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a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)
\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)
câu d
\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)
a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
b) \(\dfrac{\left(a^2-\left(b+c\right)^2\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
\(=\dfrac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(\left(a-c\right)^2-b^2\right)}\)
\(=\dfrac{\left(a-c-b\right)\left(a-c+b\right)}{\left(a-c-b\right)\left(a-c+b\right)}=1\)
c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^3-x\left(x+1\right)\left(x-1\right)+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{x^3-3x^2+3x-1-x^3+x+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{4x-1}{x^3\left(x-1\right)^2}\)
d) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)
\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)
\(=\dfrac{x}{x+y}\)
a) Ta có: \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1}{x\left(x+1\right)}+\dfrac{x+2}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\dfrac{x^2+2x+1}{x\left(x+1\right)}:\dfrac{x^2-2x+1}{x}\)
\(=\dfrac{\left(x+1\right)^2}{x\left(x+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x+1}{\left(x-1\right)^2}\)
b) Ta có: \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
\(=\dfrac{3x\left(3x+1\right)+2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)
\(=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)
\(=\dfrac{3x^2+5x}{\left(1-3x\right)\left(1+3x\right)}\cdot\dfrac{\left(1-3x\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{x\left(3x+5\right)}{1+3x}\cdot\dfrac{1-3x}{2x\left(3x+5\right)}\)
\(=\dfrac{2\left(1-3x\right)}{3x+1}\)
c) Ta có: \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)
\(=\dfrac{x^2-3x+9}{x-3}\cdot\dfrac{3}{-\left(x^2-3x+9\right)}\)
\(=\dfrac{-3}{x-3}\)
1a.
ĐKXĐ: \(x\ne\left\{1;3\right\}\)
\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)
\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)
\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)
b.
ĐKXĐ: \(x\ne\left\{-1;2\right\}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)
\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)
1c.
ĐKXĐ: \(x\ne\left\{2;5\right\}\)
\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)
\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)
\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)
2a.
\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)
\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
2b.
\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)
a) PT \(\Leftrightarrow\dfrac{x^2-x+2}{\left(x-1\right)^3}=\dfrac{A+B\left(x-1\right)+C\left(x-1\right)^2}{\left(x-1\right)^3}\)
\(\Leftrightarrow x^2-x+2=A+Bx-B+Cx^2-2Cx+C\)
\(\Leftrightarrow x^2-x+2=Cx^2+x\left(B-2C\right)+\left(A+C-B\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}C=1\\B-2C=-1\\A+C-B=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A=2\\B=1\\C=1\end{matrix}\right.\)
b: \(\Leftrightarrow\dfrac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{A\cdot x^2+A+\left(Bx+C\right)\left(x-1\right)}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow x^2\cdot A+A+x^2\cdot B-x\cdot B+x\cdot C-C=x^2+2x-1\)
\(\Leftrightarrow x^2\left(A+B\right)+x\left(-B+C\right)+A-C=x^2+2x-1\)
=>A+B=1; -B+C=2; A-C=-1
=>A+C=3; A-C=-1; A+B=1
=>A=1; C=2; B=1-A=0
1: \(MTC=2\left(x-y\right)\left(x+y\right)\)
\(\dfrac{x-y}{2x^2-4xy+2y^2}=\dfrac{x-y}{2\left(x-y\right)^2}=\dfrac{1}{2\left(x-y\right)}=\dfrac{1\cdot\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{2\left(x-y\right)\left(x+y\right)}\)
\(\dfrac{x+y}{2x^2+4xy+2y^2}\)
\(=\dfrac{x+y}{2\left(x^2+2xy+y^2\right)}\)
\(=\dfrac{x+y}{2\left(x+y\right)^2}=\dfrac{1}{2\left(x+y\right)}=\dfrac{x-y}{2\left(x+y\right)\left(x-y\right)}\)
\(\dfrac{1}{x^2-y^2}=\dfrac{2}{2\left(x^2-y^2\right)}=\dfrac{2}{2\left(x-y\right)\left(x+y\right)}\)
2: \(\dfrac{1}{x^2+8x+15}=\dfrac{1}{\left(x+3\right)\left(x+5\right)}=\dfrac{x+3}{\left(x+3\right)^2\cdot\left(x+5\right)}\)
\(\dfrac{1}{x^2+6x+9}=\dfrac{1}{\left(x+3\right)^2}=\dfrac{x+5}{\left(x+3\right)^2\cdot\left(x+5\right)}\)
3: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}=\dfrac{1\cdot\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{a-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(\dfrac{1}{\left(c-b\right)\left(c-a\right)}=\dfrac{1}{\left(b-c\right)\left(a-c\right)}=\dfrac{a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(\dfrac{1}{\left(b-a\right)\left(a-c\right)}=\dfrac{-1}{\left(a-b\right)\left(a-c\right)}=\dfrac{-\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
1 ) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=-15\)
\(\Leftrightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=-15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=-15\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(6x^2-6x^2\right)+\left(12x+12x\right)+\left(27+6-8\right)=-15\)
\(\Leftrightarrow24x+25=-15\)
\(\Leftrightarrow24x=-40\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
Vậy \(x=-\dfrac{5}{3}\)