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cầu cứu a
BT
Bùi Thế Nghị
Giáo viên
24 tháng 12 2020
x2 - x + 1 =x2 - 2.x.\(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)= (x -\(\dfrac{1}{2}\))2 + \(\dfrac{3}{4}\)
Mà (x -\(\dfrac{1}{2}\))2 ≥ 0 với mọi x => (x -\(\dfrac{1}{2}\))2 + \(\dfrac{3}{4}\)≥ \(\dfrac{3}{4}\) với mọi X .
=> A > ∀ x
15 tháng 11 2021
\(\Rightarrow\left\{{}\begin{matrix}m\ge n+4\\n+6\ge m\end{matrix}\right.\Rightarrow n+6\ge m\ge n+4\Rightarrow n+5=m\\ \Rightarrow2\left(m-n\right)+3=2\left(n+5-n\right)+3=13\)
1 tháng 1 2022
a: Xét tứ giác ABCM có
MC//AB
MC=AB
Do đó: ABCM là hình bình hành
28 tháng 1
Bài 10:
1: \(\left(\dfrac{5x+y}{x^2-5xy}+\dfrac{5x-y}{x^2+5xy}\right)\cdot\dfrac{x^2-25y^2}{x^2+y^2}\)
\(=\left(\dfrac{5x+y}{x\left(x-5y\right)}+\dfrac{5x-y}{x\left(x+5y\right)}\right)\cdot\dfrac{\left(x-5y\right)\cdot\left(x+5y\right)}{x^2+y^2}\)
\(=\dfrac{\left(5x+y\right)\left(x+5y\right)+\left(5x-y\right)\left(x-5y\right)}{x\left(x-5y\right)\left(x+5y\right)}\cdot\dfrac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\)
\(=\dfrac{5x^2+25xy+xy+5y^2+5x^2-25xy-xy+5y^2}{x\left(x^2+y^2\right)}\)
\(=\dfrac{10x^2+10y^2}{x\left(x^2+y^2\right)}=\dfrac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\dfrac{10}{x}\)
2: \(\dfrac{4xy}{y^2-x^2}:\left(\dfrac{1}{x^2+2xy+y^2}-\dfrac{1}{x^2-y^2}\right)\)
\(=\dfrac{-4xy}{\left(x-y\right)\left(x+y\right)}:\left(\dfrac{1}{\left(x+y\right)^2}-\dfrac{1}{\left(x-y\right)\left(x+y\right)}\right)\)
\(=\dfrac{-4xy}{\left(x-y\right)\left(x+y\right)}:\dfrac{x-y-x-y}{\left(x-y\right)\left(x+y\right)^2}\)
\(=\dfrac{-4xy}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x-y\right)\left(x+y\right)^2}{-2y}\)
\(=2x\left(x+y\right)\)
Bài 11:
1: ĐKXĐ: \(x\notin\left\{0;3;-3\right\}\)
\(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{-\left(x^2-3x+9\right)}\)
\(=\dfrac{-3}{x-3}\)
2: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(\left(\dfrac{2}{x-2}-\dfrac{2}{x+2}\right)\cdot\dfrac{x^2+4x+4}{8}\)
\(=\left(\dfrac{2x+4-2x+4}{\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{\left(x+2\right)^2}{8}\)
\(=\dfrac{8\left(x+2\right)^2}{8\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{x-2}\)
3: ĐKXĐ: \(x\notin\left\{\dfrac{1}{3};-\dfrac{1}{3};0;-\dfrac{5}{3}\right\}\)
\(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
\(=\left(\dfrac{-3x}{3x-1}+\dfrac{2x}{3x+1}\right)\cdot\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{-3x\left(3x+1\right)+2x\left(3x-1\right)}{\left(3x-1\right)\left(3x+1\right)}\cdot\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(=\dfrac{-9x^2-3x+6x^2-2x}{\left(3x+1\right)}\cdot\dfrac{3x-1}{2x\left(3x+5\right)}\)
\(=\dfrac{-x\left(3x+5\right)}{\left(3x+1\right)}\cdot\dfrac{3x-1}{2x\left(3x+5\right)}=\dfrac{-3x+1}{2\left(3x+1\right)}\)
4: ĐKXĐ: \(x\notin\left\{0;5;-5\right\}\)
\(\left(\dfrac{x}{x^2-25}-\dfrac{x-5}{x^2+5x}\right):\dfrac{2x-5}{x^2+5x}+\dfrac{x}{5-x}\)
\(=\left(\dfrac{x}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{x\left(x+5\right)}\right)\cdot\dfrac{x\left(x+5\right)}{2x-5}+\dfrac{x}{5-x}\)
\(=\dfrac{x^2-\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x\left(x+5\right)}{2x-5}-\dfrac{x}{x-5}\)
\(=\dfrac{\left(x-x+5\right)\left(x+x-5\right)}{\left(x-5\right)\left(2x-5\right)}-\dfrac{x}{x-5}\)
\(=\dfrac{5}{x-5}-\dfrac{x}{x-5}=\dfrac{5-x}{x-5}=-1\)
VH
1
17 tháng 11 2021
\(\dfrac{1}{-2x^2+4x-2}=\dfrac{x-2}{-2\left(x-1\right)^2\left(x-2\right)}\\ \dfrac{1}{2x^2-6x+4}=\dfrac{x-1}{2\left(x-1\right)^2\left(x-2\right)}\)
17 tháng 12 2021
\(B=\dfrac{x^2-2x+1+x^2+2x+1-3x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2x^2-3x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{2x-1}{x+1}\)
a: Để đây là phương trình bậc nhất một ẩn thì m-2<>0
hay m<>2
b: 2x+5=x+7-1
=>2x+5=x+6
=>x=1
Thay x=1 vào (1), ta được:
\(2\left(m-2\right)+3=m-2\)
=>2m-4+3-m+2=0
=>m+1=0
hay m=-1
;-; hay m <> 2 là gì ạ?