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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
\(x+y=2\\ \Rightarrow\left(x+y\right)^2=4\\ \Rightarrow x^2+2xy+y^2=4\\ \Rightarrow2xy=-6\Rightarrow xy=-3\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=2^3-3\cdot\left(-3\right)\cdot2=8-\left(-18\right)=26\)
b,
\(x+y=a\\ \Rightarrow\left(x+y\right)^2=a^2\\ \Rightarrow x^2+2xy+y^2=a^2\\ \Rightarrow2xy=a^2-b\Rightarrow xy=\dfrac{a^2-b}{2}\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=a^3-3\cdot\dfrac{a^2-b}{2}\cdot a=a^3-\dfrac{3a\left(a^2-b\right)}{2}=a^3-\dfrac{3a^3-3ab}{2}=a^3-1,5a^3+3ab=\left(1-1,5\right)a^3+3ab=0,5a^3+3ab=0,5a\left(a^2+6b\right)\)
a) x2 + y2
= (x2 + 2xy + y2) - 2xy
= (x + y)2 - 2xy
= m2 - 2n
b) x3 + y3
= (x + y)(x2 - xy + y2)
= m (x2 + 2xy + y2 - 3xy)
= m [(x + y)2 - 3xy]
= m . [ m2 - 3n ]
\(x+y=2\)
\(\Rightarrow\)\(\left(x+y\right)^2=4\)
\(\Leftrightarrow\)\(x^2+y^2+2xy=4\)
\(\Leftrightarrow\)\(2xy=-6\) do x2 + y2 = 10
\(\Leftrightarrow\)\(xy=-3\)
\(T=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=2^3-3.\left(-3\right).2=26\)
Vì \(\left(x+y\right)=2\Rightarrow\left(x+y\right)^2=4\Leftrightarrow x^2+y^2+2xy=4\Leftrightarrow2xy=-6\Leftrightarrow xy=-3\)
\(T=x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(\Rightarrow T=2.\left(10-xy\right)\)
\(\Rightarrow T=20-2xy=20+6=26\)
a) A = 5(x + 3)(x - 3) + (2x + 3)2 + (x - 6)2 = 5(x2 - 9) + (4x2 + 12x + 9) + (x2 - 12x + 36) = 10x2
Tại x = -2,A = 10.(-2)2 = 40
b) x2 + y2 = x2 + 2xy + y2 - 2xy = (x + y)2 - 2.(-25) = 102 + 50 = 150
a) Vì x + y = 2
\(\Rightarrow\left(x+y\right)^2=4\)
\(\Rightarrow x^2+y^2+2xy=4\)
\(\Rightarrow10+2xy=4\)
\(\Rightarrow2\left(5+xy\right)=4\)
\(\Rightarrow5+xy=2\)
\(\Rightarrow xy=-3\)
Do đó:
\(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\)
\(\Rightarrow x^3+y^3=2\left[10-\left(-3\right)\right]\)
\(\Rightarrow x^3+y^3=2\left(10+3\right)\)
\(\Rightarrow x^3+y^3=2.13\)
\(\Rightarrow x^3+y^3=26\)
b) x + y = a
\(\Rightarrow\left(x+y\right)^2=a^2\)
\(\Rightarrow x^2+y^2+2xy=a^2\)
\(\Rightarrow b+2xy=a^2\)
\(\Rightarrow2xy=a^2-b\)
\(\Rightarrow xy=\dfrac{a^2-b}{2}\)
Do đó:
\(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\)
\(\Rightarrow x^3+y^3=a\left[b-\left(\dfrac{a^2-b}{2}\right)\right]\)
1) Ta có x + y = 2
=> ( x + y )2 = 4
=> x2 + 2xy + y2 = 4
=> 2xy = 4 - ( x2 + y2 )
=> 2xy = 4 - 10
=> 2xy = -6
=> xy = -3
Lại có x3 + y3 = ( x + y ) . ( x2 - xy + y2 ) = 2 . ( 10 - -3 ) = 2 . 13 = 26
2 ) Tương tự công thức trên ta có x3 + y3 = a . ( b - \(\dfrac{a^2-b}{2}\) )