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a: \(\left(\dfrac{1}{\left(2x-y\right)^2}+\dfrac{2}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{1}{\left(2x+y\right)^2}\right)\cdot\dfrac{\left(2x+y\right)^2}{16x}\)
\(=\dfrac{4x^2+4xy+y^2+2\left(4x^2-y^2\right)+4x^2-4xy+y^2}{\left(2x-y\right)^2\cdot\left(2x+y\right)^2}\cdot\dfrac{\left(2x+y\right)^2}{16x}\)
\(=\dfrac{8x^2+2y^2+8x^2-2y^2}{\left(2x-y\right)^2}\cdot\dfrac{1}{16x}\)
\(=\dfrac{16x^2}{16x}\cdot\dfrac{1}{\left(2x-y\right)^2}=\dfrac{x}{\left(2x-y\right)^2}\)
b: \(\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)\)
\(=\dfrac{2x+4-4}{\left(x+2\right)^2}:\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right)\)
\(=\dfrac{2x}{\left(x+2\right)^2}:\dfrac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x}{\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{-x}=\dfrac{-2\left(x-2\right)}{x+2}\)
câu này post hồi học lớp 8 = )) giờ tốt nghiệp c3 thì có người trả lời :'))
khbiet nên cười hay khóc đây
Bài 1:
\(B=\dfrac{4\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\dfrac{\left(x^2-25\right)}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)
\(=\dfrac{4\left(x+3\right)^2}{\left(3x+5-2x\right)\left(3x+5+2x\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{\left(4x+15-x\right)\left(4x+15+x\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{5\left(x-5\right)\left(x+1\right)}-\dfrac{3\left(x+3\right)\left(x+1\right)}{15\left(x+5\right)\left(x+3\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{x+5}{5\left(x+1\right)}-\dfrac{x+1}{5\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+5\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+1\right)^2}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{4\left(x^2+6x+9\right)-\left(x^2+10x+25\right)-\left(x^2+2x+1\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{4x^2+24x+36-x^2-10x-25-x^2-2x-1}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2x^2+12x+10}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x^2+6x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x^2+5x+x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x+5\right)\left(x+1\right)}{5\left(x+5\right)\left(x+1\right)}=\dfrac{2}{5}\)
đc bn , nhg mà đề bài câu a b2 sao tự nhiên lại có " n "
bn xem lại đề đi
\(a,\dfrac{x^2-2x}{x^2-4}=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)
b) \(\dfrac{x^2+5x+4}{x^2-1}=\dfrac{x^2+x+4x+4}{x^2-1}=\dfrac{\left(x+1\right)\left(x+4\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+4}{x-1}\)
c) \(\dfrac{x^4+4}{x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1}\)
\(=\dfrac{x^4+4x^2-4x^2+4}{x^3+2x-2x^2-x^2+2x-1-1}\)
\(=\dfrac{\left(x^2+2\right)^2-4x^2}{\left(x^3+2x-2x^2\right)-\left(x^2-2x+2\right)}\)
\(=\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x\left(x^2+2-2x\right)-\left(x^2+2-2x\right)}\)
\(=\dfrac{x^2+2+2x}{x-1}\)
Bài 2:
a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
\(=\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{10}{2x+1}\)
b) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\dfrac{1-2x+x^2}{x\left(x+1\right)}:\dfrac{1+x^2-2x}{x}\)
\(=\dfrac{1}{x+1}\)
c) Trong ngoặc giữa hai phân số là dấu gì vậy ?
a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)
\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)
a: \(C=\dfrac{1}{x+2}-\dfrac{x\left(x-2\right)\left(x+2\right)}{x^2+4}\cdot\left(\dfrac{1}{\left(x+2\right)^2}+\dfrac{1}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\dfrac{1}{x+2}-\dfrac{x\left(x-2\right)\left(x+2\right)}{x^2+4}\cdot\dfrac{x-2+x+2}{\left(x-2\right)\left(x+2\right)^2}\)
\(=\dfrac{1}{x+2}-\dfrac{x}{x^2+4}\cdot\dfrac{2x}{x+2}\)
\(=\dfrac{x^2+4-2x^2}{\left(x+2\right)\left(x^2+4\right)}\)
\(=\dfrac{4-x^2}{\left(x+2\right)\cdot\left(x^2+4\right)}=\dfrac{2-x}{x^2+4}\)
b: Để C=1 thì \(x^2+4=2-x\)
\(\Leftrightarrow x^2+x+2=0\)
hay \(x\in\varnothing\)