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\(a,\)\(đkxđ\Leftrightarrow\)\(\hept{\begin{cases}x+3\ne0\\x-3\ne0\end{cases}}\)\(\Rightarrow x\ne\pm3\)
\(b,\)\(B=\frac{5}{x+3}+\frac{3}{x-3}-\frac{5x+3}{x^2-9}\)
\(=\frac{5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{5x+3}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{5x-15+3x+9-5x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x-9}{\left(x-3\right)\left(x+3\right)}=\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3}{x+3}\)
\(c,\)Tại x = 6, ta có :
\(B=\frac{3}{x+3}=\frac{3}{6+3}=\frac{3}{9}=\frac{1}{3}\)
Vậy tại x = 6 thì B = 3
\(d,\)Để \(B\in Z\Rightarrow\frac{3}{x+3}\in Z\Rightarrow x+3\inƯ_3\)
Mà \(Ư_3=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\)TH1 : \(x+3=1\Rightarrow x=-2\)
Th2: \(x+3=-1\Rightarrow x=-4\)
Th3 : \(x+3=3\Rightarrow x=0\)
TH4 \(x+3=-3\Rightarrow x=-6\)
Vậy để \(B\in Z\)thì \(x\in\left\{-6;-4;-2;0\right\}\)
a)Để B đc xác định thì :x+3 khác 0
x-3 khác 0
x^2-9 khác 0
=>x khác -3
x khác 3
b) Kết Qủa BT B là:3/x+3
1/ Thay x=-4 vao A -> A= \(\frac{-4}{-4+3}\)= 4
2/ B=\(\frac{2}{x-3}\)+\(\frac{x-15}{x^2-9}\)
B= \(\frac{2\left(x+3\right)+x-15}{\left(x-3\right)\left(x+3\right)}\)
B= \(\frac{2x+6+x-15}{\left(x-3\right)\left(x+3\right)}\)= \(\frac{3x-9}{\left(x-3\right)\left(x+3\right)}\)= \(\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)= \(\frac{3}{x+3}\)
c, B>A <=> \(\frac{3}{x+3}\)> \(\frac{x}{x+3}\)
<=> \(\frac{3}{x+3}\)- \(\frac{x}{x+3}\)> 0
<=> \(\frac{3-x}{x+3}\)>0
<=> 3-x <0 / >0 ( Đkxd x khác -3 )
x+3 <0 / >0
..............
...............................
Vậy ...
1) \(A=\frac{x}{x+3}\)( ĐKXĐ : \(x\ne-3\))
Với x = -4 ( tmđk ) thì giá trị của A là
\(A=\frac{-4}{-4+3}=\frac{-4}{-1}=4\)
2) \(B=\frac{2}{x-3}+\frac{x-15}{x^2-9}\)( ĐKXĐ : \(x\ne\pm3\))
\(B=\frac{2}{x-3}+\frac{x-15}{\left(x-3\right)\left(x+3\right)}\)
\(B=\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x-15}{\left(x-3\right)\left(x+3\right)}\)
\(B=\frac{2x+6+x-15}{\left(x-3\right)\left(x+3\right)}\)
\(B=\frac{3x-9}{\left(x-3\right)\left(x+3\right)}\)
\(B=\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3}{x+3}\)
3) Để B > A
=> \(\frac{3}{x+3}>\frac{x}{x+3}\)( ĐKXĐ : \(x\ne-3\))
<=> \(\frac{3}{x+3}-\frac{x}{x+3}>0\)
<=> \(\frac{3-x}{x+3}>0\)
Xét hai trường hợp :
1.\(\hept{\begin{cases}3-x>0\\x+3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}-x>-3\\x>-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 3\\x>-3\end{cases}}\Leftrightarrow-3< x< 3\)( tmđk )
2. \(\hept{\begin{cases}3-x< 0\\x+3< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}-x< -3\\x< -3\end{cases}}\Leftrightarrow\hept{\begin{cases}x>3\\x< -3\end{cases}}\)( loại )
Vì x nguyên => x ∈ { -2 ; -1 ; 0 ; 1 ; 2 ; 3 }
Vậy ...
\(\text{a, ĐKXĐ: }\hept{\begin{cases}x+3\ne0\\x-3\ne0\\3x^2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne\mp3\\x\ne0\end{cases}}\)
\(A=\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\left[\frac{\left(3-x\right)\left(x+3\right)^2}{\left(x+3\right)\left(x+3\right)\left(x-3\right)}+\frac{x}{x+3}\right]\cdot\frac{x+3}{3x^2}\)
\(=\frac{x-x-3}{x+3}\cdot\frac{x+3}{3x^2}\)
\(=-\frac{1}{x^2}\)
b, với x=\(-\frac{1}{2}\)ta có:
\(A=-\frac{1}{\left(-\frac{1}{2}\right)^2}=-4\)
c, Để A<0 thì \(-\frac{1}{x^2}< 0\text{ mà }x^2>0\left(\text{vì x khác 0 ĐKXĐ}\right)\)
Với x khác 0 thì thỏa mãn!
a) ĐKXĐ: \(x\ne\pm3\)
\(A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\left(\frac{3-x}{x+3}.\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\left(\frac{3-x}{x-3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\frac{\left(3-x\right)\left(x+3\right)+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{3x^2}\)
\(=\frac{3\left(3-x\right)}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{3x^2}\)
\(=-\frac{1}{x^2}\)
\(ĐKXĐ:\hept{\begin{cases}x\ne\pm3\\x\ne0\end{cases}}\)
a) \(B=\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}\right):\frac{3x^2}{x+3}\)
\(\Leftrightarrow B=\left(\frac{3-x}{x+3}\cdot\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)
\(\Leftrightarrow B=\frac{\left(3-x\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{3x^2}\)
\(\Leftrightarrow B=-\frac{x+3}{3x^2}\)
b) Khi \(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=3\left(ktm\right)\end{cases}}\)
\(\Leftrightarrow x=1\)
\(\Leftrightarrow B=-\frac{1+3}{3.1^2}=-\frac{4}{3.}\)
c) Để B > 0
\(\Leftrightarrow-\frac{x+3}{3x^2}>0\)
\(\Leftrightarrow\frac{x+3}{3x^2}< 0\)
\(\Leftrightarrow x+3< 0\) (Do 3x2 > 0; loại giá trị = 0)
\(\Leftrightarrow x< -3\)
Vậy để \(B>0\Leftrightarrow x< -3\)
a) Ta thấy x=-2 thỏa mãn ĐKXĐ của B.
Thay x=-2 và B ta có :
\(B=\frac{2\cdot\left(-2\right)+1}{\left(-2\right)^2-1}=\frac{-3}{3}=-1\)
b) Rút gọn :
\(A=\frac{3x+1}{x^2-1}-\frac{x}{x-1}\)
\(=\frac{3x+1-x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{-x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)
Xấu nhỉ ??
\(ĐKXĐ:x\ne-3;2\)
\(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{1}{x-2}\)
\(=\frac{x^2+4x+4}{\left(x+3\right)\left(x+2\right)}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{x+3}{\left(x+2\right)\left(x+3\right)}\)
\(=\frac{x^2+4x+4-5-x-3}{\left(x+2\right)\left(x+3\right)}=\frac{x^2+3x-4}{\left(x+3\right)\left(x+2\right)}=\frac{\left(x+4\right)\left(x-1\right)}{\left(x+3\right)\left(x+2\right)}\)
\(x^2-9=0\Leftrightarrow x=3\left(vì:x\ne-3\right)\)
\(\Rightarrow P=\frac{7}{15}\)
\(P\inℤ\Leftrightarrow x^2+3x-4⋮x^2+5x+6\Leftrightarrow2x+10⋮x^2+5x+6\Leftrightarrow12⋮x^2+5xx+6\)
\(................\left(dễ\right)\)
P/s: shitbo sai rồi nha bạn!Nếu không tin thì thay x = 3 vào P ban đầu và giá trị P sau khi rút gọn sẽ thấy sự khác biệt =)
ĐK: \(x\ne-3;x\ne2\)
a) \(P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}\)
\(=\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)
b) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow x=\pm3\)
Thay vào điều kiện,tìm loại x = -3 .Tìm được x =3
Ta có: \(P=\frac{x-4}{x-2}=\frac{3-4}{3-2}=-1\)
c)Ta có: \(P=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\)
Để P có giá trị nguyên thì \(\frac{2}{x-2}\) nguyên hay \(x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Suy ra \(x=\left\{0;1;3;4\right\}\)
a
\(ĐKXĐ:x\ne3;x\ne-3;x\ne0\)
b
\(A=\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)
\(=\left[\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right]:\left[\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right]\)
\(=\frac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\frac{3x-9-x^2}{3x\left(x+3\right)}\)
\(=\frac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\cdot\frac{3x\left(x+3\right)}{-\left(9-3x+x^2\right)}=\frac{-3}{x-3}\)
c
Với \(x=4\Rightarrow A=-3\)
d
Để A nguyên thì \(\frac{3}{x-3}\) nguyên
\(\Rightarrow3⋮x-3\)
Làm nốt.
