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\(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}\ge\dfrac{2}{1+xy}\)
⇔ \(\left(\dfrac{1}{1+x^2}-\dfrac{1}{1+xy}\right)+\left(\dfrac{1}{1+y^2}-\dfrac{1}{1+xy}\right)\ge0\)
⇔ \(\left(\dfrac{1+xy-\left(1+x^2\right)}{\left(1+x^2\right)\left(1+xy\right)}\right)+\left(\dfrac{1+xy-\left(1+y^2\right)}{\left(1+y^2\right)\left(1+xy\right)}\right)\ge0\)
⇔ \(\left(\dfrac{1+xy-1-x^2}{\left(1+x^2\right)\left(1+xy\right)}\right)+\left(\dfrac{1+xy-1-y^2}{\left(1+y^2\right)\left(1+xy\right)}\right)\ge0\)
⇔ \(\dfrac{-x\left(x-y\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{-y\left(y-x\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
⇔ \(\dfrac{-x\left(x-y\right)\left(1+y^2\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}+\dfrac{y\left(x-y\right)\left(1+x^2\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)
=> -x(x-y)(1+y2)+y(x-y)(1+x2) ≥ 0
⇔ (x-y)[-x(1+y2)+y(1+x2)]≥0
⇔ (x-y)(-x-xy2+y+x2y) ≥0
⇔ (x-y)[-(x-y)+(x2y-y2x)] ≥ 0
⇔ (x-y)[-(x-y)+xy(x-y) ]≥ 0
⇔ (x-y)(x-y)(xy-1)≥ 0
⇔ (x-y)2 (xy-1) ≥0 (luôn đúng ∀ xy ≥ 1)
=> đpcm
bạn pải giả sử trước chứ nếu ntn thì người chấm hỏi ai cho lôi phần chứng minh ra làm phần mục đề
Bài 2:
a: \(A=\dfrac{3}{2\left(x+1\right)}+\dfrac{10x}{2\left(x-1\right)\left(x+1\right)}-\dfrac{5}{2\left(x-1\right)}\)
\(=\dfrac{3x-3+10x-5x-5}{2\left(x-1\right)\left(x+1\right)}=\dfrac{8x-8}{2\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x+1}\)
b: Để P/2=3/x^2+2 thì \(\dfrac{4}{2x+2}=\dfrac{3}{x^2+2}\)
\(\Leftrightarrow\dfrac{2}{x+1}=\dfrac{3}{x^2+2}\)
=>\(2x^2+4-3x-3=0\)
=>2x^2-3x+1=0
=>(x-1)(2x-1)=0
=>x=1/2(nhận) hoặc x=1(loại)
a,\(\frac{x^2+y^2-xy}{x^2-y^2}:\frac{x^3+y^3}{x^2+y^2-2xy} =\frac{x^2+y^2-xy}{(x-y)(x+y)}\frac{(x+y)^2}{(x+y) (x^2-xy+y^2)}=\frac{1}{x-y} \)
b,\(\frac{x^3y+xy^3}{x^4y}:(x^2+y^2)=\frac{xy(x^2+y^2)}{x^4y(x^2+y^2)}=\frac{1}{x^3} \)
c,\(\frac{x^2-xy}{y}:\frac{x^2-xy}{xy+y}:\frac{x^2-1}{x^2+y} =\frac{x(x-y)y(x+y)(x^2+y)}{yx(x-y)(x^2-1)} =\frac{(x^2+y)(x+y)}{x^2-1} \)
d,\(\frac{x^2+y}{y}:(\frac{z}{x^2}:\frac{xy}{x^2y})=\frac{x^2+y}{ y}:(\frac{z}{x^2}\frac{x^2y}{xy})=\frac{x^2+y}{y}\frac{z}{x} \)
Sửa đề: \(\dfrac{2}{xy}:\left(\dfrac{1}{x}-\dfrac{1}{y}\right)^2:\dfrac{x^2+y^2}{\left(x-y\right)^2}=\dfrac{2xy}{x^2+y^2}\)
Ta có: \(\dfrac{2}{xy}:\left(\dfrac{1}{x}-\dfrac{1}{y}\right)^2:\dfrac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\dfrac{2}{xy}:\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{2}{xy}\right):\dfrac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\dfrac{2}{xy}:\left(\dfrac{x^2+y^2}{x^2y^2}-\dfrac{2xy}{x^2y^2}\right):\dfrac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\dfrac{2}{xy}:\dfrac{x^2-2xy+y^2}{\left(xy\right)^2}:\dfrac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\dfrac{2}{xy}\cdot\dfrac{\left(xy\right)^2}{\left(x-y\right)^2}:\dfrac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\dfrac{2xy}{\left(x-y\right)^2}:\dfrac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\dfrac{2xy}{\left(x-y\right)^2}\cdot\dfrac{\left(x-y\right)^2}{x^2+y^2}\)
\(=\dfrac{2xy}{x^2+y^2}\)
Thật đấy ạ, nãy giờ ngồi nháp mãi vẫn không hiểu sao đề bắt chứng minh nó bằng 1 được:(