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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 \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
\(A=\left(\frac{2X-1}{x^2-4}+\frac{x+2}{x^2-x-2}\right):\frac{x-2}{x^2+3x+2}ĐK:x\ne\left\{2,-2,-1\right\}\)
a) \(A=\left[\frac{\left(2x-1\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x+2}{\left(x+1\right)\left(x-2\right)}\right]:\frac{x-2}{\left(x+2\right)\left(x+1\right)}\)
\(A=\left[\frac{\left(2x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}\frac{\left(x+2\right)\left(x+2\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}\right].\frac{\left(x+2\right)\left(x+1\right)}{x-2}\)
\(A=\frac{2x^2+x-1+x^2+4x.4}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}.\frac{\left(x+2\right)\left(x+1\right)}{\left(x-2\right)}\)
\(A=\frac{3x^2+5x+3}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}.\frac{\left(x+2\right)\left(x+1\right)}{\left(x-2\right)}\)
\(A=\frac{3x^2+5x+3}{\left(x-2\right)^2}\)
Ta có :\(3x^2+5x+3\)
\(=3\left(x^2+\frac{5}{3}x+1\right)\)
\(=3\left[x^2+2.\frac{5}{6}x+\frac{25}{36}+\frac{9}{36}\right]\)
\(=3\left[\left(x+\frac{5}{6}\right)^2+\frac{9}{36}\right]>0\)
Mà \(\left(x-2\right)^2>0\)
\(\Rightarrow A>0\left(dpcm\right)\)
\(b,A=11\Leftrightarrow\frac{3x^2+5x+3}{\left(x-2\right)^2}=11\)
\(\Rightarrow3x^2+5x+3=11.\left(x-2\right)^2\)
\(\Rightarrow3x^2+5x+3=11.\left(x^2-4x+4\right)\)
\(\Rightarrow8x^2-49x+41=0\)
\(\Rightarrow8x^2-8x-41x+41=0\)
\(\Rightarrow8x\left(x-1\right)-41\left(x-1\right)=0\)
\(\Rightarrow\left(8x-41\right)\left(x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}8x-41=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{41}{8}\\x=1\end{cases}}}\)(Thỏa mãn)
a: ĐKXĐ: x<>-1
b: \(P=\left(1-\dfrac{x+1}{x^2-x+1}\right)\cdot\dfrac{x^2-x+1}{x+1}\)
\(=\dfrac{x^2-x+1-x-1}{x^2-x+1}\cdot\dfrac{x^2-x+1}{x+1}=\dfrac{x^2-2x}{x+1}\)
c: P=2
=>x^2-2x=2x+2
=>x^2-4x-2=0
=>\(x=2\pm\sqrt{6}\)
a) \(A=\left(\frac{2+x}{2-x}-\frac{2-x}{2+x}-\frac{4x^2}{x^2-4}\right):\frac{x^2-6x+9}{\left(2-x\right)\left(x-3\right)}\)(ĐKXĐ: \(\hept{\begin{cases}x\ne\pm2\\x\ne3\end{cases}}\))\(=\left[\frac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{4-x^2}\right]:\frac{\left(x-3\right)^2}{\left(2-x\right)\left(x-3\right)}\)\(=\frac{4x\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}.\frac{2-x}{x-3}=\frac{4x}{x-3}\)
b) l\(x-5\)l\(=2\Leftrightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\left(n\right)\\x=3\left(l\right)\end{cases}\Rightarrow A=\frac{4.7}{7-3}=\frac{28}{4}=7}\)
c)
* Để A có giá trị là một số nguyên thì \(A=\frac{4x}{x-3}=\frac{4x-12+12}{x-3}=4+\frac{12}{x-3}\)là một số nguyên hay \(\frac{12}{x-3}\)là một số nguyên \(\Rightarrow x-3\inƯ\left(12\right)\Rightarrow S=\left(-9;-3;-1;0;1;4;5;6;7;9;15\right)\)(1)
* Để \(A=4+\frac{12}{x-3}< 4\Leftrightarrow\frac{12}{x-3}< 0\) thì \(x-3< 0\Leftrightarrow x< 3\)(2)
(1)(2) \(\Rightarrow S=\left(-9;-3;-1;0;1\right)\)
thiếu đề : \(\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}.\)
Bài 2 :
a, Để \(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\frac{4^2-4}{5}\)
\(\Rightarrow\hept{\begin{cases}2x-2\ne0\\x^2-1\ne0\\2x+2\ne0\end{cases}}\Rightarrow\orbr{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)
b,\(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\frac{4x^2-4}{5}\)
\(B=\left[\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x+1\right)\left(x-1\right)}-\frac{x+3}{2\left(x+1\right)}\right].\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(B=\left[\frac{x^2+2x+1}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+2x-3}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(B=\left[\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(B=\frac{4}{2\left(x-1\right)\left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(B=\frac{8}{5}\)
=> giá trị của B ko phụ thuộc vào biến x
bài 1
=\(^{\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x+1\right)^2}\)
=\(\left(2x+1+2x-1\right)^2\)
=\(\left(4x\right)^2\)
=\(16x^2\)
Tại x=100 thay vào biểu thức trên ta có:
16*100^2=1600000
a) \(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\) (ĐKXĐ: \(x\ne\pm1\) )
\(=\left(\frac{x+1+2\left(1-x\right)-5+x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\left(\frac{x+1+2-2x-5+x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\left(\frac{-2}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\frac{2}{x^2-1}.\frac{x^2-1}{1-2x}=\frac{2}{1-2x}\)
b) Để x nhận giá trị nguyên <=> 2 chia hết cho 1 - 2x
<=> 1-2x thuộc Ư(2) = {1;2;-1;-2}
Nếu 1-2x = 1 thì 2x = 0 => x= 0
Nếu 1-2x = 2 thì 2x = -1 => x = -1/2
Nếu 1-2x = -1 thì 2x = 2 => x =1
Nếu 1-2x = -2 thì 2x = 3 => x = 3/2
Vậy ....
