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a) Ta có:
x + y = 2
=> ( x + y)2 = 4
=> x2 + 2xy + y2 = 4
=> 10 + 2xy = 4
=> 2xy = 4 - 10 = -6
=> xy = -6/2 = -3
Ta có:
A = x3 + y3
A = (x + y)(x2 - xy + y2)
A = 2(10 + 3)
A = 26
b) Ta có:
x + y = a
=> (x + y)2 = a2
=> x2 + 2xy + y2 = a2
=> b + 2xy = a2
=> xy = (a2 - b)/2
Ta có:
B = x3 + y3
B = (x + y)(x2 + xy + y2)
B = a[b + (a2 - b )/2]
B = ab + (a3 - b)/2
cho x+y=2(=)(x+y)^2=4(=)x^2+y^2+2xy=4
(=)10+2xy=4(=)2xy=-6(=)xy=-3
mà x^3+y^3=(x+y)(x^2+y^2-xy)
=2(10+3)=26
vậy x^3+y^3=26
có: \(x-y=5\)=>\(\left(x-y\right)^2=25\)<=> \(x^2-2xy+y^2=25\)=> \(xy=\frac{x^2+y^2-25}{2}=\frac{15-25}{2}=-\frac{10}{2}=-5\)
\(x^3-y^3=\left(x-y\right)\left(x^2+y^2+xy\right)=5.\left(15-5\right)=5.10=50\)
Có: \(\left(x-y\right)^2=25\)
\(\Leftrightarrow-2xy=25-\left(x^2+y^2\right)=25-15=10\)
\(\Leftrightarrow xy=-5\)
\(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=5\cdot\left(15-5\right)=5\cdot10=50\)
x - y = 5
=>(x - y)2 = 25
=> x2 - 2xy + y2 = 25
=> 15 - 2xy = 25
=> 2xy = 15 - 25
=> 2xy = - 10
=> xy = -10 : 2
=> xy = -5
x3 - y3 = ( x - y ) ( x 2 - xy + y2)
=> x3- y3 = 5 . ( -5 . 15 )
= >x 3 - y 3 = - 375.
1/
a,\(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{5}{-2}=\frac{-5}{2}\)
b, \(x^2+y^2=\left(x+y\right)^2-2xy=5^2-2.\left(-2\right)=25+4=29\)
c,\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=5^3-3.\left(-2\right).5=125+30=155\)
d,thiếu dữ kiện
2.
Ta có: a chia 7 dư 3 => a=7k+3 (k thuộc N)
=>\(a^2=\left(7k+3\right)\left(7k+3\right)=7k\left(7k+3\right)+3\left(7k+3\right)=7k\left(7k+3\right)+3.7k+3.3=7k\left(7k+3\right)+3.7k+7+2\)chia 7 dư 2
Vậy...
ta có : \(x^2+y^2=5\Leftrightarrow\left(x+y\right)^2-2xy=5\Leftrightarrow3^2-2xy=5\)
\(\Leftrightarrow9-2xy=5\Leftrightarrow2xy=9-5=4\Leftrightarrow xy=2\)
ta có : \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)
\(=\left(3\right)^3-3.2.3=27-18=9\)
vậy \(x^3+y^3=9\) khi \(x+y=3\) và \(x^2+y^2=5\)
ta có : \(\left(x+y\right)=3\)=> \(\left(x+y\right)^2=9\)
<=> \(x^2+2xy+y^2=9\)
=> \(xy=\frac{9-\left(x^2+y^2\right)}{2}=\frac{9-5}{2}=2\)
ta có : \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=5.\left(5-2\right)=5.3=15\)
Tính: a, x2+y2
Ta có: x+y=2 => (x+y)2=4
<=> x2+2xy+y2=4
<=> x2+y2=4-2xy=4-2.(-15)=34 (vì x.y=-15)
vậy x2+y2=34
b, x3+y3
Ta có: x+y=2 => (x+y)3=8
<=>x3+3xy(x+y) + y3 = 8
<=> x3+y3 =8 - 3xy(x+y) = 8 - 3 ( -15) . 2 =98
Vậy x3+y3 = 98
c, x5 + y5
Ta có: ( x2+y2)(x3+y3)=34.98=3332
<=> x5+x3y2+x2y3+y5=3332
<=> x5+y5+x2y2(x+y)=3332
<=> x5+y5 + (xy)2(x+y)=3332
<=> x5+y5 = 3332 - (xy)2(x+y)=3332 - (-15)2 . 2 =2882
Vậy x5+y5=2882
a) \(x+y=3\)
\(\Rightarrow\)\(\left(x+y\right)^2=9\)
\(\Leftrightarrow\)\(x^2+y^2+2xy=9\)
\(\Leftrightarrow\)\(2xy=4\) do x2 + y2 = 5
\(\Leftrightarrow\)\(xy=2\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=3^3-3.2.3=9\)
b) bạn làm tương tự
\(a,x+y=3\Rightarrow\left(x+y\right)^2=9\Rightarrow x^2+2xy+y^2=9\Rightarrow2xy=4\Leftrightarrow xy=2\)
Vì \(\left(x+y\right)=3\Rightarrow\left(x+y\right)^3=27\)
\(\Rightarrow x^3+3x^2y+3xy^2+y^3=27\)
\(\Rightarrow x^3+y^3+3xy\left(x+y\right)=27\)
\(\Rightarrow x^3+y^3+3.2.3=27\)
\(\Rightarrow x^3+y^3=27-18=9\)
\(b,x-y=5\Rightarrow\left(x-y\right)^2=25\Rightarrow x^2-2xy+y^2=25\Rightarrow2xy=-10\Leftrightarrow xy=-5\)
\(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=5.10=50\)