\(B=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+4x-5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+4x\right)^2-25\ge-25\)
\(\Rightarrow A_{min}=-25\)
Tìm GTNN của \(A=\frac{x^3-2x^2+4x-8}{\left(x-2\right)\left(x^2+1\right)+2x\left(x-2\right)}\)
Bài 1: Thực hiện phép tính:
\(3x.\left(x^2-5x+\dfrac{1}{3}\right)\\ \left(x-2\right).\left(5x-1\right)\\ 5x.\left(3x^2-4x+1\right)\\ \left(x+3\right)\left(x^2+3x-5\right)\)
Rút gọn phân thức sau:
A = \(\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1}{x^2+5x+5}\)
B = \(\dfrac{\left|x-1\right|+\left|x\right|+x}{3x^2-4x+1}\) với x < 0
thực hiện phép tính: \(\dfrac{4\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\dfrac{x^2-25}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(3x-3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)
Tính
a)\(\left(\dfrac{\left(x-1\right)^2}{\left(3x+x-1\right)^2}-\dfrac{1-2x^2+4x}{x^3-1}+\dfrac{1}{x-1}\right):\dfrac{x^2+x}{x^2+1}\)
b)\(\left(\dfrac{3\left(x+2\right)}{2\left(x^3+x^2+x+1\right)}+\dfrac{2x^2-x+10}{2\left(x^3+x^2+x+1\right)}\right):\left(\dfrac{5}{x^2+1}+\dfrac{3}{2\left(x+1\right)}-\dfrac{3}{2\left(x-1\right)}\right).\dfrac{2}{x-1}\)
c)\(\left(\dfrac{x^2}{x^2-5x+6}+\dfrac{x^2}{x^2-3x+2}\right):\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
Cho \(M=\left[\dfrac{\left(x-1\right)^2}{3x+\left(x+1\right)^2}-\dfrac{1-2x^2+4x}{x^3-1}+\dfrac{1}{x-1}\right]:\dfrac{2x}{x^3+x}\)
a, Rút gọn biểu thức M
b, Tìm giá trị của x để M đạt GTNN
Giải pt sau: \(10\left(x+\dfrac{1}{x}\right)^2+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x-5\right)^2-5\)
giải phương trình
a.\(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
b.\(x\left(2x-9\right)=3x\left(x-5\right)\)
c.\(3x-15=2x\left(x-5\right)\)
d.\(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
e.\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
B1: Tính:
\(B=\dfrac{4.\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\dfrac{x^2-25}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)
B2: Xác định a, b, c:
a, \(\dfrac{10x-4}{x^3-4x}=\dfrac{a}{x}+\dfrac{b}{1-2}+\dfrac{c}{n+2}\) với mọi x khác 0, x khác \(\pm2\)
b, \(\dfrac{1}{x^3-1}=\dfrac{a}{x-1}+\dfrac{bx+c}{x^2+x+1}\)
Help me!!!
