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\(\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{x+y}.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\right]:\dfrac{x^3+y^3}{x^2y^2}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{x+y}.\dfrac{x+y}{xy}\right].\dfrac{x^2y^2}{x^3+y^3}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{xy}\right].\dfrac{x^2y^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\dfrac{y^2+x^2+2xy}{x^2y^2}.\dfrac{x^2y^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\dfrac{\left(x+y\right)^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}-\dfrac{x+y}{x^2-xy+y^2}\)
=\(=\dfrac{x+y}{x^2-xy+y^2}-\dfrac{x+y}{x^2-xy+y^2}=0\)
\(M=\left(\dfrac{x}{y}+1-5\right)^3=\left(\dfrac{x}{y}-4\right)^3\)
\(=\left(\dfrac{12}{2}-4\right)^3=\left(6-4\right)^3=2^3=8\)
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
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
Lời giải:
a) Ta có:
\(Q=\left[\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{x+y}\left(\frac{1}{x}+\frac{1}{y}\right)\right].\frac{x^2y^2}{x^3+y^3}\)
\(=\left[\frac{x^2+y^2}{x^2y^2}+\frac{2}{x+y}.\frac{x+y}{xy}\right].\frac{x^2y^2}{x^3+y^3}\)
\(=\left[\frac{x^2+y^2}{x^2y^2}+\frac{2}{xy}\right].\frac{x^2y^2}{x^3+y^3}\)
\(=\frac{x^2+y^2}{x^2y^2}.\frac{x^2y^2}{x^3+y^3}+\frac{2x^2y^2}{xy(x^3+y^3)}\)
\(=\frac{x^2+y^2}{x^3+y^3}+\frac{2xy}{x^3+y^3}=\frac{x^2+y^2+2xy}{x^3+y^3}\)
\(=\frac{(x+y)^2}{x^3+y^3}=\frac{(x+y)^3}{(x+y)(x^2-xy+y^2)}=\frac{x+y}{x^2-xy+y^2}\)
b)
Khi \(x=1,y=2\Rightarrow Q=\frac{1+2}{1^2-1.2+2^2}=1\)
\(=\dfrac{y^2+xy+x^2}{y^2}:\dfrac{x^2-y^2}{x^2}\cdot\dfrac{y^2}{x^2-y^2}\)
\(=\dfrac{x^2+xy+y^2}{y^2}\cdot\dfrac{x^2}{x^2-y^2}\cdot\dfrac{y^2}{x^2-y^2}\)
\(=\dfrac{x^2+xy+y^2}{\left(x^2-y^2\right)^2}\cdot x^2\)