Đcm học ngu k biết xài caskov

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)

\(\Leftrightarrow x>-4\)

\(a.P=1+\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)=\dfrac{x+3}{x+2}:\left(\dfrac{2}{x-2}-\dfrac{3}{x^2-4}-\dfrac{1}{x+2}\right)=\dfrac{x+3}{x+2}.\dfrac{\left(x+2\right)\left(x-2\right)}{2x+4-3-x+2}=\left(x+3\right).\dfrac{x-2}{x+3}=x-2\left(x\ne\pm2;x\ne-3\right)\)

\(b.P=0\Leftrightarrow x-2=0\Leftrightarrow x=2\left(KTM\right)\)

\(P=1\Leftrightarrow x-2=1\Leftrightarrow x=3\left(TM\right)\)

\(c.P>0\Leftrightarrow x-2>0\Leftrightarrow x>2\)

a: ĐKXĐ: x<>0; x<>-2; x<>2; x<>-3

b: \(P=1+\dfrac{1}{x+2}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{1}{x+2}:\left(\dfrac{2}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{1}{x+2}:\dfrac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)

\(=1+\dfrac{1}{x+2}\cdot\dfrac{\left(x+2\right)\left(x-2\right)}{6}=1+\dfrac{x-2}{6}=\dfrac{6+x-2}{6}=\dfrac{x-4}{6}\)

c: Để P=0 thì x-4=0

=>x=4(nhận)

Khi P=1 thì x-4=6

=>x=10

d Để P>0 thì x-4>0

=>x>4

\(P=1+\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}+\dfrac{3x}{12-3x^2}-\dfrac{1}{2}\right)\)\(=1+\dfrac{x+3}{x^2+3x+2x+6}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}+\dfrac{3x}{3\left(4-x^2\right)}-\dfrac{1}{x+2}\right)\)\(=1+\dfrac{x+3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{2}{x-2}-\dfrac{3x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\)\(=1+\dfrac{1}{x+2}:\left(\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3x}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right)\)\(=1+\dfrac{1}{x+2}:\left(\dfrac{2x+4-3x-x+2}{\left(x-2\right)\left(x+2\right)}\right)\)

\(=1+\dfrac{1}{x+2}:\left(\dfrac{-2x+6}{\left(x-2\right)\left(x+2\right)}\right)\)

\(=1+\dfrac{1}{x+2}.\dfrac{\left(x-2\right)\left(x+2\right)}{-2x+6}\)

\(=1+\dfrac{x-2}{-2x+6}\)

\(=\dfrac{-2x+6+x-2}{-2x+6}=\dfrac{4-x}{-2\left(x-3\right)}\)

a,\(P=1+\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}+\dfrac{3x}{12-3x^2}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{x+3}{x^2+3x+2x+6}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}+\dfrac{-3x}{3\left(x^2-4\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{1}{x+2}:\left(\dfrac{2}{x-2}+\dfrac{-x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{1}{\left(x+2\right)}:\left(\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{-x}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\)

\(=1+\dfrac{1}{x+2}:\dfrac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)

\(=1+\dfrac{1}{x+2}:\dfrac{6}{\left(x-2\right)\left(x+2\right)}\)

\(=1+\dfrac{1}{x+2}.\dfrac{\left(x-2\right)\left(x+2\right)}{6}=\dfrac{x-2}{6}\)

b, Để P = 0 ⇔ \(\dfrac{x-2}{6}=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Để \(P=1\Leftrightarrow\dfrac{x-2}{6}=1\Leftrightarrow x-2=6\Leftrightarrow x=8\)

c, Để P > 0 \(\Leftrightarrow\dfrac{x-2}{6}>0\Leftrightarrow x-2>6\Leftrightarrow x>8\)

Bn ơi x² + 5x + 6 hay x² + 5x - 6

\(a,\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\left(l\right)\\x=-2\left(l\right)\end{matrix}\right.\Leftrightarrow x\in\varnothing\Leftrightarrow A\in\varnothing\\ b,\text{ý bạn là rút gọn A hả?}\\ A=\dfrac{x-2+2x+3x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{6x+4}{\left(x-2\right)\left(x+2\right)}\)

a, ĐK : \(x\ne1;2;3;4;5\)

b, \(\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\dfrac{1}{x}-\dfrac{1}{x-1}+\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}\)

\(=\dfrac{1}{x}-\dfrac{1}{x-5}=\dfrac{x-5-x}{x\left(x-5\right)}=\dfrac{-5}{x\left(x-5\right)}\)

a: ĐKXĐ: \(x\notin\left\{0;1;2;3;4;5\right\}\)

b: \(P=\dfrac{1}{x^2-x}+\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}\)

\(=\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\dfrac{-1}{x}+\dfrac{1}{x-1}-\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}-\dfrac{1}{x-4}+\dfrac{1}{x-5}\)

\(=\dfrac{1}{x-5}-\dfrac{1}{x}\)

\(=\dfrac{x-\left(x-5\right)}{x\left(x-5\right)}=\dfrac{5}{x\left(x-5\right)}\)

c: \(x^3-x^2+2=0\)

=>\(x^3+x^2-2x^2+2=0\)

=>\(x^2\cdot\left(x+1\right)-2\left(x-1\right)\left(x+1\right)=0\)

=>\(\left(x+1\right)\left(x^2-2x+2\right)=0\)

=>x+1=0

=>x=-1

Khi x=-1 thì \(P=\dfrac{5}{\left(-1\right)\left(-1-5\right)}=\dfrac{5}{\left(-1\right)\cdot\left(-6\right)}=\dfrac{5}{6}\)