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Bài 1:
a) x≠2x≠2
Bài 2:
a) x≠0;x≠5x≠0;x≠5
b) x2−10x+25x2−5x=(x−5)2x(x−5)=x−5xx2−10x+25x2−5x=(x−5)2x(x−5)=x−5x
c) Để phân thức có giá trị nguyên thì x−5xx−5x phải có giá trị nguyên.
=> x=−5x=−5
Bài 3:
a) (x+12x−2+3x2−1−x+32x+2)⋅(4x2−45)(x+12x−2+3x2−1−x+32x+2)⋅(4x2−45)
=(x+12(x−1)+3(x−1)(x+1)−x+32(x+1))⋅2(2x2−2)5=(x+12(x−1)+3(x−1)(x+1)−x+32(x+1))⋅2(2x2−2)5
=(x+1)2+6−(x−1)(x+3)2(x−1)(x+1)⋅2⋅2(x2−1)5=(x+1)2+6−(x−1)(x+3)2(x−1)(x+1)⋅2⋅2(x2−1)5
=(x+1)2+6−(x2+3x−x−3)(x−1)(x+1)⋅2(x−1)(x+1)5=(x+1)2+6−(x2+3x−x−3)(x−1)(x+1)⋅2(x−1)(x+1)5
=[(x+1)2+6−(x2+2x−3)]⋅25=[(x+1)2+6−(x2+2x−3)]⋅25
=[(x+1)2+6−x2−2x+3]⋅25=[(x+1)2+6−x2−2x+3]⋅25
=[(x+1)2+9−x2−2x]⋅25=[(x+1)2+9−x2−2x]⋅25
=2(x+1)25+185−25x2−45x=2(x+1)25+185−25x2−45x
=2(x2+2x+1)5+185−25x2−45x=2(x2+2x+1)5+185−25x2−45x
=2x2+4x+25+185−25x2−45x=2x2+4x+25+185−25x2−45x
=2x2+4x+2+185−25x2−45x=2x2+4x+2+185−25x2−45x
=2x2+4x+205−25x2−45x=2x2+4x+205−25x2−45x
c) tự làm, đkxđ: x≠1;x≠−1
\(B=\frac{x^2-2}{x^2+1}=\frac{x^2+1-3}{x^2+1}=1-\frac{3}{x^2+1}\)
\(B_{min}\Rightarrow\left(\frac{3}{x^2+1}\right)_{max}\Rightarrow\left(x^2+1\right)_{min}\)
\(x^2+1\ge1\). dấu = xảy ra khi x2=0
=> x=0
Vậy \(B_{min}\Leftrightarrow x=0\)
ta có: \(x^2+2x-2=x^2+2x+1^2-3=\left(x+1\right)^2-3\ge-3\)
dấu = xảy ra khi \(x+1=0\)
\(\Rightarrow x=-1\)
Vậy\(\left(x^2+2x-2\right)_{min}\Leftrightarrow x=-1\)
\(x\ne+-3\)
\(3\left(x-3\right)+1\left(x+3\right)+18\)
3x-9+x+3+18
4x+15
x=-15/4
a,ĐK: \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)
c, Với x = 4 thỏa mãn ĐKXĐ thì
\(A=\frac{-3}{4-3}=-3\)
d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)
Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
\(a,ĐKXĐ\hept{\begin{cases}x-3\ne0\\x+3\ne0\end{cases}\Leftrightarrow x\ne\pm3}\)
Ta có: \(M=\frac{3}{x-3}-\frac{6x}{9-x^2}+\frac{x}{x+3}\)
\(=\frac{3}{x-3}+\frac{6x}{x^2-9}+\frac{x}{x+3}\)
\(=\frac{3\left(x+3\right)+6x+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x+3}{x-3}\)
\(b,x=\frac{1}{2}\Rightarrow M=\frac{\frac{1}{2}+3}{\frac{1}{2}-3}=-\frac{7}{5}\)
a ) ĐKXĐ : \(x\ne6\)
b ) \(Q=\dfrac{3x+8}{x+6}-\dfrac{x-4}{x+6}\)
\(Q=\dfrac{3x+8-x+4}{x+6}=\dfrac{2\left(x+6\right)}{x+6}=2\)
\(a,Đkxđ:x\ne\pm2\)
\(A=\frac{1}{x-2}+\frac{1}{x+2}+\frac{x^2+1}{x^2-4}\)
\(=\frac{x+2+x-2+x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x+1\right)^2}{x^2-4}\)
b, Ta có: \(\left(x-2\right)\left(x+2\right)< 0;\forall-2< 2< 2;x\ne-1\)
Mà: \(\left(x+1\right)^2>0\left(\forall x\ne-1\right)\)
\(\Rightarrow\frac{\left(x+1\right)^2}{\left(x+2\right)\left(x-2\right)}< 0;\forall-2< x< 2;x\ne-1\)
Vậy ............
bài 1.a. điều kiện xác định của phân thức là \(x^3-8\ne0\Leftrightarrow x\ne2\)
b .ta có \(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x+2}\)
bài 2.
\(A=\left(\frac{1}{x-1}-\frac{x}{1-x^3}.\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)
\(A=\left(\frac{1}{x-1}-\frac{x}{\left(1-x\right)\left(x^2+x+1\right)}.\frac{x^2+x+1}{x+1}\right).\frac{\left(x+1\right)^2}{2x+1}\)
\(A=\left(\frac{1}{x-1}-\frac{x}{\left(1-x\right)\left(x+1\right)}\right).\frac{\left(x+1\right)^2}{2x+1}\)
\(\Leftrightarrow A=\left(\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x+1\right)^2}{2x+1}=\frac{x+1}{x-1}\)
khi \(x=\frac{1}{2}\Rightarrow A=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=-3\)
a,\(A\) xác định \(\Leftrightarrow\left[{}\begin{matrix}x+1\ne0\\1-x^2\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ne-1\\x\ne1\end{matrix}\right.\)
Vậy...
b,\(A=\dfrac{x}{x+1}:\dfrac{1-3x^2}{1-x^2}\)
\(=\dfrac{x}{x+1}:\dfrac{-\left(3x^2-1\right)}{-\left(x^2-1\right)}\)
\(=\dfrac{x}{x+1}:\dfrac{3x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x}{x+1}.\dfrac{\left(x-1\right)\left(x+1\right)}{3x^2-1}\)
\(=\dfrac{x}{1}.\dfrac{x-1}{3x^2-1}\)
\(=\dfrac{x^2-x}{3x^2-1}\)
Câu a :
Để phân thức được xác định thì :
\(\left\{{}\begin{matrix}x+1\ne0\\1-x^2\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne1\end{matrix}\right.\)
Câu b :
\(\dfrac{x}{x+1}:\dfrac{1-3x^2}{1-x^2}\)
\(=\dfrac{x}{x+1}:\dfrac{-\left(1-3x^2\right)}{x^2-1}\)
\(=\dfrac{x}{x+1}:\dfrac{3x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x}{x+1}\times\dfrac{\left(x-1\right)\left(x+1\right)}{3x^2-1}\)
\(=\dfrac{x\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(3x^2-1\right)}=\dfrac{x\left(x-1\right)}{3x^2-1}\)