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a ) Gọi \(A=\dfrac{3x^2-x}{9x^2-6x+1}\)
Ta có : \(A=\dfrac{x\left(3x-1\right)}{\left(3x\right)^2-2.3x.1+1}=\dfrac{x\left(3x-1\right)}{\left(3x-1\right)^2}=\dfrac{x}{3x-1}\)
Thay x = - 8 và biểu thức A ta được :
\(A=\dfrac{-8}{3.\left(-8\right)-1}=\dfrac{8}{25}\)
Vậy giá trị của biểu thức A là \(\dfrac{8}{25}\) tại x = - 8
b ) Gọi \(B=\dfrac{x^2+3x+2}{x^3+2x^2-x-2}\)
Ta có \(B=\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{x^2\left(x+2\right)-\left(x+2\right)}=\dfrac{x\left(x+1\right)+2\left(x+1\right)}{\left(x^2-1\right)\left(x+2\right)}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}=\dfrac{1}{x-1}\)
Thay x = 1000001 và biểu thức B ta được :
\(B=\dfrac{1}{1000001-1}=\dfrac{1}{100000}\)
Vậy giá trị của biểu thức B là \(\dfrac{1}{1000000}\) tại x = 1000001
1. ĐKXĐ: \(x\ne\pm1\)
2. \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x+3}{x+1}\right)\cdot\dfrac{x+1}{2}\)
\(=\dfrac{\left(x+1\right)^2-\left(x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{x^2+2x+1-x^2+4x-3}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{6x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{2\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x-3}{x-1}\)
3. Tại x = 5, A có giá trị là:
\(\dfrac{5-3}{5-1}=\dfrac{1}{2}\)
4. \(A=\dfrac{x-3}{x-1}\) \(=\dfrac{x-1-3}{x-1}=1-\dfrac{3}{x-1}\)
Để A nguyên => \(3⋮\left(x-1\right)\) hay \(\left(x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(tmđk\right)\\x=0\left(tmđk\right)\\x=4\left(tmđk\right)\\x=-2\left(tmđk\right)\end{matrix}\right.\)
Vậy: A nguyên khi \(x=\left\{2;0;4;-2\right\}\)
1, a)
Ta có:
\(x^2+2x+1=\left(x+1\right)^2\)
Thay x=99 vào ta có:
\(\left(99+1\right)^2=100^2=10000\)
b) Ta có:
\(x^3-3x^2+3x-1=\left(x-1\right)^3\)
Thay x=101 vào ta có:
\(\left(101-1\right)^3=100^3=1000000\)
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{x^2-4+10-x^2}{x+2}\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-1}{x-2}\)
b: Khi x=1/2 thì \(B=\dfrac{-1}{\dfrac{1}{2}-2}=\dfrac{2}{3}\)
Khi x=-1/2 thì B=2/5
c: Để B nguyên thì \(x-2\in\left\{1;-1\right\}\)
hay \(x\in\left\{3;1\right\}\)
a, đk : x khác -2 ; 2
\(B=\left(\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}:\dfrac{6}{x+2}=\dfrac{1}{2-x}\)
b, Ta có \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2};x=-\dfrac{1}{2}\)
Với x = 1/2 ta được \(B=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{2}{3}\)
Với x = -1/2 ta được \(B=\dfrac{1}{2+\dfrac{1}{2}}=\dfrac{2}{5}\)
c, \(\dfrac{1}{2-x}\Rightarrow2-x\inƯ\left(1\right)=\left\{\pm1\right\}\)
| 2-x | 1 | -1 |
| x | 1 | 3 |
a:
ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
b: \(A=\left(\dfrac{x-2}{2x-2}+\dfrac{3}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(1-\dfrac{x-3}{x+1}\right)\)
\(=\left(\dfrac{x-2}{2\left(x-1\right)}+\dfrac{3}{2\left(x-1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right):\dfrac{x+1-x+3}{x+1}\)
\(=\dfrac{\left(x-2\right)\left(x+1\right)+3\left(x+1\right)-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{x^2-x-2+3x+3-x^2-2x+3}{2\left(x-1\right)}\cdot\dfrac{1}{2}\)
\(=\dfrac{-2}{4\left(x-1\right)}=\dfrac{-1}{2\left(x-1\right)}\)
Khi x=2005 thì \(A=\dfrac{-1}{2\cdot\left(2005-1\right)}=-\dfrac{1}{4008}\)
Vì x=1 không thỏa mãn ĐKXĐ
nên khi x=1 thì A không có giá trị
c: Để A=-1002 thì \(\dfrac{-1}{2\left(x-1\right)}=-1002\)
=>\(2\left(x-1\right)=\dfrac{1}{1002}\)
=>\(x-1=\dfrac{1}{2004}\)
=>\(x=\dfrac{1}{2004}+1=\dfrac{2005}{2004}\left(nhận\right)\)
a) \(\left(\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\right):\left(\frac{1-x}{x+1}\right)\)
= \(\left(\frac{x+3}{x-2}-\frac{x+2}{x-3}+\frac{x+2}{x^2-2x-3x+6}\right):\left(\frac{1-x}{x+1}\right)\)
= \(\left(\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\right):\left(\frac{1-x}{x+1}\right)\)
= \(\left(\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}\right).\frac{x+1}{1-x}\)
=\(\frac{-3+x}{\left(x-2\right)\left(x-3\right)}.\frac{x+1}{1-x}\)
=\(\frac{1}{\left(x-2\right)}.\frac{x+1}{1-x}\)
=\(\frac{x+1}{\left(x-2\right)\left(1-x\right)}\)
b) Để A >1 \(\Leftrightarrow\frac{x+1}{\left(x-2\right)\left(1-x\right)}>1\)
\(\Leftrightarrow\frac{-\left(1-x\right)\left(3-x\right)}{\left(x-2\right)\left(1-x\right)}\)
\(\Leftrightarrow\frac{x-3}{x-2}>0\)
\(\Rightarrow\orbr{\begin{cases}x-3\ge0\\x-2>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\ge3\\x>2\end{cases}\Leftrightarrow}x\ge3}\)
\(\Rightarrow\orbr{\begin{cases}x-3< 0\\x-2< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 3\\x< 2\end{cases}\Leftrightarrow}x< 2}\)
Vậy ...
Ta có: x 3 + 2 x 2 - x - 2 = x 2 x + 2 - x + 2 = x 2 - 1 . x + 2 = x + 1 x - 1 x + 2
Do đó, để x 3 + 2 x 2 - x - 2 ≠ 0 khi (x + 2)(x – 1)(x + 1) ≠ 0 ⇒ x ≠ - 2 và x ≠ ± 1
Ta có: x = 1000001 thỏa mãn điều kiện.
Thay x = 1000001 vào biểu thức ta được: