Lời giải:
Điều kiện: \(x\ne0\)
\(\left(x+1+\dfrac{1}{x}\right)=\left(x-1-\dfrac{1}{x}\right)\)
\(\Leftrightarrow1+\dfrac{1}{x}=-1-\dfrac{1}{x}\\ \Leftrightarrow\dfrac{1}{x}+\dfrac{1}{x}=-1-1\\ \Leftrightarrow\dfrac{2}{x}=-2\\ \Leftrightarrow-2x=2\\ \Leftrightarrow x=-1\left(tm\right)\)
Vậy ...
Tính
\(A=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+9\right)\left(x+10\right)}\)
f, \(x^2-x+25\)
\(=x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2+25\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{99}{4}\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\) ≥ 0 nên \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{99}{4}\ge\dfrac{99}{4}\) với mọi x
Dấu "=" xảy ra ⇔ \(x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy GTNN của đa thức là \(\dfrac{99}{4}\) tại \(x=\dfrac{1}{2}\)
CMR các biểu thức sau không phụ thuộc vào biến:
\(\left(x-3\right)\left(x+2\right)+\left(x-1\right)\left(x+1\right)-\left(x-\dfrac{1}{2}\right)\left(x-\dfrac{1}{2}\right)-x^2\)
Thực hiện phép tính các đa thức sau
a) \(\left(3x^2-2x+5\right)\left(2x^2-3x+1\right)\)
b) \(\left(\dfrac{3}{2}x^2-\dfrac{2}{3}x-\dfrac{5}{3}\right)\left(4x^2-\dfrac{3}{2}x-3\right)\)
c) \(\left(\dfrac{3}{4}x^2+2x-\dfrac{1}{3}\right)\left(4x^2-\dfrac{3}{2}x-3\right)\)
d) \(\left(-\dfrac{1}{3}+2x-x^2\right)\left(-2x^2-\dfrac{1}{2}x+2\right)\)
e) \(\left(3xy+\dfrac{1}{2}x\right)\left(3x^{2y}-3xy^2-1\right)\)
Rút gọn \(\left[\dfrac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\dfrac{1-2x^2+4x}{x^3-1}+\dfrac{1}{x-1}\right]:\dfrac{x^2+x}{x^3+x}\)
Thực hiện phép tính\(\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2\left(x+1\right)}\right):\dfrac{x}{x^2-1}\)
Rút gọn:
\(C=\left[\left(1+\dfrac{1}{x}\right)\cdot\dfrac{2}{x^3+3x^2+3x+1}+\left(1+\dfrac{1}{x^2}\right)\cdot\dfrac{1}{1+2x+x^2}\right]:\dfrac{x-1}{x^3}\)
Cho biểu thức:
A\(=\left(\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\dfrac{2x^2+4x-1}{x^3+1}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)
a/ Rút gọn A
b/ Tìm x ∈ Z để A nguyên
\(B=\left[\dfrac{3}{x+1}+\left(\dfrac{3}{x+1}-\dfrac{x}{x^2+2x+1}\right):\dfrac{2x^2+3x}{x+1}\right]:\dfrac{1+3x}{x^2+x}\)
