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\(P=\frac{2\left(x-2\right)\left(x+2\right)}{x^2+x+5}.\frac{5\left(x^2+x+5\right)}{\left(x-4\right)\left(x+3\right)}.\frac{\left(x-1\right)\left(x-4\right)}{10\left(x-2\right)\left(x+2\right)}=\frac{x-1}{x+3}\)
ĐK: \(x\ne\left\{4;-3;1;2;-2\right\}\)
b, \(P\in Z\Rightarrow\frac{x-1}{x+3}\in Z\Rightarrow x-1⋮\left(x+3\right)\Rightarrow-4⋮\left(x+3\right)\Rightarrow\left(x+3\right)\in\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow x\in\left\{-7;-5;-4;-2;-1;1\right\}\)
\(\Rightarrow P\in\left\{2;3;5;-3;-1;0\right\}\)
a) \(A=\frac{x}{x-5}-\frac{10x}{x^2-25}-\frac{5}{x+5}\left(x\ne\pm5\right)\)
\(=\frac{x}{x-5}-\frac{10x}{\left(x-5\right)\left(x+5\right)}-\frac{5}{x+5}\)
\(=\frac{x\left(x+5\right)}{x\left(x-5\right)}-\frac{10x}{\left(x-5\right)\left(x+5\right)}-\frac{5\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{x^2+5x}{\left(x-5\right)\left(x+5\right)}-\frac{10x}{\left(x-5\right)\left(x+5\right)}-\frac{5x-25}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{x^2+5x-10x-5x+25}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}=\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\frac{x-5}{x+5}\)
Vậy \(A=\frac{x-5}{x+5}\left(x\ne\pm5\right)\)
b) Ta có \(A=\frac{x-5}{x+5}\left(x\ne\pm5\right)\)
Để A nhận giá trị nguyên thì \(\frac{x-5}{x+5}\)phải nhận giá trị nguyên
=> \(x-5⋮\)x+5
Ta có x-5=(x+5)-10
Thấy x+5 \(⋮\)x+5 => 10 \(⋮\)x+5 thì \(\left(x+5\right)-10⋮x+5\)
mà x nguyên => x+5 nguyên
=> x+5\(\inƯ\left(10\right)=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)
ta có bảng
| x+5 | -10 | -5 | -2 | -1 | 1 | 2 | 5 | 10 |
| x | -15 | -10 | -7 | -6 | -4 | -3 | 0 | 5 |
| ĐCĐK | tm | tm | tm | tm | tm | tm | tm | ktm |
Vậy x={-15;-10;-7;-6;-4;-3;0} thì \(A=\frac{x-5}{x+5}\)nhận giá trị nguyên
a) ĐKXĐ: \(x\notin\left\{3;-3;-2\right\}\)
Ta có: \(P=\left(\dfrac{2x-1}{x+3}-\dfrac{x}{3-x}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\)
\(=\left(\dfrac{\left(2x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3-10x}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{x+2}{x-3}\)
\(=\dfrac{2x^2-6x-x+3+x^2+3x-3+10x}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+2}{x-3}\)
\(=\dfrac{3x^2+6x}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+2}{x-3}\)
\(=\dfrac{3x\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x+2}\)
\(=\dfrac{3x}{x+3}\)
b) Ta có: \(x^2-7x+12=0\)
\(\Leftrightarrow x^2-3x-4x+12=0\)
\(\Leftrightarrow x\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
Thay x=4 vào biểu thức \(P=\dfrac{3x}{x+3}\), ta được:
\(P=\dfrac{3\cdot4}{4+3}=\dfrac{12}{7}\)
Vậy: Khi \(x^2-7x+12=0\) thì \(P=\dfrac{12}{7}\)
a) ĐKXĐ: \(x\notin\left\{5;-5\right\}\)
b) Ta có: \(A=\dfrac{2x}{x^2-25}+\dfrac{5}{5-x}-\dfrac{1}{x+5}\)
\(=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5}{x-5}-\dfrac{1}{x+5}\)
\(=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{2x-5x-25-x+5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4}{x-5}\)
Để A nguyên thì \(-4⋮x-5\)
\(\Leftrightarrow x-5\inƯ\left(-4\right)\)
\(\Leftrightarrow x-5\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{6;4;7;3;9;1\right\}\)(nhận)
Vậy: Để A nguyên thì \(x\in\left\{6;4;7;3;9;1\right\}\)
\(a,P=\left(\dfrac{2x-1}{x+3}-\dfrac{x}{3-x}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\left(x\ne\pm3;x\ne-2\right)\\ P=\dfrac{2x^2-7x+3+x^2+3x-3+10x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x+2}\\ P=\dfrac{3x^2+6x}{\left(x-3\right)\left(x+2\right)}=\dfrac{3x\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}=\dfrac{3x}{x-3}\\ b,x^2-7x+12=0\\ \Leftrightarrow\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow x=4\left(x\ne3\right)\\ \Leftrightarrow A=\dfrac{3\cdot4}{4-3}=12\\ c,P=\dfrac{3\left(x-3\right)+9}{x-3}=3+\dfrac{9}{x-3}\in Z\\ \Leftrightarrow x-3\inƯ\left(9\right)=\left\{-9;-3;-1;1;3;9\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;4;6;12\right\}\)
a) Ta có: \(A=\left(1+\dfrac{x^2}{x^2+1}\right):\left(\dfrac{1}{x-1}-\dfrac{2x}{x^3+x-x^2-1}\right)\)
\(=\dfrac{2x^2+1}{x^2+1}:\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}\)
\(=\dfrac{2x^2+1}{x^2+1}\cdot\dfrac{\left(x-1\right)\left(x^2+1\right)}{\left(x-1\right)^2}\)
\(=\dfrac{2x^2+1}{x-1}\)
b) Thay \(x=-\dfrac{1}{2}\) vào A, ta được:
\(A=\left(2\cdot\dfrac{1}{4}+1\right):\left(\dfrac{-1}{2}-1\right)\)
\(=\dfrac{3}{2}:\dfrac{-3}{2}=-1\)
c) Để A<1 thì A-1<0
\(\Leftrightarrow\dfrac{2x^2+1}{x-1}-1< 0\)
\(\Leftrightarrow\dfrac{2x^2+1-x+1}{x-1}< 0\)
\(\Leftrightarrow\dfrac{2x^2-x+2}{x-1}< 0\)
\(\Leftrightarrow x-1< 0\)
hay x<1
a: \(N=\dfrac{x^2-5x+5x+25+10x}{\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x-5}{x}\)
\(=\dfrac{\left(x+5\right)^2}{x+5}\cdot\dfrac{1}{x}=\dfrac{x+5}{x}\)
b: N=3/2
=>x+5/x=3/2
=>2x+10=3x
=>-x=-10
=>x=10
c: N nguyên thì x+5 chia hêt cho x
=>5 chia hết cho x
=>\(x\in\left\{1;-1\right\}\)