\(x^2\left(x+1\right)-x\left(x+1\right)\)
\(=x^3+x^2-x^2-x\)
\(=x^3-x\)
\(=x\left(x^2-1\right)\)
\(=x\left(x-1\right)\left(x+1\right)\)
Tìm x biết :
a) \(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)
b) \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^3+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x+4\right)\left(x-4\right)+3x^2\)
c) \(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2\)
d) \(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
Rút gọn các biểu thức sau:
A= \(\left(x+1\right).\left(x^2-x+1\right)+2.\left(x+1\right)-x.\left(x^2+2\right).\)
B= \(\left(5x+1\right).\left(x+7\right)-5x.\left(x-1\right).\)
Tính
\(A=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+9\right)\left(x+10\right)}\)
Rút gọn biểu thức sau:
a, \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^2+3\left(x-1\right)\left(x+1\right)\)
b, \(\left(x^4-5x^2+25\right)\left(x^2+5\right)-\left(2+x^2\right)^2+3\left(1+x^2\right)^2\)
Tìm \(x\)
a) \(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)
tìm x
\(\left[\left(x+1\right)\left(2-x\right)-\left(x^2-1\right)\right]:\left(x+1\right)=2\)
1. \(\frac{1}{2}x^2-\left(\frac{1}{2}x-4\right)\frac{1}{2}x=-14\)
2. \(3\left(1-4x\right)\left(x-1\right)+4\left(3x-2\right)\left(x+3\right)=-27\)
3. \(6x\left(5x+3\right)+3x\left(1-10x\right)=7\)
4. \(\left(3x-3\right)\left(5-21x\right)+\left(7x+4\right)\left(9x-5\right)=44\)
5. \(\left(-2+x^3\right)\left(-2+x^2\right)\left(-2+x^2\right)=1\)
CMR các biểu thức sau không phụ thuộc vào biến:
\(\left(x-3\right)\left(x+2\right)+\left(x-1\right)\left(x+1\right)-\left(x-\dfrac{1}{2}\right)\left(x-\dfrac{1}{2}\right)-x^2\)
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 \(x=\dfrac{1}{2}\)
Phân tích đa thức thành nhân tử
a) \(\left(x^2+x\right)^2+3\left(x^2+x\right)+2\)
b) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
c) \(\left(x^2+x+1\right)\left(x^2+3x+1\right)+x^2\)
d) \(\left(1+x^2\right)^2-4x\left(1-x^2\right)\)
e) \(\left(x^2-8\right)^2+36\)
f) \(81x^4+4\)
