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\(\)\(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\rightarrow\left(a;b;c\right)\)
Viết lại đề: \(\left\{{}\begin{matrix}a+b+c=2\\2ab-c^2=4\end{matrix}\right.\) . Tính \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^{2018}\)
\(\Leftrightarrow\left(a+b+c\right)^2-2ab+c^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-2ab+c^2=0\)
\(\Leftrightarrow a^2+b^2+2c^2+2bc+2ac=0\)
\(\Leftrightarrow\left(a^2+c^2+2ac\right)+\left(b^2+c^2+2bc\right)=0\)
\(\Leftrightarrow\left(a+c\right)^2+\left(b+c\right)^2=0\)
\(\Leftrightarrow....\)
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)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)
=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)
=\(\dfrac{1}{6}\)
\(B=\dfrac{1}{x}+\dfrac{1}{y}\\ =\dfrac{x+y}{xy}=\dfrac{5}{6}\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\\ =5^3-3.6.5\\ =125-90\\ =35\)
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
Hằng đẳng thức mà tương ạ! :v
a, \(\dfrac{8x^3-\dfrac{1}{125}y^3}{4x^2+\dfrac{1}{25}y^2+\dfrac{2}{5}xy}\)
\(=\dfrac{\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)}{4x^2+\dfrac{1}{25}y^2+\dfrac{2}{5}xy}=2x-\dfrac{1}{5}y\)
b, \(\dfrac{x^3-6x^2+2x+15}{x-5}\)
\(=\dfrac{x^3-5x^2-x^2+5x-3x+15}{x-5}\)
\(=\dfrac{x^2\left(x-5\right)-x\left(x-5\right)-3\left(x-5\right)}{x-5}\)
\(=\dfrac{\left(x-5\right)\left(x^2-x-3\right)}{\left(x-5\right)}=x^2-x-3\)
Rồi ạ :v!
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...
Lời giải:
Đặt \(\left\{\begin{matrix} (x+y)^2=a\neq 0\\ xy=b\end{matrix}\right.\)
Dùng cách biến đổi tương đương.
Ta có: \(A=x^2+y^2+\left(\frac{xy+1}{x+y}\right)^2=(x+y)^2-2xy+\frac{(xy+1)^2}{(x+y)^2}\)
\(A=a-2b+\frac{(b+1)^2}{a}\)
\(A\geq 2\Leftrightarrow a-2b+\frac{(b+1)^2}{a}\geq 2\)
\(\Leftrightarrow a^2-2ab+(b+1)^2\geq 2a\)
\(\Leftrightarrow a^2+b^2+1-2ab+2b-2a\geq 0\)
\(\Leftrightarrow (-a+b+1)^2\geq 0\) (luôn đúng)
Do đó ta có đpcm.
Dấu bằng xảy ra khi \(-a+b+1=0\Leftrightarrow x^2+y^2+xy=1\)
mn ơi tl giúp mik vs