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\(P=\left(\frac{8}{\left(x+4\right)\left(x-4\right)}+\frac{1}{x+4}\right):\frac{1}{x^2-2x-8}\)
\(P=\left(\frac{8}{\left(x+4\right)\left(x-4\right)}+\frac{x-4}{\left(x-4\right)\left(x+4\right)}\right)\cdot\frac{x^2-2x-8}{1}\)
\(P=\left(\frac{x+4}{\left(x+4\right)\left(x-4\right)}\right)\cdot x^2-2x-8\)
\(P=\frac{1}{x-4}\cdot x^2-2x-8\)
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 : x khác -4;4;-2
P =[ 8+x-4/(x-4).(x+4) ] : 1/(x+2).(x-4)
= x+4/(x+4).(x-4) . (x+2).(x-4)
= x+2
b, x^2-9x+20 = 0
<=> (x^2-4x)-(5x-20)=0
<=> (x-4).(x-5)=0
<=> x-4=0 hoặc x-5=0
<=> x=4 hoặc x=5
+, Với x=4 thì P = 4+2 = 6
+, Với x=5 thì P = 5+2 = 7
k mk nha
\(\text{Giải}\)
\(A=\left(\frac{x+2}{2x-4}-\frac{2-x}{2x+4}+\frac{32}{4x^2-16}\right):\frac{x-1}{x-2}\)
\(A=\left(\frac{x+2}{2x-4}-\frac{2-x}{2x+4}+\frac{32}{\left(2x-4\right)\left(2x+4\right)}\right):\frac{x-1}{x-2}\)
\(A=\left(\frac{\left(x+2\right)\left(2x+4\right)}{\left(2x-4\right)\left(2x+4\right)}-\frac{\left(2-x\right)\left(2x-4\right)}{\left(2x-4\right)\left(2x+4\right)}+\frac{32}{\left(2x-4\right)\left(2x+4\right)}\right):\frac{x-1}{x-2}\)
\(A=\left(\frac{2x^2+8x+8}{\left(2x-4\right)\left(2x+4\right)}-\frac{4x^2-8+4x}{\left(2x-4\right)\left(2x+4\right)}+\frac{32}{\left(2x-4\right)\left(2x+4\right)}\right):\frac{x-1}{x-2}\)
\(A=\frac{2x^2+8x+8-4x^2+8-4x+32}{\left(2x-4\right)\left(2x+4\right)}:\frac{x-1}{x-2}\)
\(A=\frac{4x-2x^2+48}{\left(2x-4\right)\left(2x+4\right)}:\frac{x-1}{x-2}\)
\(A=\frac{2\left(2x-x^2+24\right)}{\left(2x-4\right)\left(2x+4\right)}:\frac{x-1}{x-2}=\frac{2\left(2x-x^2+24\right)\left(x-2\right)}{\left(2x-4\right)\left(2x+4\right)\left(x-1\right)}\)
\(=\frac{2\left(2x-x^2+24\right)\left(x-2\right)}{4\left(x-2\right)\left(x+2\right)\left(x-1\right)}=\frac{2x-x^2+24}{\left(x-2\right)\left(x-1\right)}\)
c, Bạn tự giải hệ pt nhé :)
bài1 A=\(\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
=\(\left(-\frac{x-3\cdot\left(x+3\right)^2}{\left(x+3\right)^2\cdot\left(x-3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
=\(-\frac{x}{x+3}\cdot\frac{x+3}{3x^2}=\frac{-1}{3x}\)
b) thế \(x=-\frac{1}{2}\)vào biểu thức A
\(-\frac{1}{3\cdot\left(-\frac{1}{2}\right)}=\frac{2}{3}\)
c) A=\(-\frac{1}{3x}< 0\)
VÌ (-1) <0 nên 3x>0
x >0
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 \(P\in\left\{7;6\right\}\)