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\(A=\dfrac{6x^2+8x+7+x^2-x-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x-1}\)
Để 4A=x-1 thì \(\dfrac{4}{x-1}=x-1\)
=>x-1=2 hoặc x-1=-2
=>x=3(loại) hoặc x=-1(nhận)
\(A=\frac{6x^2+8x+7}{x^3-1}+\frac{x}{x^2+x+1}+\frac{6}{1-x}\)
<=>\(A=\frac{6x^2+8x+7}{x^3-1}+\frac{\left(x-1\right)x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(-6\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
<=>\(A=\frac{6x^2+8x+7}{x^3-1}+\frac{x^2-x}{x^3-1}+\frac{-6x^2-6x-6}{x^3-1}\)
<=>\(A=\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)<=>\(A=\frac{1}{x-1}\)<=>\(4A=\frac{4}{x-1}\)
Theo đề bài 4A=x-1 => \(4A=\frac{4}{x-1}=x-1\Rightarrow\left(x-1\right)^2=4\Rightarrow\orbr{\begin{cases}x-1=-2\\x-1=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vì x<0 nên x=-1
\(P=\left(\frac{8}{\left(x+4\right)\left(x-4\right)}+\frac{1}{x+4}\right):\frac{1}{x^2-2x-8}\)
\(P=\left(\frac{8}{\left(x+4\right)\left(x-4\right)}+\frac{x-4}{\left(x-4\right)\left(x+4\right)}\right)\cdot\frac{x^2-2x-8}{1}\)
\(P=\left(\frac{x+4}{\left(x+4\right)\left(x-4\right)}\right)\cdot x^2-2x-8\)
\(P=\frac{1}{x-4}\cdot x^2-2x-8\)
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 ,\(x^2-9x+20=0\)
\(\Rightarrow x^2-4x-5x+20=0\)
\(\Rightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
\(\orbr{\begin{cases}x=5\Rightarrow\\x=4\Rightarrow\end{cases}}\orbr{\begin{cases}P=7\\P=6\end{cases}}\)
2/ \(\frac{1}{2}x2y5z3=\left(\frac{1}{2}.2.5.3\right)xyz\)\(=15xyz\)
\(\Rightarrow\frac{1}{2}x2y5z3\)có bậc là 3
3/ \(\frac{x}{4}=\frac{9}{x}\Leftrightarrow x^2=9.4\Rightarrow x^2=36\) mà \(x>0\Rightarrow x=6\)
4/ \(\left|2x-\frac{1}{2}\right|+\frac{3}{7}=\frac{38}{7}\Rightarrow\left|2x+\frac{1}{2}\right|=\frac{35}{7}=5\Rightarrow\hept{\begin{cases}2x+\frac{1}{2}=5\Rightarrow2x=\frac{9}{2}\Rightarrow x=\frac{9}{4}\\2x+\frac{1}{2}=-5\Rightarrow2x=\frac{-11}{2}\Rightarrow x=\frac{-11}{4}\end{cases}}\)
\(A=\left(\frac{3-x}{x+3}\times\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\) \(\left(ĐKXĐ:x\ne\pm3\right)\)
\(A=\left(\frac{3-x}{x+3}\times\frac{x+3}{x-3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{3-x}{x-3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left[\frac{\left(3-x\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right]:\frac{3x^2}{x+3}\)
\(A=\left(\frac{9-3x}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)
\(A=\frac{-3}{x+3}\times\frac{x+3}{3x^2}\)
\(A=\frac{-1}{x^2}\)
Ta có :\(x^2+x-6=0\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\left(L\right)\\x=2\left(tm\right)\end{cases}}\)
\(\Rightarrow A=\frac{-1}{2^2}\)
\(A=\frac{-1}{4}\)
a)Với x \(\ne\)-1
Ta có: x2 + x = 0
=> x(x + 1) = 0
=> \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-1\left(ktm\right)\end{cases}}\)
Với x = 0 => A = \(\frac{0-3}{0+1}=-3\)
b) Ta có: B = \(\frac{3}{x-3}+\frac{6x}{9-x^3}+\frac{x}{x+3}\)
B = \(\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
B = \(\frac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
B = \(\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)
B = \(\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\)
B = \(\frac{x+3}{x-3}\)
c) Với x \(\ne\)\(\pm\)3; x \(\ne\)-1
Ta có: P = AB = \(\frac{x-3}{x+1}\cdot\frac{x+3}{x-3}=\frac{x+3}{x+1}=\frac{\left(x+1\right)+2}{x+1}=1+\frac{2}{x+1}\)
Để P \(\in\)Z <=> 2 \(⋮\)x + 1
<=> x + 1 \(\in\)Ư(2) = {1; -1; 2; -2}
<=> x \(\in\){0; -2; 1; -3}
1, \(=\left[\frac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}-x\right]:\frac{1-x^2}{\left(1-x\right)-x^2\left(1-x\right)}\)
\(=\left(1+x+x^2-x\right):\frac{1-x^2}{\left(1-x\right)\left(1-x^2\right)}\)\(=\left(x^2+1\right)\left(1-x\right)\)
2, để B<0 <=> (x2+1)(1-x)<0
vì x^2+1 > 0 với mọi x
=> \(\hept{\begin{cases}x^2+1>0\\1-x< 0\end{cases}\Leftrightarrow x>1}\)
3, \(\left|x-4\right|=5\Leftrightarrow\orbr{\begin{cases}x=9\\x=-1\left(loại\right)\end{cases}}\)
Thay x=9 vào B ta có: B=(92+1)(1-9)=82.(-8)=-656