right)^2}{3\left(x-3\right)\left(x+3\right)}\) = \(\dfrac{3x-x\left(x^2-6x+9\right)}{3\left(x-3\right)\left(x+3\right)}\)
M = \(\dfrac{3x-x^3+6x^2-9x}{3\left(x-3\right)\left(x+3\right)}\) = \(\dfrac{-x^3+6x^2-6x}{3\left(x-3\right)\left(x+3\right)}\)
Mk đang mệt sai thì bạn thông cảm cho mk.
12 tháng 12 2022
a: \(M=\dfrac{x}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x+3}:\dfrac{x-x+3}{x\left(x-3\right)}\)
\(=\dfrac{x}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x+3}\cdot\dfrac{x\left(x-3\right)}{3}\)
\(=\dfrac{x}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x-3\right)}{3\left(x+3\right)}\)
\(=\dfrac{3x-x\left(x^2-6x+9\right)}{3\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x-x^3+6x^2-9x}{3\left(x-3\right)\left(x+3\right)}=\dfrac{-x^3+6x^2-6x}{3\left(x-3\right)\left(x+3\right)}\)
b: Để M>1/2 thì M-1/2>0
=>\(\dfrac{-x^3+6x^2-6x}{3\left(x^2-9\right)}-\dfrac{1}{2}>0\)
=>\(\dfrac{-2x^3+12x^2-12x-3x^2+9}{6\left(x^2-9\right)}>0\)
=>\(\dfrac{-2x^3+9x^2-12x+9}{x^2-9}>0\)
TH1: \(\left\{{}\begin{matrix}-2x^3+9x^2-12x+9>0\\x^2-9>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 3\\\left[{}\begin{matrix}x>3\\x< -3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x< -3\)
TH2: \(\left\{{}\begin{matrix}-2x^3+9x^2-12x+9< 0\\x^2-9< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>3\\-3< x< 3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
C
30 tháng 10 2019
a) \(\frac{1}{m+1}+\frac{1}{\left(m+1\right)\left(2m+1\right)}\)
\(=\frac{2m+1}{\left(m+1\right)\left(2m+1\right)}+\frac{1}{\left(m+1\right)\left(2m+1\right)}\)
\(=\frac{2m+2}{\left(m+1\right)\left(2m+1\right)}\)
\(=\frac{2\left(m+1\right)}{\left(m+1\right)\left(2m+1\right)}\)
\(=\frac{2}{2m+1}=\frac{4}{4m+2}\left(đpcm\right)\)
C
30 tháng 10 2019
b) \(\frac{1}{m+2}+\frac{1}{\left(m+1\right)\left(m+2\right)}+\frac{1}{\left(m+1\right)\left(4m+3\right)}\)
\(=\frac{m+1}{\left(m+1\right)\left(m+2\right)}+\frac{1}{\left(m+1\right)\left(m+2\right)}+\frac{1}{\left(m+1\right)\left(4m+3\right)}\)
\(=\frac{m+2}{\left(m+1\right)\left(m+2\right)}+\frac{1}{\left(m+1\right)\left(4m+3\right)}\)
\(=\frac{1}{m+1}+\frac{1}{\left(m+1\right)\left(4m+3\right)}\)
\(=\frac{4m+3}{\left(m+1\right)\left(4m+3\right)}+\frac{1}{\left(m+1\right)\left(4m+3\right)}\)
\(=\frac{4m+4}{\left(m+1\right)\left(4m+3\right)}\)
\(=\frac{4\left(m+1\right)}{\left(m+1\right)\left(4m+3\right)}\)
\(=\frac{4}{4m+3}\left(đpcm\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}m+1=1\\m+1=-1\end{matrix}\right.\Leftrightarrow m\in\left\{0;-2\right\}\)