l i m x → 0 x + 1 - x 2 + x + 1 3 x bằng:
A. 0
B. -1/6
C. (-1)/2
D. -∞
K
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16 tháng 5 2020
Bài 1:
a) Ta có: 4x-20=0
\(\Leftrightarrow4\left(x-5\right)=0\)
mà \(4\ne0\)
nên x-5=0
hay x=5
Vậy: x=5
b) Ta có: 3-2x=3(x+1)-x-2
\(\Leftrightarrow3-2x=3x+3-x-2\)
\(\Leftrightarrow3-2x=2x+1\)
\(\Leftrightarrow3-2x-2x-1=0\)
\(\Leftrightarrow2-4x=0\)
\(\Leftrightarrow2\left(1-2x\right)=0\)
mà \(2\ne0\)
nên 1-2x=0
\(\Leftrightarrow2x=1\)
hay \(x=\frac{1}{2}\)
Vậy: Tập nghiệm \(S=\left\{\frac{1}{2}\right\}\)
c) Ta có: \(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Leftrightarrow\frac{x+2}{2008}+1+\frac{x+3}{2007}+1+\frac{x+4}{2006}+1+\frac{x+2028}{6}-3=0\)
\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
mà \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\)
nên x+2010=0
hay x=-2010
Vậy: Tập nghiệm S={-2010}
d) Ta có: 2x(x+3)+5(x+3)=0
\(\Leftrightarrow\left(x+3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{-5}{2}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{-3;\frac{-5}{2}\right\}\)
Bài 2:
ĐKXĐ: \(x\notin\left\{-1;1\right\}\)
Bài 6:
ĐKXĐ: \(x\notin\left\{1;2;-1;-2\right\}\)
Ta có: \(\frac{1}{x-1}+\frac{1}{x-2}=\frac{1}{x+2}+\frac{1}{x+1}\)
\(\Leftrightarrow\frac{x-2}{\left(x-1\right)\left(x-2\right)}+\frac{x-1}{\left(x-1\right)\left(x-2\right)}=\frac{x+1}{\left(x+1\right)\left(x+2\right)}+\frac{x+2}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow\frac{x-2+x-1}{\left(x-1\right)\left(x-2\right)}=\frac{x+1+x+2}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow\frac{2x-3}{\left(x-1\right)\left(x-2\right)}-\frac{2x+3}{\left(x+1\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{\left(2x-3\right)\left(x^2+3x+2\right)}{\left(x^2-1\right)\left(x^2-4\right)}-\frac{\left(2x+3\right)\left(x^2-3x+2\right)}{\left(x^2-1\right)\left(x^2-4\right)}=0\)
\(\Leftrightarrow2x^3+3x^2-5x-6-\left(2x^3-3x^2-5x+6\right)=0\)
\(\Leftrightarrow2x^3+3x^2-5x-6-2x^3+3x^2+5x-6=0\)
\(\Leftrightarrow6x^2-12=0\)
\(\Leftrightarrow6x^2=12\)
\(\Leftrightarrow x^2=2\)
hay \(x=\pm\sqrt{2}\)
Vậy: Tập nghiệm \(S=\left\{\sqrt{2};-\sqrt{2}\right\}\)
24 tháng 7 2017
\(\left|x-y-2\right|+\left|y+3\right|=0\)
\(\left\{{}\begin{matrix}\left|x-y-2\right|\ge0\forall x;y\\\left|y+3\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|x-y-2\right|+\left|y+3\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-y-2\right|=0\Rightarrow x-\left(-3\right)-2=0\Rightarrow x+1=0\Rightarrow x=-1\\\left|y+3\right|=0\Rightarrow y+3=0\Rightarrow y=-3\end{matrix}\right.\)
\(\left|x-2007\right|+\left|y-2008\right|=0\)
\(\left\{{}\begin{matrix}\left|x-2007\right|\ge0\forall x\\\left|y-2008\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|x-2007\right|+\left|y-2008\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-2007\right|=0\Rightarrow x-2007=0\Rightarrow x=2007\\\left|y-2008\right|=0\Rightarrow y-2008=0\Rightarrow y=2008\end{matrix}\right.\)
\(\left|\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{3}{4}x\right|+\left|1,5-\dfrac{11}{17}+\dfrac{23}{13}y\right|=0\)
\(\left\{{}\begin{matrix}\left|\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{3}{4}x\right|\ge0\forall x\\\left|1,5-\dfrac{11}{17}+\dfrac{23}{13}y\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{3}{4}x\right|+\left|1,5-\dfrac{11}{17}+\dfrac{23}{13}x\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{3}{4}x\right|=0\Rightarrow\dfrac{1}{6}+\dfrac{3}{4}x=0\Rightarrow\dfrac{3}{4}x=-\dfrac{1}{6}\Rightarrow x=-\dfrac{2}{9}\\\left|1,5-\dfrac{11}{17}+\dfrac{23}{13}x\right|=0\Rightarrow\dfrac{29}{34}+\dfrac{23}{13}x=0\Rightarrow\dfrac{23}{13}x=-\dfrac{29}{34}\Rightarrow x=-\dfrac{377}{782}\end{matrix}\right.\)
\(\left|x-y-5\right|+\left|y-2\right|\le0\)
\(\left\{{}\begin{matrix}\left|x-y-5\right|\ge0\forall x;y\\\left|y-2\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|x-y-5\right|+\left|y-2\right|\ge0\)
Lúc này ta có:
\(\left\{{}\begin{matrix}\left|x-y-5\right|+\left|y-2\right|\le0\\\left|x-y-5\right|+\left|y-2\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x-y-5\right|+\left|y-2\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-y-5\right|=0\Rightarrow x-2-5=0\Rightarrow x=7\\\left|y-2=0\right|\Rightarrow y=2\end{matrix}\right.\)
\(\left|3x+2y\right|+\left|4y-1\right|\le0\)
\(\left\{{}\begin{matrix}\left|3x+2y\right|\ge0\forall x;y\\ \left|4y-1\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|3x+2y\right|+\left|4y-1\right|\ge0\)
Lúc này ta có:
\(\left\{{}\begin{matrix}\left|3x+2y\right|+\left|4y-1\right|\ge0\\\left|3x+2y\right|+\left|4y-1\right|\le0\end{matrix}\right.\)
\(\Rightarrow\left|3x+2y\right|+\left|4y-1\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|3x+2y\right|=0\Rightarrow3x+\dfrac{1}{2}=0\Rightarrow3x=-\dfrac{1}{2}\Rightarrow x=-\dfrac{1}{6}\\\left|4y-1\right|=0\Rightarrow4y=1\Rightarrow y=\dfrac{1}{4}\end{matrix}\right.\)
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