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Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
1) \(\dfrac{x}{x+1}-\dfrac{2x}{x-1}+\dfrac{x+3}{x^2-1}\)
\(=\dfrac{x}{x+1}-\dfrac{2x}{x-1}+\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}\) MTC: \(\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x-1\right)-2x\left(x+1\right)+\left(x+3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-x-2x^2-2x+x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2-2x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+x-3x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-\left(x^2-x\right)-\left(3x-3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x\left(x-1\right)-3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)\left(-x-3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x-3}{x+1}\)
2) \(\dfrac{5}{x+1}-\dfrac{10}{x-x^2-1}-\dfrac{15}{x^3+1}\)
\(=\dfrac{5}{x+1}-\dfrac{10}{-\left(x^2-x+1\right)}-\dfrac{15}{x^3+1}\)
\(=\dfrac{5}{x+1}+\dfrac{10}{\left(x^2-x+1\right)}-\dfrac{15}{\left(x+1\right)\left(x^2-x+1\right)}\) MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)
\(=\dfrac{5\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{10\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{15}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5\left(x^2-x+1\right)+10\left(x+1\right)-15}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5x^2+5x}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5x}{x^2-x+1}\)
3) \(\dfrac{2}{2x+1}-\dfrac{1}{2x-1}-\dfrac{2}{1-4x^2}\)
\(=\dfrac{2}{2x+1}-\dfrac{1}{2x-1}+\dfrac{2}{4x^2-1}\)
\(=\dfrac{2}{2x+1}-\dfrac{1}{2x-1}+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\) MTC: \(\left(2x-1\right)\left(2x+1\right)\)
\(=\dfrac{2\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{2x+1}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{2\left(2x-1\right)-\left(2x+1\right)+2}{\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{4x-2-2x-1+2}{\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{2x-1}{\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{1}{2x+1}\)
4) \(\dfrac{3x^2+5x+14}{x^3+1}+\dfrac{x-1}{x^2-x+1}-\dfrac{4}{x+1}\)
\(=\dfrac{3x^2+5x+14}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x-1}{x^2-x+1}-\dfrac{4}{x+1}\) MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)
\(=\dfrac{3x^2+5x+14}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{4\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(3x^2+5x+14\right)+\left(x-1\right)\left(x+1\right)-4\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{3x^2+5x+14+x^2-1-4x^2+4x-4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{9x+9}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{9\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{9}{x^2-x+1}\)
a) \(A = \frac{2x^2 - 16x+43}{x^2-8x+22}\) = \(\frac{2(x^2-8x+22)-1}{x^2-8x+22}\) = \(2 - \frac{1}{x^2-8x+22}\)
Ta có : \(x^2-8x+22 \) = \(x^2-8x+16+6 = ( x-4)^2 +6 \)
Vì \((x-4)^2 \ge 0 \) với \( \forall x\in R\) Nên \(( x-4)^2 +6 \ge 6 \)
\(\Rightarrow \) \(x^2-8x+22 \) \( \ge 6\)\(\Rightarrow \) \(\frac{1}{x^2-8x+22} \) \(\le \frac{1}{6}\) \(\Rightarrow \) - \(\frac{1}{x^2-8x+22} \) \(\ge - \frac{1}{6}\)
\(\Rightarrow \) A = \(2 - \frac{1}{x^2-8x+22}\) \( \ge 2-\frac{1}{6}\) = \(\frac{11}{6}\) Dấu "=" xảy ra khi và chỉ khi x=4
Vậy GTNN của A = \(\frac{11}{6}\) khi và chỉ khi x=4
4)a)\(\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\)(1)
ĐKXĐ:\(\left\{{}\begin{matrix}x-5\ne0\\x+5\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne5\\x\ne-5\end{matrix}\right.\)
(1)\(\Rightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=20\)
\(\Leftrightarrow x^2+10x+25-\left(x^2-10x+25\right)=20\)
\(\Leftrightarrow x^2+10x+25-x^2+10x-25=20\)
\(\Leftrightarrow x^2-x^2+10x+10x=-25+25=20\)
\(\Leftrightarrow20x=20\)
\(\Leftrightarrow x=1\left(nh\text{ậ}n\right)\)
S=\(\left\{1\right\}\)
\(A=-\dfrac{4}{x^2-4x+10}\\ =-\dfrac{4}{\left(x^2-2.x.2+4+6\right)}\\ =-\dfrac{4}{\left(x-2\right)^2+6}\)
\(\left(x-2\right)^2\ge0\\ \Rightarrow\left(x-2\right)^2+6\ge6\\ \Rightarrow\dfrac{4}{\left(x-2\right)^2+6}\le\dfrac{2}{3}\\ \Rightarrow A=-\dfrac{4}{\left(x-2\right)^2+6}\ge-\dfrac{2}{3}\)
Min A=-2/3 khi x=2
\(C=\dfrac{2}{x^2+4x+5}=\dfrac{2}{\left(x+2\right)^2+1}\)
Vì \(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+1\ge1\)
\(\Rightarrow C\le2\)
Dấu ''='' xảy ra \(\Leftrightarrow x=-2\)
Vậy Min C = 2 kjhi x = -2
a) \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\) MTC: \(2x\left(x+3\right)\)
\(=\dfrac{3x}{2x\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-\left(x-6\right)}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
b) \(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5x+6}{4-x^2}\)
\(=\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x+6}{x^2-4}\)
\(=\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x+6}{\left(x-2\right)\left(x+2\right)}\) MTC: \(\left(x-2\right)\left(x+2\right)\)
\(=\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5x+6}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)+2\left(x+2\right)-\left(5x+6\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x-8+2x+4-5x-6}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x-10}{\left(x-2\right)\left(x+2\right)}\)
c) \(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)
\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{4x^2-2x}\)
\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{2x\left(2x-1\right)}\) MTC: \(2x\left(2x-1\right)\)
\(=\dfrac{\left(1-3x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{2x\left(3x-2\right)}{2x\left(2x-1\right)}-\dfrac{3x-2}{2x\left(2x-1\right)}\)
\(=\dfrac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)-\left(3x-2\right)}{2x\left(2x-1\right)}\)
\(=\dfrac{2x-1-6x^2+3x+6x^2-4x-3x+2}{2x\left(2x-1\right)}\)
\(=\dfrac{-2x+1}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(2x-1\right)}{2x\left(2x-1\right)}\)
\(=\dfrac{-1}{2x}\)
d) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(=\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\) MTC: \(\left(x-1\right)\left(x^2+x+1\right)\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2\left(x-2\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x^2+2\right)+2\left(x-2\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+2+2x-4-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-3+x}{\left(x-1\right)\left(x^2+x+1\right)}\)
câu nào cũng ghi lại đề nha
a) \(x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b)\(x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
c) \(\left(x+1\right)\left(x+2\right)+\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\end{matrix}\right.\)
d) \(\dfrac{1}{x-2}+3-\dfrac{3-x}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3\left(x-2\right)-\left(3-x\right)}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3x-6-3+x}{x-2}=0\) ( đk \(x\ne2\) )
\(\Leftrightarrow4x-8=0\Rightarrow x=2\)
đ) \(\dfrac{8-x}{x-7}-8-\dfrac{1}{x-7}=0\)
\(\Leftrightarrow\dfrac{8-x-8\left(x-7\right)-1}{x-7}=0\) (đk \(x\ne7\))
\(\Leftrightarrow8-x-8x+56-1=0\)
\(\Leftrightarrow-9x+63=0\)
\(\Leftrightarrow x=7\)
ko có c hả bạn?