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a) Phân thức M xác định khi và chỉ khi :
+) \(2x-2\ne0\Leftrightarrow x\ne1\)
+) \(2x+2\ne0\Leftrightarrow x\ne-1\)
+) \(1-\frac{x-3}{x+1}\ne0\)
\(\Leftrightarrow x-3\ne x+1\)
\(\Leftrightarrow0x\ne4\left(\text{luôn đúng}\right)\)
Vậy \(x\ne\left\{1;-1\right\}\)
b) \(M=\left(\frac{x-2}{2x-2}-\frac{x+3}{2x+2}+\frac{3}{2x-2}\right):\left(1-\frac{x-3}{x+1}\right)\)
\(M=\left(\frac{\left(x-2\right)\left(2x+2\right)}{\left(2x-2\right)\left(2x+2\right)}-\frac{\left(x+3\right)\left(2x-2\right)}{\left(2x-2\right)\left(2x+2\right)}+\frac{3\left(2x+2\right)}{\left(2x-2\right)\left(2x+2\right)}\right):\left(\frac{x+1-x+3}{x+1}\right)\)
\(M=\left(\frac{2x^2-2x-4-2x^2-4x+6+6x+6}{\left(2x-2\right)\left(2x+2\right)}\right):\left(\frac{4}{x+1}\right)\)
\(M=\frac{8}{2\left(x-1\right)2\left(x+1\right)}\cdot\frac{x+1}{4}\)
\(M=\frac{8\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)\cdot4}\)
\(M=\frac{8\left(x+1\right)}{8\left(x+1\right)\left(x-1\right)}\)
\(M=\frac{1}{x-1}\)
\(M=\left(\frac{x-2}{2x-2}-\frac{x+3}{2x+2}+\frac{3}{2x-2}\right):\left(1-\frac{x-3}{x+1}\right)\)
\(=\left(\frac{x+1}{2x-2}-\frac{x+3}{2x+2}\right):\left(\frac{4}{x+1}\right)=\left[\frac{\left(x+1\right)\left(2x+2\right)-\left(x+3\right)\left(2x-2\right)}{\left(2x-2\right)\left(2x+2\right)}\right]:\left(\frac{4}{x+1}\right)\)
\(=\left[\frac{2x^2+4x+2-2x^2+2x+6-6x+6}{4x^2-4}\right]:\left(\frac{4}{x+1}\right)\)
\(=\left[\frac{6x+8-6x+6}{4x^2-4}\right]:\left(\frac{4}{x+1}\right)\)
\(=\frac{14}{4x^2-4}:\left(\frac{4}{x+1}\right)=\frac{14x+14}{16x^2-16}=\frac{7x+7}{8x^2-8}\)
ĐK \(\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)
a, \(P=\frac{x^2+26x-19}{\left(x+3\right)\left(x-1\right)}-\frac{2x}{x-1}+\frac{x-3}{x+3}\)\(=\frac{x^2+26x-19-2x\left(x+3\right)+\left(x-3\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}\)
\(=\frac{x^2+26x-19-2x^2-6x+x^2-4x+3}{\left(x+3\right)\left(x-1\right)}\)\(=\frac{16\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}=\frac{16}{x+3}\)
b. Với \(x=3\Rightarrow P=\frac{16}{3+3}=\frac{8}{3}\)
Với \(x=-1\Rightarrow P=\frac{16}{-1+3}=8\)
c. \(P=4\Rightarrow\frac{16}{x+3}=4\Rightarrow x+3=4\Rightarrow x=1\)
d. \(P\in Z\Rightarrow x+3\inƯ\left(16\right)\)
\(\Rightarrow x+3\in\left\{-16;-8;-4;-2;-1;1;2;4;8;16\right\}\)
\(\Rightarrow x\in\left\{-19;-11;-7;-5;-4;-2;-1;1;5;13\right\}\)
\(P=\frac{x^2+26x-19}{\left(x-1\right)\left(x+3\right)}-\frac{2x}{x-1}+\frac{x-3}{x+3}=\)
\(P=\frac{x^2+26x-19-2x\left(x+3\right)+\left(x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=\)
\(P=\frac{x^2+26x-19-2x^2-6x+x^2-4x+4}{\left(x-1\right)\left(x+3\right)}=\)
\(P=\frac{16x-15}{\left(x-1\right)\left(x+3\right)}\)
\(e ) Để \) \(M\)\(\in\)\(Z \) \(thì\) \(1 \)\(⋮\)\(x +3\)
\(\Leftrightarrow\)\(x + 3 \)\(\in\)\(Ư\)\((1)\)\(= \) { \(\pm\)\(1 \) }
\(Lập\) \(bảng :\)
| \(x +3\) | \(1\) | \(- 1\) |
| \(x\) | \(-2\) | \(- 4\) |
\(Vậy : Để \) \(M\)\(\in\)\(Z\) \(thì\) \(x\)\(\in\){ \(- 4 ; - 2\) }
e) Để M \(\in\)Z <=> \(\frac{1}{x+3}\in Z\)
<=> 1 \(⋮\)x + 3 <=> x + 3 \(\in\)Ư(1) = {1; -1}
Lập bảng:
| x + 3 | 1 | -1 |
| x | -2 | -4 |
Vậy ....
f) Ta có: M > 0
=> \(\frac{1}{x+3}\) > 0
Do 1 > 0 => x + 3 > 0
=> x > -3
Vậy để M > 0 khi x > -3 ; x \(\ne\)3 và x \(\ne\)-3/2
\(M=\frac{x^3+26x-19}{x^2+2x-3}-\frac{2x}{x-1}+\frac{x-3}{x+3}\)
\(=\frac{x^3+26x-19}{\left(x-1\right)\left(x+3\right)}-\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x+3\right)}\)
\(=\frac{x^3+26x-19-2x^2-6x+x^2-4x+3}{\left(x-1\right)\left(x+3\right)}\)
\(=\frac{x^3-x^2+16x-16}{\left(x-1\right)\left(x+3\right)}=\frac{x^2\left(x-1\right)+16\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}\)
\(=\frac{\left(x^2+16\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+16}{x+3}\)