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A\(\dfrac{x^3-9x}{2x+3}\left(\dfrac{x+3}{x^2-3x}-\dfrac{x}{x^2-9}\right)=\dfrac{x\left(x^2-9\right)}{2x+3}\left(\dfrac{x+3}{x\left(x-3\right)}-\dfrac{x}{\left(x+3\right)\left(x-3\right)}\right)=\dfrac{x\left(x+3\right)\left(x-3\right)}{2x+3}\left(\dfrac{\left(x+3\right)\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}-\dfrac{x.x}{\left(x+3\right)\left(x-3\right)x}\right)=\dfrac{x\left(x+3\right)\left(x-3\right)}{2x+3}\left(\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}-\dfrac{x^2}{\left(x+3\right)\left(x-3\right)x}\right)\)
\(=\dfrac{x\left(x+3\right)\left(x-3\right)}{2x+3}.\dfrac{\left(x+3\right)^2-x^2}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x\left(x+3\right)\left(x-3\right)}{2x+3}.\dfrac{\left(x+3+x\right)\left(x+3-x\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x\left(x+3\right)\left(x-3\right)}{2x+3}.\dfrac{\left(2x+3\right).3}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x\left(x+3\right)\left(x-3\right)\left(2x+3\right).3}{\left(2x+3\right)x\left(x-3\right)\left(x+3\right)}=3\)
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)
a: =2x^2+6x-2x^2+x
=7x
b: =2x^2-3x-2x+3-x^2+4x-4
=x^2-x-1
c: \(=9x^2-6x+1+2x^2-x+6x-3=11x^2-x-2\)
d: \(=x^3+2x^2-x-2-x^3+8=2x^2-x+6\)
1.
a, \(\left(x+3\right)\left(x-3\right)-\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3-x+3\right)\)
\(=9\left(x-3\right)=9x-27\)
b, \(\left(2x+1\right)^2+2\left(2x+1\right)\left(x-1\right)+\left(x-1\right)^2\)
\(=\left(2x+1+x-1\right)^2=9x^2\)
c, \(x\left(x-3\right)\left(x+3\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-9\right)-\left(x^4-1\right)\)
\(=x^3-9x-x^4+1=-x^4+x^3-9x+1\)
bài 2a)
\(\left(x+2\right)^2-x^2-3x=21\)
\(\left(x+2-x\right).\left(x+2+x\right)-3x=21\)
\(\left(x-x+2\right).\left(x+x+2\right)-3x=21\)
\(2.\left(2x+2\right)-3x=21\)
\(4x+2.2-3x=21\)
\(4x+4-3x=21\)
\(4x-3x=21-4\)
\(x=17\)
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
a) \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{9x^2-6x+1}\)
\(=-\dfrac{9x^2+3x+2x-6x^2}{\left(3x-1\right)\left(3x+1\right)}.\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(=-\dfrac{x\left(3x+5\right)}{\left(3x-1\right)^2}.\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{-1}{2}\)
b) \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left(\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{3x-9-x^2}{3x\left(x+3\right)}\right)\)
\(=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\dfrac{3x\left(x+3\right)}{-x^2+3x-9}\)
\(=\dfrac{x^2-3x+9}{x-3}.\dfrac{3}{-\left(x^2-3x+9\right)}\)
\(=-\dfrac{3}{x-3}\)
1)
ĐKXĐ: x\(\ne\)3
ta có :
\(\frac{x^2-6x+9}{2x-6}=\frac{\left(x-3\right)^2}{2\left(x-3\right)}=\frac{x-3}{2}\)
để biểu thức A có giá trị = 1
thì :\(\frac{x-3}{2}\)=1
=>x-3 =2
=>x=5(thoả mãn điều kiện xác định)
vậy để biểu thức A có giá trị = 1 thì x=5
1)
\(A=\frac{x^2-6x+9}{2x-6}\)
A xác định
\(\Leftrightarrow2x-6\ne0\)
\(\Leftrightarrow2x\ne6\)
\(\Leftrightarrow x\ne3\)
Để A = 1
\(\Leftrightarrow x^2-6x+9=2x-6\)
\(\Leftrightarrow x^2-6x-2x=-6-9\)
\(\Leftrightarrow x^2-8x=-15\)
\(\Leftrightarrow x=3\) (loại vì không thỏa mãn ĐKXĐ)
