( x^2 +x ^3 -xy^2 +3 ) + ( x^3 +xy ^2 - xy -6)
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B-(\(3x^6-4xy^5+\dfrac{1}{3}xy^2\))=
B= \(\left(7x^6-\dfrac{1}{2}xy^5-xy^2-\dfrac{1}{3}\right)+\left(3x^6-4xy^5+\dfrac{1}{3}xy^2-\dfrac{3}{2}\right)\)
B= \(7x^6-\dfrac{1}{2}xy^5-xy^2-\dfrac{1}{3}+3x^6-4xy^5+\dfrac{1}{3}xy^2-\dfrac{3}{2}\)
B= \(7x^6+3x^6-\dfrac{1}{2}xy^5-4xy^5-xy^2+\dfrac{1}{3}xy^2-\dfrac{1}{3}+\dfrac{2}{3}\)
B= \(10x^6-\dfrac{9}{2}xy^5-\dfrac{2}{3}xy^2+\dfrac{1}{3}\)
a) A = (x + 2)³ + (x - 2)³ - 2x(x² + 12)
= x³ + 6x² + 12x + 8 + x³ - 6x² + 12x - 8 - 2x² - 24x
= (x³ + x³) + (6x² - 6x² - 2x²) + (12x + 12x - 24x) + (8 - 8)
= 2x³ -2x²
b) B = (xy + 2)³ - 6(xy + 2)² + 12(xy + 2) - 8
= (xy + 2 - 2)³
= (xy)³
= x³y³
a, x=1; y=2 => 12
x=2; y=1 => 21
b, x=1; y=5 => 15
x=5; y=1 => 51
c, x=1; y=6 => 16
x=6;y=1 => 61
x=2; y=3=> 23
x=3; y=2 => 32
d, x=1; y=8 => 18
x=2; y=4 => 24
x=4; y=2 => 42
x=8; y=1 => 81
\(\Leftrightarrow\left\{{}\begin{matrix}3xy-3\left(x-y\right)=-9\\x^2+y^2+xy-\left(x-y\right)=6\end{matrix}\right.\)
Trừ vế cho vế:
\(x^2+y^2-2xy+2\left(x-y\right)=15\)
\(\Leftrightarrow\left(x-y\right)^2+2\left(x-y\right)-15=0\Rightarrow\left[{}\begin{matrix}x-y=3\\x-y=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=x-3\\y=x+5\end{matrix}\right.\)
Thế vào pt đầu:
\(\Rightarrow\left[{}\begin{matrix}x\left(x-3\right)-x+x-3=-3\\x\left(x+5\right)-x+x+5=-3\end{matrix}\right.\)
\(\Leftrightarrow...\)
`a, (x-y)^2 = (x+y)^2 - 4xy = 12^2 - 35 . 4 = 144 - 140 = 4`.
`b, (x+y)^2 = (x-y)^2 + 4xy = 8^2 + 20.4 = 64 + 80 = 144`
`c, x^3 + y^3 = (x+y)^3 - 3xy(x+y) = 5^3 - 3 . 6 . 5 = 125 - 90 = 35`
`d, x^3 - y^3 = (x-y)^3 - 3xy(x-y) = 3^3 - 3 .40 . 3 = 27 - 360 = -333`.
\(=x^2+x^3-xy^2+3+x^3+xy^2-xy-6\)
\(=x^3+x^2-xy-3\)
\((x^2+x^3-xy^2+3)+(x^3+xy^2-xy-6)\)
\(=\) \(x^2+x^3-xy^2+3+x^3+xy^2-xy-6\)
\(=x^2+(x^3+x^3)+\left(-xy^2-xy\right)+\left(3-6\right)-xy\)
\(=x^2+2x^3-2xy^2-3-xy\)