Chứng minh đẳng thức :
(x+y+z)^2-x^2-y^2-z^2=2(xy+yz+zx)
CHỨNG MINH ĐẲNG THỨC (x+y+z)-x2-y2-z2=2(xy+yz+zx)
Sửa đề \(\left(x+y+z\right)^2-x^2-y^2-z^2=2\left(xy+yz+zx\right)\)
Ta có : \(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx\)(hằng đẳng thức cho 3 số )
\(\Rightarrow\left(x+y+z\right)^2-x^2-y^2-z^2=2\left(xy+yz+zx\right)\left(đpcm\right)\)
Vậy
Chứng minh đẳng thức(x+y+z)2-x2-y2-z2= 2(xy+yz+zx)
Ta có:
VT= \(\left(x+y+z\right)^2-x^2-y^2-z^2\)
\(=x^2+y^2+z^2+2xy+2yz+2zx-x^2-y^2-z^2\)
\(=2\left(xy+yz+zx\right)\) = VP
=> đpcm
\(\left(x+y+z\right)^2-x^2-y^2-z^2=2\left(xy+yz+zx\right)\)
Biến đổi vế trái:
VT\(\)\(\)\(=\left[\left(x+y\right)+z\right]^2-x^2-y^2-z^2\)
\(=\left(x+y\right)^2+2\left(x+y\right)z+z^2-x^2-y^2-z^2\)
\(=x^2+2xy+y^2+2xz+2yz+z^2-x^2-y^2-z^2\)\
\(=2xy+2yz+2zx\)
\(=2\left(xy+yz+zx\right)=\) VP
Chứng minh đẳng thức:
( x+y+z)2-x2-y2-z2= 2( xy+yz+zx)
Ta có:
\(\left(x+y+z\right)^2-x^2-y^2-z^2\)
\(=x^2+y^2+z^2+2xy+2yz+2zx-x^2-y^2-z^2\)
\(=2xy+2yz+2zx\)
\(=2\left(xy+yz+zx\right)\)
Chứng minh đẳng thức :
\(\left(x+y+z\right)^2-x^2-y^2-z^2=2\left(xy+yz+zx\right)\)
Có: \(\left(x+y+z\right)^2-x^2-y^2-z^2\)
\(=x^2+y^2+z^2+2xy+2yz+2xz-x^2-y^2-z^2\)
\(=2xy+2yz+2xz\)
\(=2\left(xy+yz+xz\right)\)
\(\left[\left(x+y\right)+z\right]^2=\left[\left(x+y\right)^2+2.\left(x+y\right)z+z^2\right]=x^2+2xy+y^2+2xz+2yz+z^2\)\(+z^2\)
Thay vào: x^2+y^2+z^2+ 2xy+2yz+2xz - x^2 - y^2 - z^2= 2(xy+yz+xz) (đpcm)
Chứng minh đẳng thức sau: \(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz\)
Ta có:
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-\left[3xy\left(x+y\right)+3xyz\right]\)
\(=\left(x+y+z\right)^3-3\left(x+y+z\right)\left(x+y\right).z-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yx-3xz-3yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
=> \(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz\)
Chứng minh hằng đẳng thức:(x+y+z)2 - x2 - y2 - z2=2(xy+yz+zx)
GIÚP VS!!!!!!!!!!!!!
\(VT=\left(x+y+z\right)^2-x^2-y^2-z^2\)
\(=\left[\left(x+y\right)+z\right]^2-x^2-y^2-z^2\)
\(=\left(x+y\right)^2+2\left(x+y\right)z+z^2-x^2-y^2-z^2\)
\(=x^2+2xy+y^2+2xz+2yz+z^2-x^2-y^2-z^2\)
\(=2xy+2yz+2zx\)
\(=2\left(xy+yz+zx\right)\)
\(=VP\)
Vậy...
Chứng minh hằng đẳng thức:(x+y+z)2 - x2 - y2 - z2=2(xy+yz+zx)
GIÚP VS!!!!!!!!!!!!!
Xin lỗi mk viết nhầm
(x+y+z)2-x2-y2-z2 =x2+y2+z2+2(xy+yz+xz)-x2-y2-z2
(x+y+z)2-x2-y2-z2
=x2+y2+2(xy+yz+xz)-x2-y2-z2
= 2(xy+yz+xz)
Vậy hằng đẳng thức được chứng minh
\(\left(x+y+z\right)^2-x^2-y^2-z^2=2\left(xy+yz+zx\right)\)
Chứng minh đẳng thức trên
(x+y+z)^2-x^2-y^2-z^2=2
=x^2+y^2+z^2+2xy+2yz+2xz-x^2-y^2-z^2
=2xy+2yz+2xz=2(xy+yz+xz) (đpcm)
(x+y+z)2-x2-y2-z2=2(xy+yz+zx)
x2+y2+z2+2xy+2yz+2zx-x2-y2-z2=2(xy+yz+zx)
\(\Rightarrow\)2xy+2yz+2zx=2(xy+yz+zx)
\(\Rightarrow\)2(xy+yz+zx)=2(xy+yz+zx)
vậy (x+y+z)2-x2-y2-z2=2(xy+yz+zx)
1)Chứng minh đẳng thức sau:
X^3 + y^3 + z^3 = ( x+ y+ z) .(x^2+ y^2 + z^2 - xy- yz- zx)
2) Chứng minh:
43^2020 + 43^2021 chia hết cho 11
Nhanh lên!Cần gấp lắm rồi!
2) \(43^{2020}+43^{2021}=43^{2020}\left(1+43\right)=43^{2020}.44\)
Mà \(44⋮11\Rightarrow43^{2020}.44⋮11\Rightarrow43^{2020}+43^{2021}⋮11\)
Phần 1 đang nghĩ -.-