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\(H=\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{1+xz}\ge\dfrac{\left(1+1+1\right)^2}{3+xy+yz+xz}=\dfrac{9}{3+xy+yz+xz}\)
Mặt khác,theo AM-GM: \(xy+yz+xz\le x^2+y^2+z^2=3\)
\(\Rightarrow\dfrac{9}{3+xy+yz+xz}\ge\dfrac{9}{3+3}=\dfrac{9}{6}=\dfrac{3}{2}\)
Dấu "=" khi: \(x=y=z=1\)
\(\)\(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\rightarrow\left(a;b;c\right)\)
Viết lại đề: \(\left\{{}\begin{matrix}a+b+c=2\\2ab-c^2=4\end{matrix}\right.\) . Tính \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^{2018}\)
\(\Leftrightarrow\left(a+b+c\right)^2-2ab+c^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-2ab+c^2=0\)
\(\Leftrightarrow a^2+b^2+2c^2+2bc+2ac=0\)
\(\Leftrightarrow\left(a^2+c^2+2ac\right)+\left(b^2+c^2+2bc\right)=0\)
\(\Leftrightarrow\left(a+c\right)^2+\left(b+c\right)^2=0\)
\(\Leftrightarrow....\)
\(B=\dfrac{1}{x}+\dfrac{1}{y}\\ =\dfrac{x+y}{xy}=\dfrac{5}{6}\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\\ =5^3-3.6.5\\ =125-90\\ =35\)
A = x2 + y2
= (x2 + 2xy + y2) - 2xy
= (x + y)2 - 2xy
= 52 - 2.6
= 25 - 12
= 13
F = x3 + y3
= (x + y)3 - 3xy(x + y)
= 53 - 3.6.5
= 125 - 90
= 35
1, Ta có: \(x+y=9\Rightarrow\left(x+y\right)^2=81\)
\(\Rightarrow x^2+2xy+y^2=81\)
\(\Rightarrow x^2+y^2=45\)
\(\Rightarrow x^2+y^2-2xy=9\)
\(\Rightarrow\left(x-y\right)^2=9\Rightarrow\left[{}\begin{matrix}x-y=3\\x-y=-3\end{matrix}\right.\)
\(A=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(\Rightarrow\left[{}\begin{matrix}A=3.63=189\\A=-3.63=-189\end{matrix}\right.\)
Vậy...
a) \(\left(6x^3y^2-4x^2y^3-10x^2y^2\right):2xy\)
=\(\left(6x^3y^2:2xy\right)-\left(4x^2y^3:2xy\right)-\left(10x^2y^2:2xy\right)\)
\(=3x^2y-2xy^2-5xy\)
b) \(\dfrac{2y}{x-2}+\dfrac{5y}{x-2}\)
=\(\dfrac{2y+5y}{x-2}\)
=\(\dfrac{7y}{x-2}\)
c)\(\dfrac{xy}{3x-y}+\dfrac{3x^2}{y-3x}\)
\(=\dfrac{xy}{3x-y}-\dfrac{3x^2}{3x-y}\)
=\(\dfrac{x\left(y-3x\right)}{3x-y}\)
=\(\dfrac{-x\left(3x-y\right)}{3x-y}\)
=-x
d)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)
=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)
=\(\dfrac{1}{6}\)
a, \(xy\left(x+y\right)-x^2\left(x+y\right)-y^2\left(x-y\right)\)
\(=x^2y+xy^2-x^3-x^2y-xy^2+y^3\)
\(=y^3-x^3\)
b, \(x^2-x^2\left(5x+1\right)+x\left(x-3\right)\)
\(=x^2-5x^3-x^2+x^2-3x\)
\(=-5x^3+x^2-3x\)
Chúc bạn học tốt!!!
c, \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=\left(3x^2-5x^2-8x^2\right)+\left(-6x-5x\right)+24\)
\(=-10x^2-11x+24\)
d, \(\dfrac{1}{2}\left(x+4\right)+\dfrac{1}{2}x^2\left(6x-3\right)-x\left(x^2+\dfrac{1}{2}\right)\)
\(=\dfrac{1}{2}x+2+3x^3-\dfrac{3}{2}x^2-x^3-\dfrac{1}{2}x\)
\(=-x^3+\left(3x^2-\dfrac{3}{2}x^2\right)+\left(\dfrac{1}{2}x-\dfrac{1}{2}x\right)+2\)
\(=-x^3+\dfrac{3}{2}x^2+2\)
\(=-\left(x^3-\dfrac{3}{2}x^2-2\right)=-\left(x^3-2x^2+\dfrac{1}{2}x^2-x+x-2\right)\)
\(=-\left[\left(x^3-2x^2\right)+\left(\dfrac{1}{2}x^2-x\right)+\left(x-2\right)\right]\)
\(=-\left[x^2.\left(x-2\right)+\dfrac{1}{2}x.\left(x-2\right)+\left(x-2\right)\right]\)
\(=-\left[\left(x-2\right).\left(x^2+\dfrac{1}{2}x+1\right)\right]\)
Chúc bạn học tốt!!!
a ) \(\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)
b ) \(\left(x^2-2xy+y^2\right)\left(x-y\right)=\left(x-y\right)^2\left(x-y\right)=\left(x-y\right)^3\)
c ) \(\left(x^2y^2-\dfrac{1}{3}xy+3y\right)\left(x-3y\right)\)
\(=\left(x^2y^2-\dfrac{1}{3}xy+3y\right)x-3y\left(x^2y^2-\dfrac{1}{3}xy+3y\right)\)
\(=x^3y^2-\dfrac{1}{3}x^2y+3xy-3x^2y^3+xy^2-9y^2\)
d ) \(\left(\dfrac{1}{5}x-1\right)\left(x^2-5x+2\right)\)
\(=\dfrac{1}{5}x\left(x^2-5x+2\right)-x^2+5x-2\)
\(=\dfrac{1}{5}x^3-x^2+\dfrac{2}{5}x-x^2+5x-2\)
\(=\dfrac{1}{5}x^3-2x^2+\dfrac{27}{5}x-2\)
mn ơi tl giúp mik vs