PTĐTTNT: xy + x + y + 1
PTĐTTNT:
a. (xy+1)^2-(x+y)^2
b.64x^4+y^4
a. \(\left(xy+1\right)^2-\left(x+y\right)^2=\left(xy+1+x+y\right)\left(xy+1-x-y\right)=\left(x+1\right)\left(y+1\right)\left(x-1\right)\left(y-1\right)\)
b.
thêm bớt 16x^2y^2
\(64x^4+y^4=64x^4+16x^2y^2+y^4-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2+y^2+4xy\right)\left(8x^2+y^2-4xy\right)\)
PTĐTTNT
\(^{x^3-xy^2+x^2y-y^2z}\)
\(A=x^3-xy^2+x^2y-y^2z\)
\(A=\left(x^3-xy^2\right)+\left(x^2y-y^2z\right)\)
\(A=x\left(x^2-y^2\right)+x\)
hình như bạn sai đề rồi mình cũng không giải được
bạn kiểm tra lại đề đi nhé
(thanks)
Phát ơi ,đề sai rồi , bạn kiểm tra lại nhé :(
1) \(x^2-2x+1+x^2y-xy=\left(x-1\right)^2+xy\left(x-1\right)=\left(x-1\right)\left(x+xy-1\right)\)
2) \(x^2+6x+9+x^2y+3xy\)
\(=\left(x+3\right)^2+xy\left(x+3\right)\)
\(=\left(x+3\right)\left(x+xy+3\right)\)
PTĐTTNT:
x3 + xy2 - 9xz2 - 2xyz
\(x^3+xy^2-9xz^2-2xyz\)
\(=x\left(x^2+y^2-9z^2-2yz\right)\)
I : 1) PTĐTTNT : a) \(x^2-2xy+y^2-x+y\)
b) CMR giá trị của biểu thức sau không phụ thuộc vào biến
\(C=\left(x-y\right)\left(x^2+xy+y^2\right)-x\left(x^2-y\right)+y\left(y^2-x\right)\)
help me
1) x2-2xy+y2-x+y
(=) (x-y)2-(x-y)
(=) [(x-y)-1].(x-y)
(=) (x-y-1).(x-y)
C= (x-y)(x2+xy+y2)-x(x2-y)+y(y2-x)
(=) x3-y3-x3+xy+y3-xy
(=)(x3-x3)+(-y3+y3)+(xy-xy)
(=) 0
x ³ + y ³ -2(x ² - y ²)
ptđttnt
\(x^3+y^3-2(x^2-y^2)\\=(x+y)(x^2-xy+y^2)-2(x-y)(x+y)\\=(x+y)[x^2-xy+y^2-2(x-y)]\\=(x+y)(x^2-xy+y^2-2x+2y)\\=(x+y)(-x^2-xy+2y+y^2)\)
PTĐTTNT:
1) x2y2 - 1
1) \(x^2.y^2-1\)
\(=\left(x.y\right)^2-1^2\)
\(=\left(x.y-1\right).\left(x.y+1\right)\)
Chúc bạn học tốt!
PTĐTTNT:
\(\text{a) }\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(xy+yz+zx\right)^2\)
\(\text{b) }2\left(x^4+y^4+z^4\right)-\left(x^2+y^2+z^2\right)^2-2\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(x+y+z\right)^4\)
(x−y+z)2+(z−y)2+2(x−y+z)(y−z)(x−y+z)2+(z−y)2+2(x−y+z)(y−z)
=(x−y+z)2+(z−y)2+(x−y+z)(y−z)+(x−y+z)(y−z)=(x−y+z)2+(z−y)2+(x−y+z)(y−z)+(x−y+z)(y−z)
=(x−y+z)2+(x−y+z)(y−z)+(z−y)2+(x−y+z)(y−z)=(x−y+z)2+(x−y+z)(y−z)+(z−y)2+(x−y+z)(y−z)
=(x−y+z)2+(x−y+z)(y−z)+(z−y)2−(x−y+z)(z−y)=(x−y+z)2+(x−y+z)(y−z)+(z−y)2−(x−y+z)(z−y)
=(x−y+z)(x−y+y+z−z)+(z−y)[z−y−(x−y+z)]=(x−y+z)(x−y+y+z−z)+(z−y)[z−y−(x−y+z)]
=(x−y+z)x+(z−y)(z−y−x+y−z)=(x−y+z)x+(z−y)(z−y−x+y−z)
=x2−xy+xz+(z−y)(−x)=x2−xy+xz+(z−y)(−x)
=x2−xy+xz−xz+xy=x2−xy+xz−xz+xy
=x2
PTĐTTNT
\(\text{a) }\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(xy+yz+xz\right)^2\)
\(\text{b) }2\left(x^4+y^4+z^4\right)-\left(x^2+y^2+z^2\right)^2-2\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(x+y+z\right)^4\)
\(x;y;z\rightarrow q;h;p\)
\(=\left(q^2+h^2+p^2\right)\left(q^2+h^2+p^2+2qh+2hp+2qp\right)+\left(qh+hp+pq\right)^2\)
\(Dat:\hept{\begin{cases}q^2+h^2+p^2=f\\qh+hp+qp=g\end{cases}}\Rightarrow\left(p^2+h^2+q^2\right)\left(p+q+h\right)^2+\left(qh+pq+ph\right)^2\)
\(=f\left(f+2g\right)+g^2=f^2+2fg+g^2=\left(f+g\right)^2=\left(q^2+h^2+p^2+qh+hp+pq\right)^2\)
shitbo Cho đệ sửa lại bài SP chứ bài SP dài quá ạ:p
\(\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(xy+yz+zx\right)^2\)
\(=\left(x^2+y^2+z^2\right)\left(x^2+y^2+z^2+2xy+yz+zx\right)+\left(xy+yz+zx\right)^2\)
Đặt \(x^2+y^2+z^2=a;xy+yz+zx=b\)
\(\Rightarrow a\left(a+2b\right)+b^2=a^2+2ab+b^2=\left(a+b\right)^2=\left(x^2+y^2+z^2+xy+yz+zx\right)^2\)
Đặt \(x^4+y^4+z^4=a;x^2+y^2+z^2=b;x+y+z=c\)
Ta có:\(2a-b^2-2bc^2+c^4\)
\(=2a-2b^2+b^2-2bc^2+c^4\)
\(=2\left(a-b^2\right)+\left(b-c^2\right)^2\)
Lại có:
\(a-b^2=-2\left(x^2y^2+y^2z^2+z^2x^2\right);b-c^2=-2\left(xy+yz+zx\right)\)( Nhác quá hơi tắt xíu )
Thay vào ta được:
\(2\left(a-b^2\right)+\left(b-c^2\right)^2\)
\(=-4\left(x^2y^2+y^2z^2+z^2x^2\right)+4\left(x^2y^2+y^2z^2+z^2x^2+xyz\left(x+y+z\right)\right)\)
\(=4xyz\left(x+y+z\right)\)
x .(x-z)-y.(x+z)^3 PTĐTTNT