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\(C=x^2-y^2\)
Tương tự câu \(A=x^2+y^2\)
\(D=x^4+y^4\)
Thay x + y = 17; x.y = 60 vào \(\left(x+y\right)^2=x^2+2xy+y^2\):
172 = x2 + 2.60 + y2
289 = x2 + 120 + y2
\(\Leftrightarrow x^2+y^2=169\)
Lại có:
\(\left(x^2+y^2\right)^2=x^4+y^4+2x^2y^2\)
\(\left(x^2+y^2\right)^2=x^4+y^4+\left(2xy\right)^2\)
Thay \(x^2+y^2=169;x.y=60\)vào biểu thức trên:
1692 = x4 + y4 + 2 . 602
\(\Leftrightarrow x^4+y^4=28561-7200\)
\(\Leftrightarrow x^4+y^4=21361\)
a) \(2x+2y\)
\(=2\left(x+y\right)\)
b) \(5x+20y\)
\(=5\left(x+4y\right)\)
c) \(6xy-30y\)
\(=6y\left(x-5\right)\)
d) \(5x\left[x-110-10y\left(x-11\right)\right]\)
\(=5x\left(x-110-10xy+110\right)\)
\(=5x\left(x-10xy\right)\)
\(=5x^2\left(1-10y\right)\)
e) \(x^3-4x^2+x\)
\(=x\left(x^2-4x+1\right)\)
f) \(x\left(x+y\right)-\left(2x+2y\right)\)
\(=x\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-2\right)\)
h) \(5x\left(x-2y\right)+2\left(2y-x\right)\)
\(=5x\left(x-2y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\left(5x-2\right)\)
i) \(x^2y^3-\dfrac{1}{2}x^4y^8\)
\(=x^2y^3\left(1-\dfrac{1}{2}xy^5\right)\)
j) \(a^2b^4+a^3b-abc\)
\(=ab\left(ab^3+a^2-c\right)\)
\(x+y+z=0< =>x+y=-z=>\left(x+y\right)^2=\left(-z\right)^2.\)
\(< =>x^2+2xy+y^2=z^2< =>x^2+y^2-z^2=-2xy\)
\(< =>\left(x^2+y^2-z^2\right)=\left(-2xy\right)^2\)
\(< =>x^4+y^4+z^4+2x^2y^2-2x^2z^2-2y^2z^2=4x^2y^2\)
\(< =>x^4+y^4+z^4=2x^2y^2+2y^2z^2+2x^2z^2\)
\(< =>2\left(x^4+y^4+z^4\right)=x^4+y^4+z^4+2x^2y^2+2y^2z^2+2x^2z^2=\left(x^2+y^2+z^2\right)^2.\)
\(< =>x^4+y^4+z^4=\frac{\left(x^2+y^2+z^2\right)^2}{2}=\frac{a^4}{2}\)
Vậy \(x^4+y^4+z^4=\frac{a^4}{2}\)
1,Thực hiện phép tính :
a, (x + 2)9 : (x + 2)6
=(x+2)9-6
=(x+2)3
b, (x - y) 4 : (x - 2)3
=(x-y)4-3
=x-y
c, ( x2+ 2x + 4)5 : (x2 + 2x + 4)
=(x2+2x+4)5-1
=(x2+2x+4)4
d, 2(x2 + 1)3 : 1/3(x2 + 1)
=(2÷1/3).[(x2+1)3÷(x2+1)]
=6(x2+1)2
e, 5 (x - y)5 : 5/6 (x - y)2
=(5÷5/6).[(x-y)5÷(x-y)2]
=6(x-y))3