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a, ĐKXĐ :\(x\ne3;x\ne-3\)
b, \(P=\frac{3\cdot\left(x-3\right)}{\left(x-3\right)\cdot\left(x+3\right)}+\frac{x+3}{\left(x+3\right)\cdot\left(x-3\right)}+\frac{18}{\left(x+3\right)\cdot\left(x-3\right)}\)
\(=\frac{3x-9+x+3+18}{\left(x+3\right)\cdot\left(x-3\right)}\)\(=\frac{4x+12}{\left(x-3\right)\cdot\left(x+3\right)}\)
\(=\frac{4\cdot\left(x+3\right)}{\left(x+3\right)\cdot\left(x-3\right)}=\frac{4}{x-3}\)
c, Với P = 4 \(\Rightarrow\frac{4}{x-3}=4\Rightarrow4=4\cdot\left(x-3\right)\)\(\Rightarrow1=x-3\Rightarrow x=4\)
a, B=[(x+3)/(x-3)+(2x^2-6)/(9-x^2)+x/(x+3)]:[(6x-12)/(2x^2-18)]
=[(x+3)/(x-3)+ -(2x^2-6)/(x^2-9)+x/(x+3)]:[(6x-12)/(2x^2-18)]
=[(x+3)/(x-3)+ -(2x^2-6)/(x-3)(x+3)+x/(x+3)]:[(6x-12)/2(x-3)(x+3)]
={[(x+3)^2-2x^2+6+x(x-3)]/(x-3)(x+3)}:[6(x-2)/2(x-3)(x+3)]
=(x^2+6x+9-2x^2+6+x^2-3x)/(x-3)(x+3): 6(x-2)/2(x-3)(x+3)
=3x+15/(x-3)(x+3): 6(x-2)/2(x-3)(x+3)
=3(x+5)/(x-3)(x+3): 6(x-2)/2(x-3)(x+3
=3(x+5)/(x-3)(x+3).2(x-3)(x+3)/6(x-2)
=3(x+5).6/(x-2)
=6(x+5)/6(x-2)
=x+5/x-2
b,Ta thay : x=1
=>x+5/x-2=1+5/1-2=-6
Ta thay : x=-3
=>x+5/x-2=-3+5/-3-2=-2/5
c, Ta co : x+5/x-2=0
x+5=(x-2).0
x+5=0
x=-5
Vậy : x=-5
Câu 3 :
\(a,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\) ĐKXđ : \(x\ne\pm1\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{5\left(x-1\right)}\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{10}{x+1}\)
\(B=\left(\frac{x}{3x-9}+\frac{2x-3}{3x-x^2}\right).\frac{3x^2-9x}{x^2-6x+9}.\)
ĐKXđ : \(x\ne0;x\ne3\)
\(B=\left(\frac{x}{3\left(x-3\right)}+\frac{2x-3}{x\left(3-x\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\left(\frac{x^2}{3x\left(x-3\right)}+\frac{9-6x}{3x\left(x-3\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\frac{x^2-6x+9}{3x\left(x-3\right)}.\frac{3x\left(x-3\right)}{x^2-6x+9}=1\)
\(ĐKXĐ:\hept{\begin{cases}x\ne\pm3\\1-\frac{1}{x+3}\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne\pm3\\x\ne-2\end{cases}}}\)
a ) \(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)
\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{x-4}{x-3}-\frac{x-1}{x+3}\right):\left(1-\frac{1}{x+3}\right)\)
\(=\frac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}:\frac{x+3-1}{x+3}\)
\(=\frac{21+x^2-x-12-\left(x^2-4x+3\right)}{\left(x-3\right)\left(x+3\right)}:\frac{x+2}{x+3}\)
\(=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}\)
\(=\frac{3.\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}\)
\(=\frac{3}{x-3}\)
b ) \(B=-\frac{3}{5}\Leftrightarrow\frac{3}{x-3}=-\frac{3}{5}\)
\(\Leftrightarrow x-3=-5\Leftrightarrow x=-2\) ( do \(x\ne\pm3;x\ne-2\) )
c ) \(B< 0\Leftrightarrow\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow\) \(\hept{\begin{cases}x< 3\\x\ne-2\\x\ne-3\end{cases}}\)
\(a,M=1:\left(\frac{x^2+2}{x^3-1}+\frac{x+1}{x^2+x+1}-\frac{1}{x-1}\right)\)
\(=1:\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+1}{x^2+x+1}+\frac{-1}{x-1}\right]\)
\(=1:\left[\frac{\left(x^2+2\right)+\left(x+1\right)\left(x-1\right)+\left(-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\left[\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\left[\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left[\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\frac{x}{x^2+x+1}=\frac{x^2+x+1}{x}\)
ĐKXĐ:\(x\ne-3;x\ne3\)
\(A=\frac{5}{x+3}-\frac{2}{3-x}-\frac{3x^2-2x-9}{x^2-9}\)
\(=\frac{5}{x+3}+\frac{2}{x-3}-\frac{3x^2-2x-9}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{5\left(x-3\right)+2\left(x+3\right)-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{-3x^2+9x}{\left(x-3\right)\left(x+3\right)}=\frac{-3x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=-\frac{3x}{x+3}\)
b
\(\left|x-2\right|=1\Rightarrow x-2=1\left(h\right)x-2=-1\Rightarrow x=3;x=1\)
Tại \(x=3\) thì \(A=-\frac{3\cdot3}{3+3}=-\frac{9}{6}=-\frac{3}{2}\)
Tại \(x=1\) thì \(A=-1\cdot\frac{3}{1+3}=-\frac{3}{4}\)
c
Để A nguyên thì \(\frac{3x}{x+3}\) nguyên
\(\Rightarrow3x⋮x+3\)
\(\Rightarrow3\left(x+3\right)-9⋮x+3\)
\(\Rightarrow9⋮x+3\)
\(\Rightarrow x+3\in\left\{1;3;9;-1;-3;-9\right\}\)
\(\Rightarrow x\in\left\{-2;0;6;-4;-6;-12\right\}\)
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) \(ĐKXĐ:\hept{\begin{cases}x\ne0;x\ne2\\x\ne-1\end{cases}}\)
\(Q=1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)
\(\Leftrightarrow Q=1+\left(\frac{x+1}{x^3+1}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right):\frac{x^2\left(x-2\right)}{x\left(x^2-x+1\right)}\)
\(\Leftrightarrow Q=1+\frac{\left(x+1\right)+\left(x+1\right)-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{x\left(x-2\right)}{x^2-x+1}\)
\(\Leftrightarrow Q=1+\frac{x+1+x+1-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(\Leftrightarrow Q=1+\frac{-2x^2+4x}{x\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow Q=1+\frac{-2x\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow Q=1+\frac{-2}{x+1}\)
\(\Leftrightarrow Q=\frac{x-1}{x+1}\)
b) \(\left|x-\frac{3}{4}\right|=\frac{5}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\left(ktm\right)\\x=-\frac{1}{2}\left(tm\right)\end{cases}}\)
Thay \(x=-\frac{1}{2}\)vào Q, ta được :
\(Q=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}\)
\(\Leftrightarrow Q=\frac{-\frac{3}{2}}{\frac{1}{2}}\)
\(\Leftrightarrow Q=-3\)
c) Để \(Q\inℤ\)
\(\Leftrightarrow x-1⋮x+1\)
\(\Leftrightarrow x+1-2⋮x+1\)
\(\Leftrightarrow2⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(\Leftrightarrow x\in\left\{-2;0;-3;1\right\}\)
Vậy để \(Q\inℤ\Leftrightarrow x\in\left\{-2;0;-3;1\right\}\)
d) \(A>0\Leftrightarrow\frac{-1}{x-2}>0\)
\(\Leftrightarrow x-2< 0\) ( vì \(-1< 0\))
\(\Leftrightarrow x< 2\)
\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(A=\frac{-1}{x-2}\)
