a.
\(A=\dfrac{x^2-4x+1}{x^2}\)
\(\Rightarrow A=\dfrac{x^2-4x+4-3}{x^2}\)
\(\Rightarrow A=\dfrac{\left(x-2\right)^2-3}{x^2}\)
Ta có: \(\left(x-2\right)^2-3\ge-3\)
\(\Rightarrow x=2\)
Khi đó ta được Min A = \(\dfrac{\left(2-2\right)-3}{2^2}\ge\dfrac{-3}{4}\)
Vậy Min A = \(\dfrac{-3}{4}\)
Tìm GTNN của :
a) \(A=\dfrac{x^2-4x+1}{x^2}\)
b) \(B=\dfrac{4x^2-6x+1}{\left(2x-1\right)^2}\)
Giải các phương trình sau :
a)\(\dfrac{5x+2}{6}\)\(-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
b)\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
c)\(2x^3 +6x^2=x^2+3x\)
d)\(\left|x-4\right|+3x=5\)
Giải các phương trình :
a) \(\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)
b) \(\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
c) \(\dfrac{x+2}{3}+\dfrac{3\left(2x-1\right)}{4}-\dfrac{5x-3}{6}=x+\dfrac{5}{12}\)
1. Cho biểu thức:
A = \(x-2+\dfrac{6x^2-3x}{x^3+2x^2}+\left(\dfrac{x+1}{x^2-1}+\dfrac{2}{x+1}-\dfrac{3}{x}\right):\dfrac{x+2}{x^2-1}\)
a) Rút gọn A.
b) Tìm x sao cho A nhận giá trị âm.
2. Giải phương trinh: \(\dfrac{a+b-x}{c}+\dfrac{b+c-x}{a}+\dfrac{a+c-x}{b}=1-\dfrac{4x}{a+b+c}\) với \(a,b,c\ne0\); \(a+b+c\ne0\); \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ne\dfrac{4}{a+b+c}\) và x là ẩn số.
3. Giải bất phương trình: \(3x^3+4x^2+5x-6>0\).
4. Tìm x sao cho: 2 < x < 3 và \(2\left|x\right|-3\left|x-2\right|+4\left|x-3\right|=5\)
a)\(\dfrac{x}{x-1}-\dfrac{2}{x-1}\)
b)\(\dfrac{4+4x}{3x^2+6x}+\dfrac{x}{3x+6}\)
c)\(\dfrac{x^2-2x}{x-1}\cdot\dfrac{1}{x}:\dfrac{x^2-4}{x^2-2x+1}\)
Thực hiện phép tính
\(a,\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(b,\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
\(c,\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(d,\dfrac{x+1}{x+2}:\left(\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\right)\)
\(e,\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(f,\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(g,\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(h,\dfrac{x^3}{x-1}-\dfrac{x^2}{x+1}-\dfrac{1}{x-1}+\dfrac{1}{x+1}\)
Tìm x, y biết:a, \(\left[\dfrac{1}{2}x^2\left(2x-1\right)^m-\dfrac{1}{2}x^{m+2}\right]:\dfrac{1}{2}x^2=0\) (m thuộc N)
b, \(\left(2x-3\right)^6=\left(2x-3\right)^8\)
c, \(4x^2-4x+y^2-\dfrac{2}{3}y+\dfrac{10}{9}=0\)
a) Rút gọn biểu thức sau: \(A=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
rút gọn biểu thức sau A=\(\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
