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a) P xác định \(\Leftrightarrow\hept{\begin{cases}2x+10\ne0\\x\ne0\\2x\left(x+5\right)\ne0\end{cases}\Leftrightarrow x\ne\left\{-5;0\right\}}\)
b) \(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^2\left(x+2\right)}{2x\left(x+5\right)}+\frac{2\left(x-5\right)\left(x+5\right)}{2x\left(x+5\right)}+\frac{5\left(10-x\right)}{2x\left(x+5\right)}\)
\(P=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+5x^2-x^2-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^2\left(x+5\right)-x\left(x+5\right)}{2x\left(x+5\right)}\)
\(P=\frac{\left(x+5\right)\left(x^2-x\right)}{2x\left(x+5\right)}\)
\(P=\frac{x\left(x-1\right)}{2x}\)
\(P=\frac{x-1}{2}\)
c) Để P = 0 thì \(x-1=0\Leftrightarrow x=1\)( thỏa mãn ĐKXĐ )
Để P = 1/4 thì \(\frac{x-1}{2}=\frac{1}{4}\)
\(\Leftrightarrow4\left(x-1\right)=2\)
\(\Leftrightarrow4x-4=2\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\frac{3}{2}\)( thỏa mãn ĐKXĐ )
d) Để P > 0 thì \(\frac{x-1}{2}>0\)
Mà 2 > 0, do đó để P > 0 thì \(x-1>0\Leftrightarrow x>1\)
Để P < 0 thì \(\frac{x-1}{2}< 0\)
Mà 2 > 0, do đó để P < 0 thì \(x-1< 0\Leftrightarrow x< 1\)
a: ĐKXĐ: \(x\notin\left\{0;-5\right\}\)
\(C=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
Bài 1:
\(a,ĐK:x\ne\pm5\\ b,P=\dfrac{x-5+2x+10-2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x+5}\\ c,P=-3\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{16}{3}\\ d,P\in Z\Leftrightarrow x+5\inƯ\left(1\right)=\left\{-1;1\right\}\\ \Leftrightarrow x\in\left\{-6;-4\right\}\)
Bài 2:
\(a,\Leftrightarrow\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}=0\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow\dfrac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}=0\Leftrightarrow\dfrac{-x}{x+2}=0\Leftrightarrow x=0\)
a: \(P=\dfrac{x\left(x+2\right)}{2\left(x+5\right)}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50-5x+50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
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) ĐKXĐ: \(x\notin\left\{0;-5\right\}\)
Ta có: \(B=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x^2+10x}\)
\(=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2}{2x\left(x+5\right)}+\dfrac{2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}-\dfrac{5x-50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50-5x+50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+5x-x-5}{2\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)-\left(x+5\right)}{2\left(x+5\right)}\)
\(=\dfrac{\left(x+5\right)\left(x-1\right)}{2\left(x+5\right)}\)
\(=\dfrac{x-1}{2}\)
b) Để B=0 thì \(\dfrac{x-1}{2}=0\)
\(\Leftrightarrow x-1=0\)
hay x=1(nhận)
Vậy: Để B=0 thì x=1
Để \(B=\dfrac{1}{4}\) thì \(\dfrac{x-1}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow4\left(x-1\right)=2\)
\(\Leftrightarrow4x-4=2\)
\(\Leftrightarrow4x=6\)
hay \(x=\dfrac{3}{2}\)(nhận)
Vậy: Để \(B=\dfrac{1}{4}\) thì \(x=\dfrac{3}{2}\)
c) Thay x=3 vào biểu thức \(B=\dfrac{x-1}{2}\), ta được:
\(B=\dfrac{3-1}{2}=\dfrac{2}{2}=1\)
Vậy: Khi x=3 thì B=1
d) Để B<0 thì \(\dfrac{x-1}{2}< 0\)
\(\Leftrightarrow x-1< 0\)
\(\Leftrightarrow x< 1\)
Kết hợp ĐKXĐ, ta được:
\(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-5\right\}\end{matrix}\right.\)
Vậy: Để B<0 thì \(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-5\right\}\end{matrix}\right.\)
Để B>0 thì \(\dfrac{x-1}{2}>0\)
\(\Leftrightarrow x-1>0\)
hay x>1
Kết hợp ĐKXĐ, ta được: x>1
Vậy: Để B>0 thì x>1
\(\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{5x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
a ) ĐKXĐ : \(x\ne0,x\ne-5\)
b ) Rút gọn trước cái đã
\(\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{5x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+10x^2+50x-10x-50+50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+12x^2+35x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x+7\right)}{2x\left(x+5\right)}=\dfrac{x+7}{2x}\)
Khi \(A=1\), thì :
\(\dfrac{x+7}{2x}=1\Leftrightarrow x=7\)
Khi A = 3, thì :
\(\dfrac{x+7}{2x}=3\Leftrightarrow x=-1.\)
Bài 2 :
a ) ĐKXĐ : x\(\ne-3;2\)
b ) \(\dfrac{x-2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}+\dfrac{1}{2-x}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\)
c ) Khi \(A=-\dfrac{3}{4}\), thì :
\(\dfrac{x-4}{x-2}=-\dfrac{3}{4}\)
\(\Leftrightarrow4x-16=-3x+6\)
\(\Leftrightarrow x=\dfrac{22}{7}\)
d ) Ta có :
\(A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)
Để A nguyên thi \(x-2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Thay vào rồi tìm ra nếu x có trong đkxđ thì loại .
e ) \(x^2-9=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Thay từng x vào A là tìm ra
B1:
a) A = \(\dfrac{1}{x+2}+\dfrac{x^2-x-2}{x^2-7x+10}-\dfrac{2x-4}{x-5}\)
= \(\dfrac{1}{x+2}+\dfrac{\left(x^2-2x\right)+\left(x-2\right)}{\left(x^2-2x\right)-\left(5x-10\right)}-\dfrac{2\left(x-2\right)}{x-5}\)
= \(\dfrac{1}{x+2}+\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{2\left(x-2\right)}{x-5}\) [ĐKXĐ: x ≠ -2; x ≠ 5]
= \(\dfrac{x-5}{\left(x+2\right)\left(x-5\right)}+\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-5\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)\left(x+2\right)}{\left(x-5\right)\left(x+2\right)}\)
= \(\dfrac{-x^2+4x+5}{\left(x+2\right)\left(x-5\right)}\)
= \(\dfrac{-x\left(x-5\right)-\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}\)
= \(\dfrac{\left(x-5\right)\left(-x-1\right)}{\left(x-5\right)\left(x+2\right)}\)
= \(-\dfrac{x+1}{x+2}\)
b) Thay x = 3 vào A, ta có:
A = \(-\dfrac{3+1}{3+2}=-\dfrac{4}{5}\)
c) A = 1
<=> \(-\dfrac{x+1}{x+2}\)= 1 <=> -(x + 1) = x + 2 <=> -x - 1 = x + 2
<=> -2x = 3 <=> x = \(\dfrac{-3}{2}\)
d) A = \(\dfrac{-\left(x+1\right)}{x+2}\)= \(\dfrac{-\left(x+2\right)+1}{x+2}\)= -1 + \(\dfrac{1}{x+2}\)
A đạt giá trị nguyên khi 1 chia hết cho x + 2 hay x + 2 ∈ Ư(1) = {1;-1}
* x + 2 = 1 <=> x = -1
* x + 2 = -1 <=> x = -3
B2: M = \(\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
= \(\dfrac{x\left(x+2\right)}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{5\left(10-x\right)}{2x\left(x+5\right)}\)[ĐKXĐ: x ≠ 0; x ≠ -5
= \(\dfrac{x^2\left(x+2\right)+2\left(x+5\right)\left(x-5\right)+5\left(10-x\right)}{2x\left(x+5\right)}\)
= \(\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
= \(\dfrac{x^2+4x-5}{2\left(x+5\right)}\)
= \(\dfrac{\left(x^2+5x\right)-\left(x+5\right)}{2\left(x+5\right)}\)
\(\dfrac{\left(x+5\right)\left(x-1\right)}{2\left(x+5\right)}=\dfrac{x-1}{2}\)
b) Thay x = 3 vào M, ta có:
M = \(\dfrac{3-1}{2}=1\)
Thay x = 5 vào M, ta có:
M = \(\dfrac{5-1}{2}=2\)