Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) =(x-y)5+(x-y)3=(x-y)3[(x-y)2+1]
b) =33(y-2x)3:-9(y-2x)=-3(y-2x)2
c) =(x-y)2 [3(x-y)3-2(x-y)2+3]:5(x-y)2=[3(x-y)3-2(x-y)2+3]/5
Duy Nguyễn, Duy Nguyễn, Duy Nguyễn, Duy Nguyễn, Duy nguyễn, Duy Nguyễn, Duy Nguyễn, Duy Nguyễn, Duy Nguyễn
giúp bn ấy vs
\(\left(x+y\right)^2-2xy=x^2+y^2=4^2-2.1=14\)
\(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=14^2-2=196-2=194\)
\(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=4\left(14-1\right)=52\)
\(\left(x^4+y^4\right)\left(x+y\right)=194.4=776\Leftrightarrow x^5+y^5+x^4y+y^4x=\left(x^5+y^5\right)+xy\left(x^3+y^3\right)=\left(x^5+y^5\right)+1.52=\left(x^5+y^5\right)+52=776\Rightarrow x^5+y^5=724\)
\(\left\{{}\begin{matrix}x+y=4\\xy=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+2xy+y^2=16\\4xy=4\end{matrix}\right.\Rightarrow x^2+2xy-4xy+y^2=\left(x-y\right)^2=12mà:x>y\Leftrightarrow x-y>0\Rightarrow x-y=\sqrt{12}=2\sqrt{3};x+y=2.2\Rightarrow\left\{{}\begin{matrix}x=\sqrt{3}+2\\y=2-\sqrt{3}\end{matrix}\right.\)
\(x^2-y^2=\left(x-y\right)\left(x+y\right)=4.2\sqrt{3}=8\sqrt{3}\)
\(\left(x^2+y^2\right)\left(x^2-y^2\right)=8\sqrt{3}.14=112\sqrt{3}\Rightarrow x^4-y^4=112\sqrt{3}\)
\(\left(x^3-y^3\right)=\left(x-y\right)\left(x^2+xy+y^2\right);x^6-y^6=\left(x^3+y^3\right)\left(x^3-y^3\right)tựlm\)
Mạn phép bỏ câu a :))
b) a2(b2 - a2) + b2(b2 + a2)
= a2.b2 + a2.(-a2) + b2.b2 + b2.a2
= a2.b2 - a4 + b4 + a2.b2
= a4 + 2a2b2 + b2 (hđt)
c) x2(x3 + 2y - x2y) - y(x2 - x4 + y)
= x2.x3 + x2.2y + x2.(-x2y) + (-y).x2 + (-y).(-x)4 + (-y).y
= x5 + 2x2y - x4y - x2y + x4y - y2
= x5 + (2xy2 - xy2) + (-x4y + x4y) - y2
= x5 + xy2 - y2
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
a) \(\left(x+y\right)^5-x-y=\left(x+y\right)^5-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^4-1\right]\)
= \(\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\) #áp dụng hàng đẳng thức#
c) \(x^9-x^7-x^6-x^5+x^4+x^3+x^2+1\)nhóm vào là đc
b) \(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3+\left(y^2+z^2\right)^3\)
=\(\left(y^2+x^2\right)\left[\left(x^2+y^2\right)^2-\left(x^2+y^2\right)\left(z^2-x^2\right)+\left(z^2-x^2\right)^2\right]+\left(y^2+z^2\right)^3\)
= \(\left(y^2+z^2\right)\left[x^4+y^4+2x^2y^2-x^2z^2+x^4-y^2z^2+x^2y^2+z^4+x^4-2x^2z^2+y^4+z^4+2y^2z^2\right]\)
=\(=\left(y^2+z^2\right)\left(2x^4+2y^4+2z^4+3x^2y^2-3x^2z^2+y^2z^2\right)\)
