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2 tháng 10 2020
a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne2\\x\ne-4\end{cases}}\)
\(A=\frac{3}{x+4}-\frac{x\left(x-1\right)}{x+4}\times\frac{2x-5}{x\left(x-2\right)\left(x+4\right)}-\frac{17}{\left(x+4\right)^2}\)
\(=\frac{3\left(x+4\right)}{\left(x+4\right)^2}-\frac{x\left(x-1\right)\left(2x-5\right)}{\left(x+4\right)x\left(x-2\right)\left(x+4\right)}-\frac{17}{\left(x+4\right)^2}\)
\(=\frac{3x+12}{\left(x+4\right)^2}-\frac{\left(x-1\right)\left(2x-5\right)}{\left(x+4\right)^2\left(x-2\right)}-\frac{17}{\left(x+4\right)^2}\)
\(=\frac{\left(3x+12\right)\left(x-2\right)}{\left(x+4\right)^2\left(x-2\right)}-\frac{2x^2-7x+5}{\left(x+4\right)^2\left(x-2\right)}-\frac{17\left(x-2\right)}{\left(x+4\right)^2\left(x-2\right)}\)
\(=\frac{3x^2+6x-24-2x^2+7x-5-17x+34}{\left(x+4\right)^2\left(x-2\right)}\)
\(=\frac{x^2-4x+5}{\left(x+4\right)^2\left(x-2\right)}=\frac{x^2-4x+5}{x^3+6x^2-32}\)
b) \(18A=1\)
<=> \(18\times\frac{x^2-4x+5}{x^3+6x^2-32}=1\)( ĐK : \(\hept{\begin{cases}x\ne0\\x\ne2\\x\ne-4\end{cases}}\))
<=> \(\frac{x^2-4x+5}{x^3+6x^2-32}=\frac{1}{18}\)
<=> 18( x2 - 4x + 5 ) = x3 + 6x2 - 32
<=> 18x2 - 72x + 90 = x3 + 6x2 - 32
<=> x3 + 6x2 - 32 - 18x2 + 72x - 90 = 0
<=> x3 - 12x2 + 72x - 122 = 0
Rồi đến đây chịu á :)
HN
19 tháng 2 2019
a) \(-ĐKXĐ:x\ne\pm2;1\)
Rút gọn : \(A=\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)
\(=\left(\frac{1}{x+2}+\frac{-2}{x-2}+\frac{x}{x^2-4}\right).\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\left[\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{\left(-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{\left(x-2\right)\left(x+2\right)}\right]\)\(.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\left[\frac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\right].\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)\(=\frac{x+1}{\left(x+2\right)^2}\)
b) \(A>0\Leftrightarrow\frac{x+1}{\left(x+2\right)^2}>0\Leftrightarrow\orbr{\begin{cases}x+1< 0;\left(x+2\right)^2< 0\left(voly\right)\\x+1>0;\left(x+2\right)^2>0\end{cases}}\)
\(\Leftrightarrow x>1;x>-2\Leftrightarrow x>1\)
Vậy với mọi x thỏa mãn x>1 thì A > 0
c) Ta có : \(x^2+3x+2=0\Leftrightarrow x^2+x+2x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
Vậy x = -1;-2
YN
25 tháng 11 2021
Answer:
a) ĐKXĐ: \(\hept{\begin{cases}x\ne0\\x\ne-1\end{cases}}\)
\(A=\left(\frac{1}{x}+\frac{x}{x+1}\right):\frac{x}{x^2+x}\)
\(=\frac{x+1+x.x}{x.\left(x+1\right)}.\frac{x.\left(x+1\right)}{x}\)
\(=\frac{x^2+x+1}{x}\)
b) \(x=4\)
\(\Rightarrow A=\frac{4^2+4+1}{4}=\frac{21}{4}\)
c) \(A=3\)
\(\Rightarrow\frac{x^2+x+1}{x}=3\)
\(\Rightarrow x^2+x+1=3x\)
\(\Rightarrow x^2-2x+1=0\)
\(\Rightarrow\left(x-1\right)^2=0\)
\(\Rightarrow x=1\left(TMĐK\right)\)
MN
19 tháng 2 2020
\(ĐKXĐ:x\ne1\)
a) \(A=\frac{2\left(x+1\right)}{x^2+x+1}+\frac{2x^2-9x+4}{x^3-1}+\frac{1}{x-1}\)
\(\Leftrightarrow A=\frac{2\left(x+1\right)\left(x-1\right)+2x^2-9x+4+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow A=\frac{2\left(x^2-1\right)+3x^2-8x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow A=\frac{2x^2-2+3x^2-8x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow A=\frac{5x^2-8x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow A=\frac{\left(5x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow A=\frac{5x-3}{x^2+x+1}\)
b) Để \(A=1\)
\(\Leftrightarrow5x-3=x^2+x+1\)
\(\Leftrightarrow x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy để \(A=1\Leftrightarrow x=2\)