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a, x + y = 3 => (x + y)2 = 9 <=> x2 + 2xy + y2 = 9 <=> 5 + 2xy = 9 <=> 2xy = 4 <=> xy = 2
Ta có: x3 + y3 = (x + y)(x2 - xy + y2) = 3 . (5 - 2) = 3 . 3 = 9
b, x - y = 5 => (x - y)2 = 25 <=> x2 - 2xy + y2 = 25 <=> 15 - 2xy = 25 <=> -2xy = 10 <=> xy = -5
Ta có: x3 - y3 = (x - y)(x2 + xy + y2) = 5 . (15 - 5) = 5 . 10 = 50
x2+y2=5
=>x2+2xy+y2=5+2xy
=>(x+y)2=5+2xy
=>32=5+2xy
=>4=2xy =>xy=2
Suy ra: x3+y3=(x+y)(x2-xy+y2)=3.(5-2)=9
1.Tính:
[(x+y)5-2(x+y)4 ] : [-5(x+y)3]
= -5(x+y)2 + \(\dfrac{2}{5}\)(x+y)
2.Tìm a để đa thức 24x3 -14x2 +23x+2a+4 \(⋮\) 4x+1
24x3 -14x2 +23x+2a+4 \(|^{4x+1}_{6x^2-5x+7}\)
24x3 +6x2
\(\overline{-20x^2}+23x+2a+4\)
-20x2 -5x
\(\overline{28x+2a+4}\)
28x +7
\(\overline{2a+11}\)
Để 24x3 -14x2 +23x+2a+4 \(⋮\) 4x+1 thì 2a+11=0 \(\Leftrightarrow\) a= \(\dfrac{11}{2}\)
3. Phân tích đa thức thành NT :
a, 12x3 -12x2 +3x = 3x(4x2 -4x+1) = 3x (2x+1)
b, x2.(x-1)+9(1-x) = x2 (x-1) -9(x-1) = (x-1)(x2-9)
=(x-1)(x-3)(x+3)
c,8(x-y)-x3 (x-y) = (x-y)(8-x3)= (x-y)(2-x)(4+2x+x2)
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.\(x^2+y^2=\left(x-y\right)^2+2xy=1+2\times6=13\)
\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=1^3+6.3.1=19\)
\(x^5-y^5=\left(x-y\right)\left[\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\right]\)
\(=\left(x-y\right)\left[\left(x^2+y^2\right)^2-x^2y^2+xy\left(x^2+y^2\right)\right]=1.\left(13^2-6^2+6.13\right)=211\)
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)
Ta có :
\(x+y=3\)
\(x^2+y^2=5\Leftrightarrow\left(x+y\right)^2-2xy=5\Leftrightarrow9-2xy=5\Leftrightarrow2xy=4\Rightarrow xy=2\)
\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3.\left(5-2\right)=9\)
b)
Ta có :
\(x-y=5\)
\(x^2+y^2=15\Leftrightarrow\left(x-y\right)^2+2xy=15\Leftrightarrow25+2xy=15\Rightarrow xy=-5\)
=> \(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=\left(5\right)\left(15+-5\right)=50\)