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x + y = 1
<=> (x + y)2 = 12
<=> x2 + y2 + 2xy = 1
<=> x2 + y2 = 1 - 2xy
Ta có:
\(\dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)
= \(\dfrac{x\left(x^3-1\right)}{\left(y^3-1\right)\left(x^3-1\right)}-\dfrac{y\left(y^3-1\right)}{\left(y^3-1\right)\left(x^3-1\right)}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)
= \(\dfrac{x^4-x-y^4+y}{x^3y^3-y^3-x^3+1}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)-\left(x-y\right)}{x^3y^3-\left(x+y\right)\left(x^2+y^2-xy\right)+1}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\dfrac{\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)-\left(x-y\right)}{x^3y^3-\left(1-2xy-xy\right)+1}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\dfrac{\left(x-y\right)\left(1-2xy-1\right)}{x^3y^3+3xy}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\dfrac{-2xy\left(x-y\right)}{xy\left(x^2y^2+3\right)}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)
\(=-\dfrac{2\left(x-y\right)}{x^2y^2+3}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)
= 0 (đpcm)
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 \(x=1,y=2\Rightarrow Q=\frac{1+2}{1^2-1.2+2^2}=1\)
a)
\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)
c)
\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)
d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)
f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)
g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)
\(M=\dfrac{yz\sqrt{x-1}+xz\sqrt{y-2}+xy\sqrt{z-3}}{xyz}\)
\(=\dfrac{yz\sqrt{x-1}}{xyz}+\dfrac{xz\sqrt{y-2}}{xyz}+\dfrac{xy\sqrt{z-3}}{xyz}\)
\(=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{x-1}\le\dfrac{1+x-1}{2}=\dfrac{x}{2}\)\(\Rightarrow\dfrac{\sqrt{x-1}}{x}\le\dfrac{x}{2}\cdot\dfrac{1}{x}=\dfrac{1}{2}\)
\(\sqrt{y-2}=\dfrac{\sqrt{2\left(y-2\right)}}{\sqrt{2}}\le\dfrac{y}{2\sqrt{2}}\)\(\Rightarrow\dfrac{\sqrt{y-2}}{y}\le\dfrac{y}{2\sqrt{2}}\cdot\dfrac{1}{y}=\dfrac{1}{2\sqrt{2}}\)
\(\sqrt{z-3}=\dfrac{\sqrt{3\left(z-3\right)}}{\sqrt{3}}\le\dfrac{z}{2\sqrt{3}}\)\(\Rightarrow\dfrac{\sqrt{z-3}}{z}\le\dfrac{z}{2\sqrt{3}}\cdot\dfrac{1}{z}=\dfrac{1}{2\sqrt{3}}\)
Cộng theo vế 3 BĐT trên ta có:
\(M\le\dfrac{1}{2}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}\right)\) (ĐPCM)