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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=a
x2+y2=b
E=x3+y3=(x+y)(x2-xy+y2) =a(b-xy) { bh ta cần tính xy là ok}
Ta có: (x+y)2 =a2
<=>x2+xy+y2=a2
<=>x2+y2+xy=a2
<=> xy =a2-b { chỗ này thay x2+y2=b vào và chuyển vế }
E=a(b-xy)=a(b-a2-b)=-a3
x2 +y2=b\(\Leftrightarrow\)(x+y)2-2xy=b\(\Leftrightarrow\)a2-2xy=b\(\Leftrightarrow\)xy=\(\frac{a^2-b}{2}\)
E=x3+y3=(x+y)(x2-xy+y2)=a\(\times\)(b \(-\)\(\frac{a^2-b}{2}\))=\(\frac{3ab-a^3}{2}\)
\(x^3+y^3\) \(=\left(x+y\right)\left(x^2-xy+y^2\right)\) \(=a.\left(b-xy\right)\) \(=ab\) \(-\) axy
Có : \(x+y=a\Rightarrow x^2+2xy+y^2=a^2\)
\(\Leftrightarrow b+2xy=a^2\)
\(\Leftrightarrow xy=\frac{a^2-b}{2}\)
Lại có :
\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=a.\left(b-xy\right)=ab-a.\frac{a^2-b}{2}\)
a) Ta có:
x + y = 3
=> ( x + y)2 = 9
=> x2 + 2xy + y2 = 9
=> 10 + 2xy = 9
=> 2xy = 9 - 10 = -1
=> xy = -1/2
Ta có:
x3 + y3 = (x + y)(x2 - xy + y2)
= 3.(10 + 1/2) = 63/2
b) Ta có: x + y = a
=> (x + y)2 = a2
=> x2 + 2xy + y2 = a2
=> b + 2xy = a2
=> xy = (a2 - b)/2
Ta có: x3 + y3 = (x + y)(x2 + xy + y2)
= a[b + (a2 - b )/2] = ab + (a3 - b)/2.
Làm b) công thức tổng quát luôn
x+y=a => (x+y)^2 =a^2 => x^2+y^2+2xy=a^2
Thay x^2+y^2=b vào ta được:
b+2xy=a^2 => xy=(a^2-b)/2
TA có x^3+y^3 =(x+y)(x^2+y^2 -xy)= a [b+(a^2-b)/2] =ab +(a^3-ab)/2=ab/2+a^3/2
a) Vì \(x-y=1\)
\(\Rightarrow\left(x-y\right)^3=1\)
\(\Leftrightarrow x^3-y^3-3xy\left(x-y\right)=1\)
\(\Leftrightarrow x^3-y^3-3xy=1\)
b) \(B=2\left(x^3-y^3\right)-3\left(x+y\right)^2\)
\(=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(=4\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(=4x^2+4xy+4y^2-3x^2-6xy-3y^2\)
\(=x^2-2xy+y^2\)
\(=\left(x-y\right)^2\)
\(=4\)
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 \(\left(x+y\right)=3\Rightarrow\left(x+y\right)^3=27\)
\(\Rightarrow x^3+3x^2y+3xy^2+y^3=27\)
\(\Rightarrow x^3+y^3+3xy\left(x+y\right)=27\)
\(\Rightarrow x^3+y^3+3.2.3=27\)
\(\Rightarrow x^3+y^3=27-18=9\)
\(b,x-y=5\Rightarrow\left(x-y\right)^2=25\Rightarrow x^2-2xy+y^2=25\Rightarrow2xy=-10\Leftrightarrow xy=-5\)
\(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=5.10=50\)
Ta có: \(x+y=a\)
\(\Leftrightarrow\left(x+y\right)^2=a^2\)
\(\Leftrightarrow x^2+2xy+y^2=a^2\)
\(\Leftrightarrow b+2xy=a^2\)
\(\Leftrightarrow2xy=a^2-b\)
\(\Leftrightarrow xy=\frac{a^2-b}{2}\)
Ta có: \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=a\cdot\left(b-\frac{a^2-b}{2}\right)\)
\(=a\cdot\left(\frac{2b-a^2+b}{2}\right)\)
\(=a\cdot\frac{3b-a^2}{2}\)
\(=\frac{3ab-a^3}{2}\)
x + y = a => (x + y)2 = a2 => x2 + xy + y2 = a2 => 2xy = a2 - 2b => xy = 1/2(a2 - 2b)
(x + y)3 = a3 =>x3 + y3 + 3xy(x + y) = a3 => x3 + y3 = a3 - 3*1/2(a2 - 2b)*a
=> x3 + y3 = a3 - 3/2a(a2 - 2b).