P= \(\dfrac{5x+3}{9-x^2}+\dfrac{5}{3+x}+\dfrac{3}{x-3}\)
a, Rút gọn biểu thức P
b, Tính x để P= -1
c, Tìm gtri nguyên của x để P có gtri nguyên
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ĐKXĐ : \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\\9-x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
a, \(A=\dfrac{x-5}{x-3}-\dfrac{2x}{x+3}-\dfrac{2x^2-x+15}{9-x^2}\)
\(=\dfrac{\left(x-5\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x^2-x+15}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2-2x-15-2x^2+6x+2x^2-x+15}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x}{x+3}\)
b, \(\left|x-1\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\left(kot/m\right)\\x=-1\left(t/m\right)\end{matrix}\right.\)
Thay x =- 1 vào biểu thức A ,có :
\(\dfrac{-1}{-1+3}=\dfrac{-1}{2}\)
Vậy tại x = -1 gtri của bt A là -1/2
Vậy tại x = 3 biểu thức A ko có giá trị
c,\(\dfrac{x}{x+3}=\dfrac{x+3-3}{x+3}=1-\dfrac{3}{x+3}\)
Để A có giá trị nguyên
\(\Leftrightarrow\dfrac{3}{x+3}\) là số nguyên
\(\Leftrightarrow3⋮x+3\)
\(\Leftrightarrow x+3\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(x+3\) | 1 | -1 | 3 | -3 |
x | -2 (t/m) | -4(t/m) | 0 (t/m) | -6(t/m) |
Vậy \(x\in\left\{0;-2;-4;-6\right\}\) thì A có giá trị nguyên
\(B=\frac{5}{x+3}+\frac{3}{x-3}-\frac{5x+3}{x^2-9}\)
\(B=\frac{5}{x+3}+\frac{3}{x-3}-\frac{5x+3}{\left(x-3\right)\left(x+3\right)}\)
B xác định \(\Leftrightarrow\hept{\begin{cases}x-3\ne0\\x+3\ne0\end{cases}\Leftrightarrow}x\ne\pm3\)
Vậy B xác định \(\Leftrightarrow x\ne\pm3\)
\(B=\frac{5}{x+3}+\frac{3}{x-3}-\frac{5x+3}{x^2-9}\)
\(B=\frac{5}{x+3}+\frac{3}{x-3}-\frac{5x+3}{\left(x-3\right)\left(x+3\right)}\)
\(B=\frac{5\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{5x+3}{\left(x-3\right)\left(x+3\right)}\)
\(B=\frac{5x-15+3x+9-5x-3}{\left(x+3\right)\left(x-3\right)}\)
\(B=\frac{3x-9}{\left(x+3\right)\left(x-3\right)}\)
\(B=\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(B=\frac{3}{x+3}\)
a) Ta có: \(P=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x^2+\sqrt{x}}{x\sqrt{x}+x}\)
\(=\dfrac{2x+2}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
a)B = \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{7x+3}{9-x^2}\left(ĐK:x\ne\pm3\right)\)
= \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{7x+3}{x^2-9}\)
= \(\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)-7x-3}{\left(x+3\right)\left(x-3\right)}\)
= \(\dfrac{3x^2-9x}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x}{x+3}\)
b) \(\left|2x+1\right|=7< =>\left[{}\begin{matrix}2x+1=7< =>x=3\left(L\right)\\2x+1=-7< =>x=-4\left(C\right)\end{matrix}\right.\)
Thay x = -4 vào B, ta có:
B = \(\dfrac{-4.3}{-4+3}=12\)
c) Để B = \(\dfrac{-3}{5}\)
<=> \(\dfrac{3x}{x+3}=\dfrac{-3}{5}< =>\dfrac{3x}{x+3}+\dfrac{3}{5}=0\)
<=> \(\dfrac{15x+3x+9}{5\left(x+3\right)}=0< =>x=\dfrac{-1}{2}\left(TM\right)\)
d) Để B nguyên <=> \(\dfrac{3x}{x+3}\) nguyên
<=> \(3-\dfrac{9}{x+3}\) nguyên <=> \(9⋮x+3\)
x+3 | -9 | -3 | -1 | 1 | 3 | 9 |
x | -12(C) | -6(C) | -4(C) | -2(C) | 0(C) | 6(C) |
a) để A xát định thì
\(\left[{}\begin{matrix}2x+10\ne0\\x\ne0\\2x\left(x-5\right)\ne0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}2x\ne-10\\x\ne0\\\left[{}\begin{matrix}2x\ne0\\x-5\ne0\end{matrix}\right.\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x\ne-5\\x\ne0\\\left[{}\begin{matrix}x\ne0\\x\ne5\end{matrix}\right.\end{matrix}\right.\)
vậy \(\left[{}\begin{matrix}x\ne0\\x\ne-5\\x\ne5\end{matrix}\right.\) thì A được xác định
a: \(P=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)
a) Ta có: \(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)
\(=\dfrac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x^2-6x+x^2+4x+3+11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x}{x-3}\)
b)
ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)
Ta có: P=AB
\(=\dfrac{3x}{x-3}\cdot\dfrac{x-3}{x+1}\)
\(=\dfrac{3x}{x+1}\)
Để \(P=\dfrac{9}{2}\) thì \(\dfrac{3x}{x+1}=\dfrac{9}{2}\)
\(\Leftrightarrow9\left(x+1\right)=6x\)
\(\Leftrightarrow9x-6x=-9\)
\(\Leftrightarrow3x=-9\)
hay x=-3(loại)
Vậy: Không có giá trị nào của x để \(P=\dfrac{9}{2}\)
a, ĐKXĐ:\(\left\{{}\begin{matrix}x+3\ne0\\x^2+x-6\ne0\\2-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x^2+x-6\ne0\\x\ne2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)
b, \(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+3\right)}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x-4}{x-2}\)
\(c,A=\dfrac{-3}{4}\\ \Leftrightarrow\dfrac{x-4}{x-2}=\dfrac{-3}{4}\\ \Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\\ \Leftrightarrow4x-16x=-3x+6\\ \Leftrightarrow4x-16x+3x-6=0\\ \Leftrightarrow7x-22=0\\ \Leftrightarrow x=\dfrac{22}{7}\)
d, \(A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)
Để \(A\in Z\Rightarrow\dfrac{2}{x-2}\in Z\Rightarrow x-2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Ta có bảng:
x-2 | -2 | -1 | 1 | 2 |
x | 0 | 1 | 3 | 4 |
Vậy \(x\in\left\{0;1;3;4\right\}\)