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b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
Bài 1:
a) \(\dfrac{15xy}{10x^2y}\)
= \(\dfrac{3.5xy}{2.5xyx}\)
= \(\dfrac{3}{2x}\)
d) \(\dfrac{6x\left(x+5\right)^3}{2x^2\left(x+5\right)}\)
= \(\dfrac{3.2x\left(x+5\right)\left(x+5\right)^2}{x.2x\left(x+5\right)}\)
= \(\dfrac{3\left(x+5\right)^2}{x}\)
Lời giải:
a)
\(\frac{x-2}{6x^2-6x}-\frac{1}{4x^2-4}=\frac{x-2}{6x(x-1)}-\frac{1}{4(x^2-1)}=\frac{x-2}{6x(x-1)}-\frac{1}{4(x-1)(x+1)}\)
\(=\frac{2(x+1)(x-2)}{12x(x-1)(x+1)}-\frac{3x}{12x(x-1)(x+1)}=\frac{2(x+1)(x-2)-3x}{12x(x-1)(x+1)}\)
\(=\frac{2x^2-5x-4}{12x(x-1)(x+1)}=\frac{2x^2-5x-4}{12x^3-12x}\)
b) ĐK: \(x\neq \pm 1\)
\(\frac{(x+1)(x^2-2x+1)}{6x^3+6}:\frac{x^2-1}{4x^2-4x+4}\)
\(=\frac{(x+1)(x-1)^2}{6(x^3+1)}.\frac{4x^2-4x+4}{x^2-1}\)
\(=\frac{4(x+1)(x-1)^2(x^2-x+1)}{6(x+1)(x^2-x+1)(x^2-1)}\)
\(=\frac{2(x-1)}{3(x+1)}\)
a,\(\dfrac{3}{x-3}\) - \(\dfrac{6x}{9-x^2}\) + \(\dfrac{x}{x+3}\) (*)
đkxđ: x khác 3, x khác -3
(*) \(\dfrac{3(x+3)}{\left(x-3\right).\left(x+3\right)}\)- \(\dfrac{6x}{\left(x-3\right).\left(x+3\right)}\) + \(\dfrac{x\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}\)
=>3x+9 -6x + x2+3x
<=>x2 + 3x-6x+3x + 9
<=>x2 +9
<=>(x-3).(x+3)
1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)
ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\) \(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )
\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn
a) \(A=\left[\dfrac{x+3}{\left(x-3\right)^2}+\dfrac{6}{x^2-9}-\dfrac{x-3}{\left(x+3\right)^2}\right]\left[1:\left(\dfrac{24x^2}{x^4-81}-\dfrac{12}{x^2+9}\right)\right]\)
\(\left(ĐKXĐ:x\ne\pm3\right)\)
\(=\dfrac{\left(x+3\right)^3+6\left(x-3\right)\left(x+3\right)-\left(x-3\right)^3}{\left(x-3\right)^2\left(x+3\right)^2}\cdot\left[1:\dfrac{24x^2-12\left(x^2-9\right)}{\left(x^2-9\right)\left(x^2+9\right)}\right]\)
\(=\dfrac{x^3+9x^2+27x+27+6x^2-54-x^3+9x^2-27x+27}{\left(x-3\right)^2\left(x+3\right)^2}\cdot\dfrac{\left(x^2-9\right)\left(x^2+9\right)}{24x^2-12x^2+108}\)
\(=\dfrac{24x^2\left(x^2+9\right)\left(x-3\right)\left(x+3\right)}{12\left(x^2+9\right)\left(x-3\right)^2\left(x+3\right)^2}\)
\(=\dfrac{2x^2}{x^2-9}\)
b) \(B=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left[\left(x-2\right)+\dfrac{10-x^2}{x+2}\right]\)
\(=\left(\dfrac{x}{x^2-4}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\left(\dfrac{x-2}{1}+\dfrac{10-x^2}{x+2}\right)\)
\(=\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\)
\(=\dfrac{x-2x-4+x-2}{x^2-4}\cdot\dfrac{x+2}{x^2-4+10-x^2}\)
\(=\dfrac{-6\left(x+2\right)}{6\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{-1}{x-2}\)
phần b điều kiện xác định là \(x\ne\pm2\) nhé
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