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\(A=\frac{5}{x+3}-\frac{2}{3-x}-\frac{3x^2-2x-9}{x^2-9}\)
a) ĐKXĐ: \(\hept{\begin{cases}x+3\ne0\\3-x\ne0\\x^2-9\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne3\\x\ne3;x\ne-3\end{cases}}}\)
Vậy ĐKXĐ: x khác -3; x khác 3 ( b vào tcn của mìnk để thấy chi tiết)
Rút gọn:
\(A=\frac{5}{x+3}-\frac{2}{3-x}-\frac{3x^2-2x-9}{x^2-9}\)
\(\Leftrightarrow A=\frac{5}{x+3}+\frac{2}{x-3}-\frac{3x^2-2x-9}{\left(x-3\right)\left(x+3\right)}\) MTC: (x-3)(x+3)
\(\Leftrightarrow A=\frac{5\left(x-3\right)+2\left(x+3\right)-\left(3x^2-2x-9\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{9x-3x^2}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{3x\left(3-x\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{-3x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-3x}{x+3}\)
Vậy A=-3x/x+3 với x khác 3 và x khác -3
b) |x-2|=1
Bỏ dấu gt tuyệt đối ta có 2 TH: (đối chiếu đkxđ)
* x-2=1=> x=1+2=>x=3 (o t/m)
*x-2=-1=>x=-1+2=>x=1 (tm)
Thay x=1 vào phân thức A rút gọn ta có:
\(A=\frac{-3x}{x+3}=\frac{-3.1}{1+3}=\frac{-3}{4}\)
Vậy A=-3/4 khi x=1
c) Để A có gt nguyên => A thuộc Z
=> \(A=\frac{-3x}{x+3}\in Z\)
Ta có: -3x chia hết x+3
=> -3(x-3)-9 chia hết x+3
=> -9 chia hết cho x+3
=> x+3 thược Ư(-9)={1;-1;9;-9;3;-3)
Lập bảng thay vào hoặc o cần cx được
| x+3 | 1 | -1 | 9 | -9 | 3 | -3 |
| x | -2(tm) | -4(tm) | 6(tm) | -12(tm) | 0(tm) | -6(tm) |
Vậy...
a, ĐKXĐ : \(\hept{\begin{cases}2-x\ne0\\x^2-4\ne0\\2+x\ne0\end{cases}}\)hoặc \(2x^2-x^3\ne0\)hay \(x\ne\pm2;0\)
\(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)
\(=\left(-\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\right)\)
\(=\frac{-x^2-2x-1-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}:\frac{x-3}{x\left(2-x\right)}\)
\(=\frac{-4x^2-6x+3}{\left(x-2\right)\left(x+2\right)}.\frac{-x\left(x-2\right)}{x-3}=\frac{\left(-4x^2-6x+3\right)\left(-x\right)}{\left(x+2\right)\left(x-3\right)}=\frac{4x^3+6x^2-3x}{\left(x+2\right)\left(x-3\right)}\)
b, Ta có : A > 0 hay \(\frac{4x^3+6x^2-3x}{\left(x+2\right)\left(x-3\right)}>0\)
\(\Leftrightarrow x\left(4x^2+6x-3\right)>0\)
\(\Leftrightarrow4x^2+6x-3>0\) bạn xem lại bài mình có chỗ nào sai ko nhé !!!
c, Ta có : \(\left|x-7\right|=4\Rightarrow\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=3\end{cases}}}\)
TH1 : Thay x = 11 vào phân thức trên : ...
TH2 : Thay x = 3 vào phân thức trên : .... tự làm
Bài làm:
a) \(đkxd:x\ne2;x\ne-2;x\ne0;x\ne3\)
Ta có: \(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)
\(A=\left(\frac{\left(x+2\right)^2+4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\right):\left(\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\right)\)
\(A=\left[\frac{x^2+4x+4+4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\right]:\frac{x-3}{x\left(2-x\right)}\)
\(A=\frac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(2-x\right)}{x-3}\)
\(A=\frac{4x\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(2-x\right)}{x-3}\)
\(A=\frac{4x^2}{x-3}\)
b) Ta có: \(4x^2>0\left(\forall x\ne0\right)\)
=> Để A>0 thì \(x-3>0\)
\(\Rightarrow x>3\)
Vậy với \(x>3\)thì A>0
c) Ta có: \(\left|x-7\right|=4\)\(\Rightarrow\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=11\\x=3\end{cases}}\)
Mà theo điều kiện xác định, \(x\ne3\)
\(\Rightarrow x=11\)
Khi đó, \(A=\frac{4.11^2}{11-3}=\frac{121}{2}\)
Vậy \(A=\frac{121}{2}\)
Học tốt!!!!
\(A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\left[\frac{-\left(x-3\right)}{x+3}.\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right].\frac{x+3}{3x^2}\)
\(=\left[\frac{-\left(x-3\right)\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)^2}+\frac{x}{x+3}\right].\frac{x+3}{3x^2}\)
\(=\left(-1+\frac{x}{x+3}\right).\frac{x+3}{3x^2}\)
\(=\frac{-x-3+x}{x+3}.\frac{x+3}{3x^2}=\frac{-3}{x+3}.\frac{x+3}{3x^2}=\frac{-1}{x^2}\)
b ) Để \(A=-\frac{1}{x^2}< 0\forall x\ne0\)
Vậy \(x\ne0\) thì \(A< 0\)
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.
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 |
\(P=\left(\frac{9}{x^2-3x}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3-3x}\)
a,\(ĐKXĐ:x\ne0;x\ne3;x\ne1\)
\(P=\left(\frac{9}{x^2-3x}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3-3x}=\left(\frac{9}{x\left(x-3\right)}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3\left(1-x\right)}\)
\(=\left(\frac{9+\left(x-2\right)\left(x-3\right)-x.x}{x\left(x-3\right)}\right).\frac{x}{3\left(1-x\right)}=\frac{9+x^2-5x+6-x^2}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}\)
\(=\frac{-5x+15}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}=\frac{-5\left(x-3\right)}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}=-\frac{5}{3\left(1-x\right)}\)
b, \(x=\frac{1}{2}\)
\(\Rightarrow P=-\frac{5}{3\left(1-\frac{1}{2}\right)}=-\frac{5}{3.\frac{1}{2}}=-5:\frac{3}{2}=-\frac{10}{3}\)
c, Để \(P\in z\)thì \(3\left(1-x\right)\inƯ\left(5\right)=\left(-5;-1;1;5\right)\)
\(3\left(1-x\right)=-5\Rightarrow1-x=-\frac{5}{3}\Rightarrow x=\frac{8}{3}\)
\(3\left(1-x\right)=-1\Rightarrow1-x=-\frac{1}{3}\Rightarrow x=\frac{4}{3}\)
\(3\left(1-x\right)=1\Rightarrow1-x=\frac{1}{3}\Rightarrow x=\frac{2}{3}\)
\(3\left(1-x\right)=5\Rightarrow1-x=\frac{5}{3}\Rightarrow x=-\frac{2}{3}\)