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a) Điều kiện : \(x\ne\pm\dfrac{1}{3}\)
\(B=\left[\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right]:\dfrac{6x^2+10x}{1-6x+9x^2}\)
\(=\left(\dfrac{3x\left(3x+1\right)}{\left(1-3x\right)\left(3x+1\right)}+\dfrac{2x\left(1-3x\right)}{\left(1-3x\right)\left(3x+1\right)}\right):\dfrac{6x^2+10x}{ \left(3x-1\right)^2}\)
\(=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(3x+1\right)}\cdot\dfrac{\left(1-3x\right)^2}{6x^2+10x}\)
\(=\dfrac{x\left(3x+5\right)}{\left(1-3x\right)\left(3x+1\right)}\cdot\dfrac{\left(1-3x\right)^2}{2x\left(3x+5\right)}=\dfrac{1-3x}{2\left(3x+1\right)}\)
b) Sai đề = Không làm
c) B >0
=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}1-3x>0\\2\left(3x+1\right)>0\end{matrix}\right.\\\left[{}\begin{matrix}1-3x< 0\\2\left(3x+1\right)< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{1}{3}\\x>-\dfrac{1}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{1}{3}\\x< -\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)
TH1 => \(-\dfrac{1}{3}< x< \dfrac{1}{3}\)
TH2 :Vô lí
Vậy giá trị x thỏa mãn :
\(-\dfrac{1}{3}< x< \dfrac{1}{3}\)
a. Ta có :
\(x^4-x^3-2x-4\)
\(=x^4-2x^3+x^3-2x-4\)
\(=x^3\left(x-2\right)+\left(x^3-2x^2\right)+\left(x^2-4\right)+\left(x^2-2x\right)\)
\(=x^3\left(x-2\right)+x^2\left(x-2\right)+\left(x+2\right)\left(x-2\right)+x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3+x^2+x+2+x\right)\)
\(=\left(x-2\right)\left[\left(x^3+2x\right)+\left(x^2+2\right)\right]\)
\(=\left(x-2\right)\left[x\left(x^2+2\right)+\left(x^2+2\right)\right]\)
\(=\left(x-2\right)\left(x^2+2\right)\left(x+1\right)\)
Ta lại có :
\(2x^4-3x^3+2x^2-6x-4\) ... biến đổi tương tự ta được \(\left(x^2+2\right)\left(x-2\right)\left(2x+1\right)\)
Do đó với \(x\ne2;x\ne\frac{1}{2}\) thì \(P=\frac{\left(x^2+2\right)\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x^2+2\right)\left(2x+1\right)}=\frac{x+1}{2x+1}\) ( = 1/2 )
Cảm ơn Let Hate Him nha! Nhưng bạn có thể biến đổi nốt phần sau giúp mình được không?
ĐKXĐ: \(x\ne\pm3\)
\(P=\left[\dfrac{x\left(x+3\right)}{x^2\left(x+3\right)+9\left(x+3\right)}+\dfrac{3}{x^2+9}\right]:\left[\dfrac{1}{x-3}-\dfrac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right]\)
\(=\left[\dfrac{x\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}+\dfrac{3}{x^2+9}\right]:\left[\dfrac{1}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\)
\(=\dfrac{x+3}{x^2+9}:\dfrac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}=\dfrac{x+3}{x^2+9}.\dfrac{\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)^2}=\dfrac{x+3}{x-3}\)
Ý 2 mình k hiểu ý bạn lắm
\(P=\dfrac{x+3}{x-3}=\dfrac{x-3+6}{x-3}=1+\dfrac{6}{x-3}\in Z\)
\(\Leftrightarrow\left(x-3\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
Kết hợp vs ĐKXĐ \(\Rightarrow x\in\left\{0;1;2;4;5;6;9\right\}\)
\(a,P=\left[\dfrac{x+1}{3x\left(x+1\right)}-\dfrac{2x-1}{3x\left(2x-1\right)}-1\right]\cdot\dfrac{2x}{1-x}\left(x\ne1;x\ne-1;x\ne0\right)\\ P=\left(\dfrac{1}{3x}-\dfrac{1}{3x}-1\right)\cdot\dfrac{2x}{1-x}\\ P=-1\cdot\dfrac{2x}{1-x}=\dfrac{2x}{x-1}\\ b,P=2+\dfrac{2}{x-1}\in Z\\ \Leftrightarrow x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{2;3\right\}\left(x\ne-1;x\ne0\right)\\ c,P\le1\Leftrightarrow\dfrac{2x}{x-1}-1\le0\\ \Leftrightarrow\dfrac{x+1}{x-1}\le0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\le0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\ge0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-1\le x< 1\)
a: \(P=\left(\dfrac{x+1}{3x\left(x+1\right)}-\dfrac{2x-1}{3x\left(2x-1\right)}-1\right)\cdot\dfrac{2x}{x-1}\)
\(=\dfrac{1-1-3x}{3x}\cdot\dfrac{2x}{x-1}\)
\(=\dfrac{-3x}{3x}\cdot\dfrac{2x}{x-1}=\dfrac{-2x}{x-1}\)
Câu 3 :
\(a,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\) ĐKXđ : \(x\ne\pm1\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{5\left(x-1\right)}\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{10}{x+1}\)
\(B=\left(\frac{x}{3x-9}+\frac{2x-3}{3x-x^2}\right).\frac{3x^2-9x}{x^2-6x+9}.\)
ĐKXđ : \(x\ne0;x\ne3\)
\(B=\left(\frac{x}{3\left(x-3\right)}+\frac{2x-3}{x\left(3-x\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\left(\frac{x^2}{3x\left(x-3\right)}+\frac{9-6x}{3x\left(x-3\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\frac{x^2-6x+9}{3x\left(x-3\right)}.\frac{3x\left(x-3\right)}{x^2-6x+9}=1\)