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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
- Đk : \(\hept{\begin{cases}x-3\ne0\\x-2\ne0\\x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-2\\x\ne2\end{cases}}}\)
- \(P=\frac{\left(2+x\right)^2+4x^2-\left(2-x\right)^2}{\left(x-2\right)\left(x+2\right)}.\frac{x^2\left(2-x\right)}{x\left(x-3\right)}\)\(\Rightarrow P=\frac{8x+4x^2}{\left(x-2\right)\left(x+2\right)}.\frac{x\left(2-x\right)}{x-3}\)\(\Rightarrow p=\frac{4x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}.\frac{x\left(x-2\right)}{3-x}=\frac{4x^2}{3-x}\)
- \(|x-5|=2\)
- nếu \(x\ge5\)=> x-5=2 =>x=7 (TM) => \(P=\frac{4.7^2}{-7+3}=-49\)
- Nếu \(x< 5\)=> x-5 = -2 => x = 3 Loại
a: ĐKXĐ: x<>2; x<>-2; x<>0; x<>3
b: \(P=\left(\dfrac{-\left(x+2\right)}{x-2}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(=\dfrac{-x^2-4x-4+4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)
\(=\dfrac{4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{\left(x-3\right)}=\dfrac{-4x^2\left(x-2\right)}{\left(x+2\right)\left(x-3\right)}\)
c: 2(x-1)=6
=>x-1=3
=>x=4
Thay x=4 vào P, ta đc:
\(P=\dfrac{-4\cdot4^2\cdot\left(4-2\right)}{\left(4+2\right)\left(4-3\right)}=\dfrac{-64\cdot2}{6}=\dfrac{-128}{6}=-\dfrac{64}{3}\)
a: ĐKXĐ: \(x\notin\left\{2;-2;0;3\right\}\)
b: \(P=\left(\dfrac{-\left(x+2\right)}{x-2}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(=\dfrac{-x^2-4x-4+4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x-3}{x\left(2-x\right)}\)
\(=\dfrac{4x^2-8x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)
\(=\dfrac{4x}{x+2}\cdot\dfrac{-x\left(x-2\right)}{x-3}=\dfrac{-4x^2\left(x-2\right)}{\left(x+2\right)\left(x-3\right)}\)
a) ĐKXĐ: x - 2 \(\ne\)0 x \(\ne\)2
x + 2 \(\ne\)0 => x\(\ne\)-2 =>x \(\ne\)\(\pm\)2 và x \(\ne\)-10
x2 - 4 \(\ne\)0 x \(\ne\)\(\pm\)2
x + 10 \(\ne\)0 x \(\ne\)-10
b) Ta có: P = \(\left(\frac{x+5}{x-2}+\frac{3x}{x+2}-\frac{4x^2}{x^2-4}\right)\cdot\frac{x^2+2x}{x+10}\)
P = \(\left(\frac{\left(x+5\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{x\left(x+2\right)}{x+10}\)
P = \(\left(\frac{x^2+2x+5x+10+3x^2-6x-4x^2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{x\left(x+2\right)}{x+10}\)
P = \(\frac{x+10}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x\left(x+2\right)}{x+10}\)
P = \(\frac{x}{x-2}\)
c)Với x \(\ne\)\(\pm\)2 và x \(\ne\)-10
Ta có: x2 - x - 6 = 0
=> x2 - 3x + 2x - 6 = 0
=> x(x - 3) + 2(x - 3) = 0
=> (x + 2)(x- 3) = 0
=> \(\orbr{\begin{cases}x+2=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-2\left(ktm\right)\\x=3\end{cases}}\)
Với x = 3 => P = \(\frac{3}{3-2}=3\)
Câu 1 :
a) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)
b) Để \(P=1\Leftrightarrow\frac{4x^2+4x}{\left(x+1\right)\left(2x-6\right)}=1\)
\(\Leftrightarrow\frac{4x^2+4x-\left(x+1\right)\left(2x-6\right)}{\left(x+1\right)\left(2x-6\right)}=0\)
\(\Rightarrow4x^2+4x-2x^2+4x+6=0\)
\(\Leftrightarrow2x^2+8x+6=0\)
\(\Leftrightarrow x^2+4x+4-1=0\)
\(\Leftrightarrow\left(x+2-1\right)\left(x+2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=-3\left(TMĐKXĐ\right)\end{cases}}\)
Vậy : \(x=-3\) thì P = 1.