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a) \(ĐKXĐ:x\ne\pm4;x\ne-2\)
\(P=\left(\frac{8}{x^2-16}+\frac{1}{x+4}\right):\frac{1}{x^2-2x-8}\)
\(\Leftrightarrow P=\left(\frac{8}{\left(x-4\right)\left(x+4\right)}+\frac{1}{x+4}\right):\frac{1}{\left(x-4\right)\left(x+2\right)}\)
\(\Leftrightarrow P=\frac{8+x-4}{\left(x-4\right)\left(x+4\right)}:\frac{1}{\left(x-4\right)\left(x+2\right)}\)
\(\Leftrightarrow P=\frac{x+4}{\left(x-4\right)\left(x+4\right)}:\frac{1}{\left(x-4\right)\left(x+2\right)}\)
\(\Leftrightarrow P=\frac{1}{x-4}.\left(x-4\right)\left(x+2\right)\)
\(\Leftrightarrow P=\frac{\left(x-4\right)\left(x+2\right)}{\left(x-4\right)}\)
\(P=x+2\)
b) Ta có :
\(x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}P=x+2=5+2=7\\P=x+2=4+2=6\end{cases}}\)
Vậy \(P\in\left\{7;6\right\}\)
\(P=\left(\frac{8}{\left(x+4\right)\left(x-4\right)}+\frac{1}{x+4}\right):\frac{1}{x^2-2x-8}\)
\(P=\left(\frac{8}{\left(x+4\right)\left(x-4\right)}+\frac{x-4}{\left(x-4\right)\left(x+4\right)}\right)\cdot\frac{x^2-2x-8}{1}\)
\(P=\left(\frac{x+4}{\left(x+4\right)\left(x-4\right)}\right)\cdot x^2-2x-8\)
\(P=\frac{1}{x-4}\cdot x^2-2x-8\)
P\(P=\frac{x^2+2x-4x+8}{x-4}\)
\(P=\frac{x\left(x+2\right)-4\left(x+2\right)}{x-4}\)
\(P=\frac{\left(x-4\right)\left(x+2\right)}{x-4}\)
\(P=x+2\)
2 ,\(x^2-9x+20=0\)
\(\Rightarrow x^2-4x-5x+20=0\)
\(\Rightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
\(\orbr{\begin{cases}x=5\Rightarrow\\x=4\Rightarrow\end{cases}}\orbr{\begin{cases}P=7\\P=6\end{cases}}\)
a/ ĐKXĐ ....
A=\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}\)
=\(\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+...+\frac{1}{x-5}-\frac{1}{x-4}\)
=\(\frac{1}{x}-\frac{1}{x-5}\)
=\(-\frac{5}{x^2-5x}\)
b/ \(x^3-x+2=0\Leftrightarrow\left(x+1\right)\left(\left(x-1\right)^2+1\right)=0\)
<=> x=-1, thay vào tính nốt
Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
a ) Gọi \(A=\dfrac{3x^2-x}{9x^2-6x+1}\)
Ta có : \(A=\dfrac{x\left(3x-1\right)}{\left(3x\right)^2-2.3x.1+1}=\dfrac{x\left(3x-1\right)}{\left(3x-1\right)^2}=\dfrac{x}{3x-1}\)
Thay x = - 8 và biểu thức A ta được :
\(A=\dfrac{-8}{3.\left(-8\right)-1}=\dfrac{8}{25}\)
Vậy giá trị của biểu thức A là \(\dfrac{8}{25}\) tại x = - 8
b ) Gọi \(B=\dfrac{x^2+3x+2}{x^3+2x^2-x-2}\)
Ta có \(B=\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{x^2\left(x+2\right)-\left(x+2\right)}=\dfrac{x\left(x+1\right)+2\left(x+1\right)}{\left(x^2-1\right)\left(x+2\right)}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}=\dfrac{1}{x-1}\)
Thay x = 1000001 và biểu thức B ta được :
\(B=\dfrac{1}{1000001-1}=\dfrac{1}{100000}\)
Vậy giá trị của biểu thức B là \(\dfrac{1}{1000000}\) tại x = 1000001
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 \(x=-2\)
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..
Ta có :
\(P=\left(\dfrac{8}{x^2-16}+\dfrac{1}{x+4}\right):\dfrac{1}{x^2-2x-8}\)
\(P=\left(\dfrac{8+x-4}{\left(x+4\right)\left(x-4\right)}\right):\dfrac{1}{\left(x+2\right)\left(x-4\right)}\)
\(P=\dfrac{x+4}{\left(x+4\right)\left(x-4\right)}:\dfrac{1}{\left(x+2\right)\left(x-4\right)}\)
\(P=\dfrac{1}{x-4}.\left(x+2\right)\left(x-4\right)\)
\(P=\dfrac{\left(x+2\right)\left(x-4\right)}{\left(x-4\right)}\)
\(P=x+2\)
2 . Ta có :
\(x^2-9x+20=0\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
Thay \(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\) vào biểu thức \(P=x+2\) ta được :
\(\left[{}\begin{matrix}4+2=6\\5+2=7\end{matrix}\right.\)
Kết luận __________________________________
ĐKXĐ của phân thức là : \(\left\{{}\begin{matrix}x^2-16\ne0\\x+4\ne0\\x^2-2x-8\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-4\right)\left(x+4\right)\ne0\\x\ne-4\\\left(x-4\right)\left(x+2\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne4\\x\ne-4\\x\ne-2\end{matrix}\right.\)
\(P=\left(\dfrac{8}{x^2-16}+\dfrac{1}{x+4}\right):\dfrac{1}{x^2-2x-8}\) \(=\left(\dfrac{8}{\left(x-4\right)\left(x+4\right)}+\dfrac{1}{x+4}\right).\left(x^2-2x-8\right)\) \(=\dfrac{8+x-4}{\left(x-4\right)\left(x+4\right)}.\left(x^2-4x+2x-8\right)\) \(=\dfrac{x+4}{\left(x-4\right)\left(x+4\right)}.\left(x-4\right)\left(x+2\right)\) \(=x+2\) + Tính giá trị của P tại x2 - 9x + 20 = 0 \(x^2-9x+20=0\) \(\Rightarrow x^2-4x-5x+20=0\) \(\Rightarrow\left(x^2-4x\right)-\left(5x-20\right)=0\) \(\Rightarrow x\left(x-4\right)-5\left(x-4\right)=0\) \(\Rightarrow\left(x-4\right)\left(x-5\right)=0\) \(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\left(kot/m\right)\\x=5\left(t/m\right)\end{matrix}\right.\) Thay x = 5 vào biểu thức P ,có : \(5+2=7\) Vậy tại x= 5 giá trị của P là 7