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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
1.a)\(\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}\)
\(=\frac{x^3}{\left(x+2\right)\left(x-2\right)}-\frac{x}{x-2}-\frac{2}{x+2}\)
Để biểu thức được xác định thì:\(\left(x+2\right)\left(x-2\right)\ne0\)\(\Rightarrow x\ne\pm2\)
\(\left(x+2\right)\ne0\Rightarrow x\ne-2\)
\(\left(x-2\right)\ne0\Rightarrow x\ne2\)
Vậy để biểu thức xác định thì : \(x\ne\pm2\)
b) để C=0 thì ....
1, c , bn Nguyễn Hữu Triết chưa lm xong
ta có : \(/x-5/=2\)
\(\Rightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=3\end{cases}}\)
thay x = 7 vào biểu thứcC
\(\Rightarrow C=\frac{4.7^2\left(2-7\right)}{\left(7-3\right)\left(2+7\right)}=\frac{-988}{36}=\frac{-247}{9}\)KL :>...
thay x = 3 vào C
\(\Rightarrow C=\frac{4.3^2\left(2-3\right)}{\left(3-3\right)\left(3+7\right)}\)
=> ko tìm đc giá trị C tại x = 3
a) \(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)
\(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}\)
\(\Leftrightarrow A=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2\left(x+2\right)}{x-3}\)
\(\Leftrightarrow A=\frac{2x-9-\left(x-3\right)\left(x+3\right)+2\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{\left(x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{x+4}{x-3}\)
b) Để \(A\inℤ\)
\(\Leftrightarrow\frac{x+4}{x-3}\inℤ\)
\(\Leftrightarrow1+\frac{7}{x-3}\inℤ\)
\(\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Leftrightarrow x\in\left\{2;4;-4;10\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{2;4;-4;10\right\}\)
c) Để \(A=\frac{3}{5}\)
\(\Leftrightarrow\frac{x+4}{x-3}=\frac{3}{5}\)
\(\Leftrightarrow5x+20=3x-9\)
\(\Leftrightarrow2x+29=0\)
\(\Leftrightarrow x=-\frac{29}{2}\)
d) Để \(A< 0\)
\(\Leftrightarrow\frac{x+4}{x-3}< 0\)
\(\Leftrightarrow1+\frac{7}{x-3}< 0\)
\(\Leftrightarrow\frac{-7}{x-3}< 1\)
\(\Leftrightarrow-7< x-3\)
\(\Leftrightarrow x>-4\)
e) Để \(A>0\)
\(\Leftrightarrow\frac{x+4}{x-3}>0\)
\(\Leftrightarrow1+\frac{7}{x-3}>0\)
\(\Leftrightarrow\frac{-7}{x-3}>1\)
\(\Leftrightarrow-7>x-3\)
\(\Leftrightarrow x< -4\)
a: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{4}{x-2}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)
\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{6\left(x+1\right)\cdot x\left(x+2\right)}{3\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x}{x-2}\)
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 \(x-1< 0\Leftrightarrow x< 1\)
ĐKXĐ: \(x\ne-5;0\)
\(A=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x.\left(x+5\right)}\)
\(=\frac{\left(x^2+2x\right).x}{2x.\left(x+5\right)}+\frac{2.\left(x+5\right).\left(x-5\right)}{2x.\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)
\(=\frac{x^3+2x^2}{2x\left(x+5\right)}+\frac{2.\left(x^2-25\right)}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}=\frac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\frac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\frac{x-1}{2}\)
b. \(A=0\Leftrightarrow\frac{x-1}{2}=0\Rightarrow x-1=0\Leftrightarrow x=1\)
\(A=\frac{1}{4}\Leftrightarrow\frac{x-1}{2}=\frac{1}{4}\Leftrightarrow4x-4=2\Leftrightarrow4x-6=0\Leftrightarrow x=\frac{3}{2}\)
c. Với x=0 thì \(A=\frac{0-1}{2}=-\frac{1}{2}\)
Với x=2 thì: \(A=\frac{2-1}{2}=\frac{1}{2}\)
d. \(A>0\Leftrightarrow\frac{x-1}{2}>0\Rightarrow\left(x-1\right).2>0\Rightarrow x-1>0\Leftrightarrow x>1\)
\(A< 0\Leftrightarrow\frac{x-1}{2}< 0\Leftrightarrow\left(x-1\right).2< 0\Leftrightarrow x-1< 0\Leftrightarrow x< 1;x\ne-5,0\)
e. \(A=\frac{x-1}{2}\inℤ\Rightarrow x-1\in Z\Rightarrow x\inℤ\)
Và \(\left(x-1\right)⋮2\Rightarrow x:2dư1\)
Vậy \(A\in Z\Leftrightarrow x\inℤ\)và x chia 2 dư 1
d. Bổ sung x khác -5 nữa nhé