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KN
29 tháng 7 2019
a) \(A=\frac{2x}{x+3}-\frac{x+1}{3-x}-\frac{3-11x}{x^2-9}\)
\(\Leftrightarrow A=\frac{2x}{x+3}+\frac{x+1}{x-3}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{2x^2-6x}{\left(x+3\right)\left(x-3\right)}+\frac{x^2+4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{3x^2-13x}{x^2-9}\)
14 tháng 10 2020
\(A=\frac{2x}{x+3}-\frac{x+1}{3-x}-\frac{3-11x}{x^2-9}\)
a) ĐK : x ≠ ±3
\(=\frac{2x}{x+3}+\frac{x+1}{x-3}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x^2-6x}{\left(x-3\right)\left(x+3\right)}+\frac{x^2+4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x^2-6x+x^2+4x+3-3+11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3x}{x-3}\)
b) Để A < 2
=> \(\frac{3x}{x-3}< 2\)
<=> \(\frac{3x}{x-3}-2< 0\)
<=> \(\frac{3x}{x-3}-\frac{2x-6}{x-3}< 0\)
<=> \(\frac{3x-2x+6}{x-3}< 0\)
<=> \(\frac{x+6}{x-3}< 0\)
Xét hai trường hợp :
1. \(\hept{\begin{cases}x+6>0\\x-3< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>-6\\x< 3\end{cases}}\Leftrightarrow-6< x< 3\)
2. \(\hept{\begin{cases}x+6< 0\\x-3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< -6\\x>3\end{cases}}\)( loại )
Vậy -6 < x < 3
27 tháng 1 2018
a) \(ĐKXĐ:\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}\)
\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)
\(=\left[\frac{\left(x+1\right)\left(x+2\right)}{3x\left(x+1\right)}+\frac{6x}{3x\left(x+1\right)}-\frac{9x\left(x+1\right)}{3x\left(x+1\right)}\right]:\frac{2\left(1-2x\right)}{x+1}-\frac{3x+1-x^2}{3x}\)
\(=\frac{\left(x+1\right)\left(x+2\right)+6x-9x\left(x+1\right)}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(=\frac{2-8x^2}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(=\frac{1+2x-3x-1+x^2}{3x}\)
\(=\frac{x\left(x-1\right)}{3x}=\frac{x-1}{3}\)
27 tháng 1 2018
b)\(\text{Với }x\ne0,x\ne-1,x\ne\frac{1}{2}\text{ ta có:}\)
\(\text{Để A< 0\Leftrightarrow}\frac{x-1}{3}< 0\Rightarrow x-1< 0\Leftrightarrow x< 1\)
VK
20 tháng 1 2017
\(C=\frac{\left(x+3\right)^2-2x^2+6+x\left(x-3\right)}{x^2-9}.\frac{2x^2-18}{6x-12}\)\(\)
\(C=\frac{x^2+6x+9-2x^2+6+x^2-3x}{x^2-9}.\frac{2\left(x^2-9\right)}{6x-12}\)\(C=\frac{3x+15}{6x-12}.2=\frac{x+5}{x-2}=1+\frac{7}{x-2}\)
Để C nguyên =>(x-2) thuộc Ư(7) \(\Rightarrow x\in\left\{3;1;9;-5\right\}\)
24 tháng 8 2020
b, P=x+2x+3−5x2+3x−2x−6+12−xP=x+2x+3−5x2+3x−2x−6+12−x
=x+2x+3−5(x+3)(x−2)−1x−2=x+2x+3−5(x+3)(x−2)−1x−2
=(x+2)(x−2)(x+3)(x−2)−5(x+3)(x−2)−x+3(x+3)(x−2)=(x+2)(x−2)(x+3)(x−2)−5(x+3)(x−2)−x+3(x+3)(x−2)
=x2−4−5−x−3(x+3)(x−2)=x2−x−12(x+3)(x−2)=x2−4−5−x−3(x+3)(x−2)=x2−x−12(x+3)(x−2)
=x2−4x+3x−12(x+3)(x−2)=x2−4x+3x−12(x+3)(x−2)
=(x−4)(x+3)(x+3)(x−2)=x−4x−2=(x−4)(x+3)(x+3)(x−2)=x−4x−2
c, Để P=−34P=−34
⇔x−4x−2=−34⇔x−4x−2=−34
⇔4(x−4)=−3(x−2)⇔4(x−4)=−3(x−2)
⇔4x−16+3x−6=0⇔4x−16+3x−6=0
⇔7x−22=0⇔7x−22=0
⇔x=227⇔x=227
d, Để P có giá trị nguyên
⇔x−4⋮x−2⇔x−4⋮x−2
⇔(x−2)−2⋮x−2⇔(x−2)−2⋮x−2
⇔2⋮x−2⇔x−2∈Ư(2)={1;−1;2;−2}⇔2⋮x−2⇔x−2∈Ư(2)={1;−1;2;−2}
| x−2x−2 | 1 | -1 | 2 | -2 |
| x | 3 | 1 | 4 | 0 |
e,
x2−9=0x2−9=0
⇒x2=9⇒[x=3x=−3⇒x2=9⇒[x=3x=−3
Với x=3,có :
x−4x−2=3−43−2=−11=−1x−4x−2=3−43−2=−11=−1
Với x=-3,có :
x−4x−2=−3−4−3−2=75x−4x−2=−3−4−3−2=75
TT
25 tháng 2 2020
Trước tiên ta đi rút gọn biểu thức trên :
Đặt \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
ĐKXĐ : \(x\ne\pm2,x\ne0\)
Ta có : \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left(\frac{x^2}{x\left(x^2-4\right)}+\frac{6}{3\left(2-x\right)}+\frac{1}{x+2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\right)\)
\(=\left(\frac{x\cdot3-6\cdot\left(x+2\right)+3\cdot\left(x-2\right)}{3\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)\)
\(=\frac{-18}{3\left(x-2\right)\left(x+2\right)}:\left(-\frac{6}{x+2}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{\left(-6\right)}=\frac{1}{x-2}\)
Để \(A\) nhận giá trị nguyên
\(\Leftrightarrow\frac{1}{x-2}\inℤ\) \(\Leftrightarrow1⋮x-2\) \(\Leftrightarrow x-2\inƯ\left(1\right)\)
\(\Leftrightarrow x-2\in\left\{-1,1\right\}\)
\(\Leftrightarrow x\in\left\{1,3\right\}\) ( Thỏa mãn ĐKXĐ )
Vậy : \(x\in\left\{1,3\right\}\) thì A nhận giá trị nguyên.
11 tháng 12 2020
a, \(\left(\frac{x^3+1}{x^2-1}-\frac{x^2-1}{x-1}\right):\left(x+\frac{x}{x-1}\right)\)
\(=\left(\frac{x^3+1}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x\left(x-1\right)}{x-1}+\frac{x}{x-1}\right)\)
\(=\left(\frac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x\left(x-1\right)+x}{x-1}\right)\)
\(=\left(\frac{\left(x+1\right)\left[x^2-x+1-x^2+1\right]}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x^2}{x-1}\right)\)
\(=\frac{\left(x+1\right)\left(2-x\right)}{\left(x-1\right)\left(x+1\right)}.\frac{x-1}{x^2}=\frac{2-x}{x^2}\)
b, Ta có : A = 3 hay \(\frac{2-x}{x^2}=3\)
\(3x^2=2-x\Leftrightarrow3x^2+x-2=0\)
\(\Leftrightarrow3x^2+3x-2x-2=0\Leftrightarrow\left(x+1\right)\left(3x-2\right)=0\Leftrightarrow x=-1;\frac{2}{3}\)
NC
11 tháng 12 2020
a,\(A=\left(\frac{x^3+1}{x^2-1}-\frac{x^2-1}{x-1}\right)\div\left(x+\frac{x}{x-1}\right)\)
\(=\left(\frac{x^3+1}{\left(x+1\right)\left(x-1\right)}-\frac{\left(x^2-1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)\div\left(\frac{x\left(x-1\right)}{x-1}+\frac{x}{x-1}\right)\)
\(=\left(\frac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\right)\div\left(\frac{x\left(x-1\right)+x}{\left(x-1\right)}\right)\)
\(=\left(\frac{\left(x+1\right)\left(x^2-x+1-x^2+1\right)}{\left(x-1\right)\left(x+1\right)}\right)\div\left(\frac{x^2}{x-1}\right)\)
\(=\left(\frac{\left(x+1\right)\left(2-x\right)}{\left(x-1\right)\left(x+1\right)}\right)\div\frac{x^2}{x-1}\)
\(=\frac{\left(x+1\right)\left(2-x\right)}{\left(x-1\right)\left(x+1\right)}\times\frac{x-1}{x^2}\)
\(=\frac{\left(x+1\right)\left(2-x\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)x^2}=\frac{2-x}{x^2}\)
MN
25 tháng 2 2020
a) \(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)
\(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}\)
\(\Leftrightarrow A=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2\left(x+2\right)}{x-3}\)
\(\Leftrightarrow A=\frac{2x-9-\left(x-3\right)\left(x+3\right)+2\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{\left(x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{x+4}{x-3}\)
b) Để \(A\inℤ\)
\(\Leftrightarrow\frac{x+4}{x-3}\inℤ\)
\(\Leftrightarrow1+\frac{7}{x-3}\inℤ\)
\(\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Leftrightarrow x\in\left\{2;4;-4;10\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{2;4;-4;10\right\}\)
c) Để \(A=\frac{3}{5}\)
\(\Leftrightarrow\frac{x+4}{x-3}=\frac{3}{5}\)
\(\Leftrightarrow5x+20=3x-9\)
\(\Leftrightarrow2x+29=0\)
\(\Leftrightarrow x=-\frac{29}{2}\)
d) Để \(A< 0\)
\(\Leftrightarrow\frac{x+4}{x-3}< 0\)
\(\Leftrightarrow1+\frac{7}{x-3}< 0\)
\(\Leftrightarrow\frac{-7}{x-3}< 1\)
\(\Leftrightarrow-7< x-3\)
\(\Leftrightarrow x>-4\)
e) Để \(A>0\)
\(\Leftrightarrow\frac{x+4}{x-3}>0\)
\(\Leftrightarrow1+\frac{7}{x-3}>0\)
\(\Leftrightarrow\frac{-7}{x-3}>1\)
\(\Leftrightarrow-7>x-3\)
\(\Leftrightarrow x< -4\)
\(\frac{x-2}{x-3}=\frac{x-3+1}{x-3}=1+\frac{1}{x-3}\)
\(\Rightarrow x-3\inƯ\left(1\right)=\left\{\pm1\right\}\)
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