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a: ĐKXĐ: x<>0; x<>-3
b: \(=\dfrac{x^2+6x+9}{x\left(x+3\right)}\cdot\dfrac{2}{x+3}=\dfrac{2}{x}\)
c: Khi x=1/5 thì A=2:1/5=10
a,ĐK: \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)
c, Với x = 4 thỏa mãn ĐKXĐ thì
\(A=\frac{-3}{4-3}=-3\)
d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)
Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
a) ĐKXĐ: \(x\ne-10;x\ne0;x\ne-5\)
b) \(P=\dfrac{x^2+2x}{2x+20}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+2x}{2\left(x+10\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x^2+2x\right)\left(x+5\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{2\left(x-5\right)\left(x+10\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{\left(50-5x\right)\left(x+10\right)}{2x\left(x+5\right)\left(x+10\right)}\)
\(=\dfrac{x^4+7x^3+10x^2+2x^2+10x-100+500-5x^2}{2x\left(x+10\right)\left(x+5\right)}\)
\(=\dfrac{x^4+7x^3+7x^2+10x+400}{2x\left(x+10\right)\left(x+5\right)}\)
c) \(P=0\Rightarrow x^4+7x^3+7x^2+10x+400=0\Leftrightarrow...\)
Số xấu thì câu c, d làm cũng như không. Bạn xem lại đề.
a) Phân thức B xác định \(\Leftrightarrow\hept{\begin{cases}2x-2\ne0\\x^2-1\ne0\\2x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne\left\{\pm1\right\}\\x\ne-1\end{cases}\Leftrightarrow}x\ne\left\{\pm1\right\}}\)
b) \(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\cdot\frac{4x^2-4}{5}\)
\(B=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{3\cdot2}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{\left(2x\right)^2-2^2}{5}\)
\(B=\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(2x-2\right)\left(2x+2\right)}{5}\)
\(B=\frac{10\cdot2\left(x-1\right)\cdot2\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)\cdot5}\)
\(B=\frac{40\left(x-1\right)\left(x+1\right)}{10\left(x-1\right)\left(x+1\right)}\)
\(B=4\)
Vậy với mọi giá trị của x thì B luôn bằng 4
Vậy giá trị của B không phụ thuộc vào biến ( đpcm )
\(Giải:\)
\(ĐKXĐ:x\ne\pm1\)
\(B=\left[\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right]=\left[\frac{x+1}{2x-2}+\frac{12}{4x^2-4}-\frac{x+3}{2x+2}\right]\)
\(=\left[\frac{x+1}{2x-2}+\frac{12}{\left(2x+2\right)\left(2x-2\right)}-\frac{x+3}{2x+2}\right]\)
\(=\left[\frac{\left(x+1\right)\left(2x+2\right)}{\left(2x+2\right)\left(2x-2\right)}+\frac{12}{\left(2x+2\right)\left(2x-2\right)}-\frac{\left(x+3\right)\left(2x-2\right)}{\left(2x-2\right)\left(2x+2\right)}\right]\)
\(=\frac{2x^2+4x+14-2x^2+2x-6x+6}{\left(2x-2\right)\left(2x+2\right)}\)
\(=\frac{6}{\left(2x-2\right)\left(2x+2\right)}\)
1)
a) \(5x\left(x^2-3x+\dfrac{1}{5}\right)\)
\(=5x^3-15x^2+x\)
b) \(\left(x-3\right)\left(2x-1\right)\)
\(=2x^2-x-6x+3\)
\(=2x^2-7x+3\)
2)
a) \(3x^2-15xy\)
\(=3x\left(x-5y\right)\)
b) \(x^2-6x-y^2+9\)
\(=\left(x^2-6x+9\right)-y^2\)
\(=\left(x-3\right)^2-y^2\)
\(=\left(x-3-y\right)\left(x-3+y\right)\)
c) \(x^2+3x+2\)
\(=\left(x^2+x\right)+\left(2x+2\right)\)
\(=x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x+2\right)\)
bài 4
vì x2+1 >0 với mọi x , do đó GT của Q luôn xác định với mọi x
Q=\(\dfrac{2x^2-4x+5}{x^2+1}=\dfrac{\left(3x^2+3\right)+\left(2x^2-4x+2\right)}{x^2+1}\)=\(\dfrac{3\left(x^2+1\right)+2\left(x-1\right)^2}{x^2+1}=\dfrac{3\left(x^2+1\right)}{x^2+1}+\dfrac{2\left(x-1\right)^2}{x^2+1}\)=\(3+\dfrac{2\left(x-1\right)^2}{x^2+1}\)
Do (x-1)2 ≥ 0
=>2(x-1)2 ≥ 0
x2+1 ≥ 0
=>\(\dfrac{2\left(x-1\right)^2}{x^2+1}\ge0\)
=>\(3+\dfrac{2\left(x-1\right)^2}{x^2+1}\ge3\)
=> Q ≥ 3
=>GTNN của Q =3 khi
x-1=0
=>x=1
Vậy GTNN của Q =3 khi x=1