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a, sửa đề : \(C=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}+\frac{1}{2-x}\)ĐK : \(x\ne-3;2\)
\(=\frac{\left(x+2\right)\left(x-2\right)-5-x-3}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-12-x}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)
b, Ta có : \(x^2-x=2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\Leftrightarrow x=-1;x=2\)
Kết hợp với giả thiết vậy x = -1
Thay x = -1 vào biểu thức C ta được : \(\frac{-1-4}{-1-2}=-\frac{5}{-3}=\frac{5}{3}\)
c, Ta có : \(C=\frac{1}{2}\Rightarrow\frac{x-4}{x-2}=\frac{1}{2}\Rightarrow2x-8=x-2\Leftrightarrow x=6\)( tm )
d, \(C>1\Rightarrow\frac{x-4}{x-2}>1\Rightarrow\frac{x-4}{x-2}-1>0\Leftrightarrow\frac{x-4-x+2}{x-2}>0\Leftrightarrow\frac{-2}{x-2}>0\)
\(\Rightarrow x-2< 0\Leftrightarrow x< 2\)vì -2 < 0
e, tự làm nhéee
f, \(C< 0\Rightarrow\frac{x+4}{x+2}< 0\)
mà x + 4 > x + 2
\(\hept{\begin{cases}x+4>0\\x+2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-4\\x< -2\end{cases}\Leftrightarrow-4< x< -2}}\)
Vì \(x\inℤ\Rightarrow x=-3\)( ktmđk )
Vậy ko có x nguyên để C < 0
g, Ta có : \(\frac{x+4}{x+2}=\frac{x+2+2}{x+2}=1+\frac{2}{x+2}\)
Để C nguyên khi \(x+2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| x + 2 | 1 | -1 | 2 | -2 |
| x | -1 | -3 | 0 | -4 |
h, Ta có : \(D=C\left(x^2-4\right)=\frac{x+4}{x+2}.\frac{\left(x-2\right)\left(x+2\right)}{1}=x^2+2x-8\)
\(=\left(x+1\right)^2-9\ge-9\)
Dấu ''='' xảy ra khi x = -1
Vậy GTNN D là -9 khi x = -1
d) \(A>0\Leftrightarrow\frac{-1}{x-2}>0\)
\(\Leftrightarrow x-2< 0\) ( vì \(-1< 0\))
\(\Leftrightarrow x< 2\)
\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(A=\frac{-1}{x-2}\)
\(ĐKXĐ:x\ne-3;2\)
\(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{1}{x-2}\)
\(=\frac{x^2+4x+4}{\left(x+3\right)\left(x+2\right)}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{x+3}{\left(x+2\right)\left(x+3\right)}\)
\(=\frac{x^2+4x+4-5-x-3}{\left(x+2\right)\left(x+3\right)}=\frac{x^2+3x-4}{\left(x+3\right)\left(x+2\right)}=\frac{\left(x+4\right)\left(x-1\right)}{\left(x+3\right)\left(x+2\right)}\)
\(x^2-9=0\Leftrightarrow x=3\left(vì:x\ne-3\right)\)
\(\Rightarrow P=\frac{7}{15}\)
\(P\inℤ\Leftrightarrow x^2+3x-4⋮x^2+5x+6\Leftrightarrow2x+10⋮x^2+5x+6\Leftrightarrow12⋮x^2+5xx+6\)
\(................\left(dễ\right)\)
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 \(x=\left\{0;1;3;4\right\}\)
\(ĐKXĐ:x\ne1\)
a) \(A=\left(1+\frac{x^2}{x^2+1}\right):\left(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\right)\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\left[\frac{1}{x-1}-\frac{2x}{x\left(x^2+1\right)-\left(x^2+1\right)}\right]\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\left[\frac{1}{x-1}-\frac{2x}{\left(x^2+1\right)\left(x-1\right)}\right]\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\frac{x^2+1-2x}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\frac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\frac{x-1}{x^2+1}\)
\(\Leftrightarrow A=\frac{\left(2x^2+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{2x^2+1}{x-1}\)
b) Thay \(x=-\frac{1}{2}\)vào A, ta được :
\(A=\frac{2\left(-\frac{1}{2}\right)^2+1}{-\frac{1}{2}-1}\)
\(\Leftrightarrow A=\frac{\frac{3}{2}}{-\frac{3}{2}}\)
\(\Leftrightarrow A=-1\)
c) Để A < 1
\(\Leftrightarrow2x^2+1< x-1\)
\(\Leftrightarrow2x^2-x+2< 0\)
\(\Leftrightarrow2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right)+\frac{15}{8}< 0\)
\(\Leftrightarrow2\left(x-\frac{1}{4}\right)^2+\frac{15}{8}< 0\)
\(\Leftrightarrow x\in\varnothing\)
Vậy để \(A< 1\Leftrightarrow x\in\varnothing\)
d) Để A có giá trị nguyên
\(\Leftrightarrow2x^2+1⋮x-1\)
\(\Leftrightarrow2x^2-2x+2x-2+3⋮x-1\)
\(\Leftrightarrow2x\left(x-1\right)+2\left(x-1\right)+3⋮x-1\)
\(\Leftrightarrow2\left(x+1\right)\left(x-1\right)+3⋮x-1\)
\(\Leftrightarrow3⋮x-1\)
\(\Leftrightarrow x-1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow x\in\left\{2;0;4;-2\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{2;0;4;-2\right\}\)
a, ĐK : \(x\ne\pm3;\frac{1}{2}\)
\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)
\(=\left(\frac{\left(x-1\right)\left(x-3\right)+2\left(x+3\right)-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\frac{2x-1-2x-1}{2x+1}\right)\)
\(=\frac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}:\left(-\frac{2}{2x+1}\right)\)
\(=\frac{-2x+6}{\left(x+3\right)\left(x-3\right)}.\frac{-\left(2x+1\right)}{2}=\frac{2x+1}{x+3}\)
b, Ta có : \(\left|x+1\right|=\frac{1}{2}\)
TH1 : \(x+1=\frac{1}{2}\Leftrightarrow x=-\frac{1}{2}\)
Thay vào biểu thức A ta được : \(\frac{-1+1}{-\frac{1}{2}+3}=0\)
TH2 : \(x+1=-\frac{1}{2}\Leftrightarrow x=-\frac{3}{2}\)
Thay vào biểu thức A ta được : \(\frac{-3+1}{-\frac{3}{2}+3}=\frac{-2}{\frac{3}{2}}=-\frac{4}{3}\)
c, Ta có : \(P=\frac{x}{2}\Rightarrow\frac{2x+1}{x+3}=\frac{x}{2}\Rightarrow4x+2=x^2+3x\)
\(\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)
b, Ta có : \(\frac{2x+1}{x+3}=\frac{2\left(x+3\right)-5}{x+3}=2-\frac{5}{x+3}\)
\(\Rightarrow x+3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
| x + 3 | 1 | -1 | 5 | -5 |
| x | -2 | -4 | 2 | -8 |
\(C=\left(\dfrac{2x^2+1}{x^3-1}-\dfrac{1}{x-1}\right)\div\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)
ĐKXĐ: \(x\ne1\)
\(C=[\left(\dfrac{2x^2+1}{(x-1)\left(x^2+x+1\right)}-\dfrac{1}{x-1}\right)]\div\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)
\(\Leftrightarrow C=[\left(\dfrac{2x^2+1}{(x-1)\left(x^2+x+1\right)}-\dfrac{1\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}\right)]\div[\dfrac{(x-1)\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}-\dfrac{(x^2-2)(x-1)}{(x^2+x+1)\left(x-1\right)}]\)
\(\Rightarrow C=\left[2x^2+1-1\left(x^2+x+1\right)\right]\div\left[\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2\right)\right]\)
\(\Rightarrow C=(2x^2+1-x^2-x-1)\div\left[\left(x-1\right)\left(x^2+x+1-x^2+2\right)\right]\)
\(\Rightarrow C=\left(x^2-x\right)\div\left[\left(x-1\right)\left(x+3\right)\right]\)