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a)
A = \(\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
=\(\left(\dfrac{3-x}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{x}{x+3}\right):\dfrac{3x^3}{x+3}\) (đkxđ: x \(\ne\)\(\pm\)3)
= \(\left(\dfrac{x}{x+3}-1\right).\dfrac{x+3}{3x^2}\)
= \(\dfrac{x-x-3}{x+3}.\dfrac{x+3}{3x^2}\)
= -x2
b) Thay x = \(\dfrac{1}{2}\) vào A, ta có:
A = -\(\left(\dfrac{1}{2}\right)^2\)
= -\(\dfrac{1}{4}\)
c) Để A < 0 thì -x2 < 0
mà -x2 \(\le\) 0 \(\forall\)x
\(\Rightarrow\) Với mọi x (x\(\ne\)0) thì A < 0
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) A = \(\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\) ( x # 0 ; x # 3 ; x# - 3)
A = \(\left(\dfrac{-\left(x-3\right)}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A = \(\left(\dfrac{-x-3}{x+3}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A = \(\dfrac{-3}{x+3}.\dfrac{x+3}{3x^2}=\dfrac{-1}{x^2}\)
b) Với x = \(\dfrac{-1}{2}\) , ta có :
A = \(\dfrac{-1}{x^2}=\dfrac{-1}{\left(\dfrac{-1}{2}\right)^2}=-4\)
c) Để A < 0
⇔ \(\dfrac{-1}{x^2}< 0\)
⇔ x2 > 0 ( luôn đúng ∀x # 0)
KL...
a) A \(=\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
\(\)\(=\left(\dfrac{9-x^2}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
\(=\dfrac{-3}{x+3}:\dfrac{3x^2}{x+3}\)
\(=\dfrac{-1}{x^2}\)
b) \(x=\dfrac{-1}{2}\) (Thỏa mãn ĐKXĐ \(x\ne3;x\ne-3\) )
Thay \(x=\dfrac{-1}{2}\) vào biểu thức A, ta có:
\(A=\dfrac{-1}{\left(\dfrac{-1}{2}\right)^2}=-4\)
Vậy với \(x=\dfrac{-1}{2}\) giá trị của biểu thức A = -4.
c) \(\dfrac{-1}{x^2}< 0\)
\(\Rightarrow x^2>0\) (Luôn đúng)
Vậy với mọi giá trị của \(x\) để A < 0
\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{x^3+1}\)
\(A=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{3-3x}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x^3-x^2+x-3-3x+x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{1}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{1}{x^3+1}\)
1/ a, \(A=\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
Vậy \(A=x\)
b/ Khi \(x=\dfrac{1}{2}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{2}}=2\)
Vậy...
2/a,
\(A=\dfrac{5x+2}{3x^2+2x}+\dfrac{-2}{3x+2}\)
\(=\dfrac{5x+2}{x\left(3x+2\right)}-\dfrac{2x}{x\left(3x+2\right)}\)
\(=\dfrac{5x+2-2x}{x\left(3x+2\right)}\)
\(=\dfrac{3x+2}{x\left(3x+2\right)}\)
\(=\dfrac{1}{x}\)
Vậy....
b/ Với \(x=\dfrac{1}{3}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{3}}=3\)
Vậy..
a) \(A=\left(\dfrac{3-x}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\right).\dfrac{x+3}{3x^2}\)
\(\Leftrightarrow A=\left(\dfrac{-\left(x-3\right)}{x+3}.\dfrac{\left(x+3\right)}{x-3}\right).\dfrac{x+3}{3x^2}\)
\(\Leftrightarrow A=\dfrac{-x-3}{3x^2}\)
b) Khi \(x=\dfrac{2}{3}\Rightarrow\) \(A=\dfrac{-\dfrac{2}{3}-3}{3.\left(\dfrac{2}{3}\right)^2}=\dfrac{-11}{4}\)
c) Để A < 0 thì
\(\dfrac{-x-3}{3x^2}< 0\)
=> -x -3 <0
<=> -x < 3
\(\Rightarrow x>3\)