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`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`.
áp dụng hằng đẳng thức (a+b)2=a2+2ab+b2 vào (x+y)2 ta được:
\(\left(x+y\right)^2=x^2+2xy+y^2=x^2+y^2+2xy\)
ta có : xy=27
=>2xy=54
thay 2xy=54 và x2+y2=29 vào bt x2+y2+2xy ta được
\(29+54=83\)
vậy giá trị của biểu thức (x+y)2 tại x2+y2=29 và xy=27 là 83
Ta có :
\(x^3\) + \(y^3\) - xy = \(-\dfrac{1}{27}\)
⇔ \(x^3\) + \(y^3\) - xy + \(\dfrac{1}{27}\) = 0
⇔ \(x^3\) + \(y^3\) + \(\dfrac{1^3}{3^3}\) - 3xy.\(\dfrac{1}{3}\) = 0
⇔ (x + y + \(\dfrac{1}{3}\))(\(x^2\) + \(y^2\) + \(\dfrac{1}{9}\) - xy - \(\dfrac{1}{3}x-\dfrac{1}{3}y\)) = 0
TH1 :
x + y + \(\dfrac{1}{3}\) = 0
⇔ x + y = - \(\dfrac{1}{3}\) (loại vì x>0 ; y>0)
TH2 :
\(x^2+y^2+\dfrac{1}{9}-xy-\dfrac{1}{3}x-\dfrac{1}{3}y=0\)\(\dfrac{1}{3}x-\dfrac{1}{3}y\)
⇔ (\(x-\dfrac{1}{3}\))\(^2\) + (\(y-\dfrac{1}{3}\))\(^2\) + (x - y)\(^2\) = 0
⇒ \(x-\dfrac{1}{3}\) = 0
\(y-\dfrac{1}{3}\) = 0
\(x-y\) = 0
⇔ x = y = \(\dfrac{1}{3}\)
Thay x = y = \(\dfrac{1}{3}\) vào \(\dfrac{x}{y^2}\) ta được :
\(\dfrac{1}{3}\) : \(\dfrac{1}{9}\)
= \(\dfrac{1}{3}\) . 9
= 3
\(\dfrac{1}{3}\)\(x^2+y^2+\dfrac{1}{9}-xy-\dfrac{1}{3}x-\dfrac{1}{3}y=0\)
a. ta có : \(x^2+y^2=\left(x+y\right)^2-2xy=1^2-2\times\left(-6\right)=13\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=1^3-3\times\left(-6\right)\times1=19\)
\(x^5+y^5=\left(x+y\right)\left[x^4-x^3y+x^2y^2-xy^3+y^4\right]\)
\(=\left(x+y\right)\left[\left(x^2+y^2\right)^2-x^2y^2-xy\left(x^2+y^2\right)\right]=1.\left(13^2-\left(-6\right)^2-\left(-6\right).13\right)=211\)
b.\(x^2+y^2=\left(x-y\right)^2+2xy=1+2\times6=13\)
\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=1^3+6.3.1=19\)
\(x^5-y^5=\left(x-y\right)\left[\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\right]\)
\(=\left(x-y\right)\left[\left(x^2+y^2\right)^2-x^2y^2+xy\left(x^2+y^2\right)\right]=1.\left(13^2-6^2+6.13\right)=211\)
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Có x^3+y^3=(x+y)^3-3xy(x+y)
=> x^3+y^3=27^3-3.24.27
=> x^3+y^3=17739.
Vậy x^3+y^3==17739
Ta có: x3 + y3 = (x + y)(x2 - xy + y2) = (x + y)(x2 + 2xy + y2) - 3xy(x + y)
= (x + y)(x + y)2 - 3xy(x + y)
= (x + y)3 - 3xy(x + y)
= 273 - 3.24.27
= 17739
Theo bài ra ta có:
\(x^2y+xy^2+x+y=\left(x^2y+xy^2\right)+x+y\)
\(=xy\left(x+y\right)+x+y=x+y+x+y\)
\(\Rightarrow2\left(x+y\right)=16\Rightarrow x+y=16\div2=8\)
\(\Rightarrow\left(x+y\right)^2=8^2=64\)
\(\Rightarrow x^2+2xy+y^2=64\)
\(\Rightarrow x^2+2+y^2=64\)
\(\Rightarrow x^2+y^2=64-2=62\)
Vậy \(x^2+y^2=62\)
\(x^2y+xy^2+x+y=16\)
\(\Leftrightarrow2x+2y=16\)
\(\Leftrightarrow x+y=8\)
Lại có\(x^2+y^2=\left(x+y\right)^2-2xy\)
\(=8^2-2\)
\(=62\)
Vậy\(x^2+y^2=62\)
(x+y)^2 =a^2
x^2 +2xy +y^2 =a^2
x^2+y^2 =a^2-2xy =a^2 -2b
x^3 +y^3 = (x+y)(x^2 -xy +y^2)
=a(a^2-2b-b)
=a(a^2-3b)
=a^3- 3ab
(x^2 +y^2)^2=(a^2-2b)^2 ( cái này tính cho x^4 + y^4)
tương tự như câu đầu tiên
x^5+ y^5 (cái đó mình không biết)
(x + y)2 = 272
<=> x2 + y2 + 2xy = 729
<=> x2 + y2 + 2.180 = 729
<=> x2 + y2 = 369
<=> x2 + y2 - 2xy = 369 - 2xy
<=> (x - y)2 = 9
Mà x < y => x - y = -3