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Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
a.
ĐKXĐ: \(x\ne\pm4\)
\(C=\left(\dfrac{4\left(x+4\right)-4\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\right)\cdot\dfrac{\left(x+4\right)^2}{32}\) có lẽ là nhân
\(\dfrac{4x+16-4x+16}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{\left(x+4\right)^2}{32}\)
\(=\dfrac{32}{\left(x+4\right)\left(x-4\right)}\cdot\dfrac{\left(x+4\right)^2}{32}=\dfrac{x+4}{x-4}\)
b.
\(C=1\Leftrightarrow x+4=x-4\Leftrightarrow0=-8\left(vo-li\right)\)
c.
\(C=\dfrac{1}{3}\Leftrightarrow3\left(x+4\right)=x-4\Leftrightarrow2x=-16\Leftrightarrow x=-8\)
d.
\(C>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+4>0\\x-4>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+4< 0\\x-4< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>4\\x< -4\end{matrix}\right.\)
Luân Đàotran nguyen bao quanDƯƠNG PHAN KHÁNH DƯƠNG
KHUÊ VŨNguyễn Huy TúAkai HarumaAce LegonaNguyễn Thanh HằngMashiro Shiina giúp mk vs
a: \(A=\dfrac{2x-5+x^2-4+x^2-9}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2+2x-18}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x+6}{x-3}\)
b: Để A/2=x+3/x-3 là số nguyên thì \(x-3+6⋮x-3\)
=>\(x-3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(x\in\left\{4;51;6;0;9;-3\right\}\)
c: Để A=1/x-1 thì \(\dfrac{2x+6}{x-3}=\dfrac{1}{x-1}\)
=>2x^2-2x+6x-6=x-3
=>2x^2+5x-6-x+3=0
=>2x^2+4x-3=0
hay \(x=\dfrac{-2\pm\sqrt{10}}{2}\)
a) P xác định \(\Leftrightarrow\hept{\begin{cases}2x+10\ne0\\x\ne0\\2x\left(x+5\right)\ne0\end{cases}\Leftrightarrow x\ne\left\{-5;0\right\}}\)
b) \(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^2\left(x+2\right)}{2x\left(x+5\right)}+\frac{2\left(x-5\right)\left(x+5\right)}{2x\left(x+5\right)}+\frac{5\left(10-x\right)}{2x\left(x+5\right)}\)
\(P=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+5x^2-x^2-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^2\left(x+5\right)-x\left(x+5\right)}{2x\left(x+5\right)}\)
\(P=\frac{\left(x+5\right)\left(x^2-x\right)}{2x\left(x+5\right)}\)
\(P=\frac{x\left(x-1\right)}{2x}\)
\(P=\frac{x-1}{2}\)
c) Để P = 0 thì \(x-1=0\Leftrightarrow x=1\)( thỏa mãn ĐKXĐ )
Để P = 1/4 thì \(\frac{x-1}{2}=\frac{1}{4}\)
\(\Leftrightarrow4\left(x-1\right)=2\)
\(\Leftrightarrow4x-4=2\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\frac{3}{2}\)( thỏa mãn ĐKXĐ )
d) Để P > 0 thì \(\frac{x-1}{2}>0\)
Mà 2 > 0, do đó để P > 0 thì \(x-1>0\Leftrightarrow x>1\)
Để P < 0 thì \(\frac{x-1}{2}< 0\)
Mà 2 > 0, do đó để P < 0 thì \(x-1< 0\Leftrightarrow x< 1\)
a, P = \(\left(\dfrac{1}{x-2}-\dfrac{2x}{4-x^2}+\dfrac{1}{2+x}\right)\left(\dfrac{2}{x}-1\right)\)
P = (\(\dfrac{-1}{2-x}-\dfrac{2x}{\left(2+x\right)\left(2-x\right)}+\dfrac{1}{2+x}\)) ( \(\dfrac{2-x}{x}\))
P = ( \(\dfrac{-2-x-2x+2-x}{\left(2+x\right)\left(2-x\right)}\)) ( \(\dfrac{2-x}{x}\) )
P = (\(\dfrac{-4x}{\left(2+x\right)\left(2-x\right)}\)) ( \(\dfrac{2-x}{x}\))
P = \(\dfrac{-4x}{x\left(2+x\right)}\)
P = \(\dfrac{-4}{2+x}\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne-3\end{cases}}\)
b) \(P=1+\frac{x+3}{x^2+5x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{\left(x-2\right)\left(x+2\right)}{6\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{x-2}{6}\)
\(\Leftrightarrow P=\frac{x+4}{6}\)
c) Để P = 0
\(\Leftrightarrow\frac{x+4}{6}=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Để P = 1
\(\Leftrightarrow\frac{x+4}{6}=1\)
\(\Leftrightarrow x+4=6\)
\(\Leftrightarrow x=2\)
d) Để P > 0
\(\Leftrightarrow\frac{x+4}{6}>0\)
\(\Leftrightarrow x+4>0\)(Vì 6>0)
\(\Leftrightarrow x>-4\)
a: \(C=\dfrac{5x+1+\left(2x-1\right)\left(x-1\right)+2x^2+2x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{2x^2+7x+3+2x^2-2x-x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{4}{x-1}\)
b: x=4 thì C=4/(4-1)=4/3
Khi x=-4 thì C=4/(-4-1)=-4/5
c: C>0
=>x-1>0
=>x>1
camon ạaaaa<3