C/m đẳng thức:
(xy + z)\(^2\) - x\(^2\)y\(^2\)= z(2xy+z)
Chứng minh đẳng thức:
(xy+z)^2-x^2y^2=z(2xy+z)
ko cần khó khăn
\(VT=\left(xy+z\right)^2-x^2y^2=\left(xy+z\right)^2-\left(xy\right)^2\)
\(=\left(xy+z-xy\right)\left(xy+z+xy\right)\)
\(=z\left(2xy+z\right)=VP\)
Xin ghi lại đề
\(CM:\left(xy+z\right)^2-x^2y^2=z\left(2xy+z\right)\)
Và mới học lớp 7
:))
Bạn có viết sai đề ko vậy .Hãy kiểm tra lại đi
Chứng minh đẳng thức
a, (x-y-z)^2=x^2 + y^2+z^2-2xy+2yz-2zx
b, ( x+y-z)^2=x^2+y^2+z^2+2xy-2yz-2zx
c, ( x-y)(x^3+x^2y+xy^2+y^3)=5x(x+1)
d, ( x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)=x^5+y^5
Giúp mk vs ạ mk đang cần
a, b, nhân vào là ra à
c, nghe cứ là lạ
d, cũng nhân là ra hà
\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5=x^5+y^5\)
a) Ta có: \(VT=\left(x-y-z\right)^2\)
\(=\left(x-y-z\right)\left(x-y-z\right)\)
\(=x^2-xy-xz-yx+y^2+yz-zx+zy+z^2\)
\(=x^2+y^2+z^2-2xy+2yz-2xz\)
=VP(đpcm)
b) Ta có: \(VT=\left(x+y-z\right)^2\)
\(=\left(x+y-z\right)\left(x+y-z\right)\)
\(=x^2+xy-xz+yx+y^2-yz-zx-zy+z^2\)
\(=x^2+y^2+z^2+2xy-2yz-2zx\)
=VP(đpcm)
c) Sửa đề: Chứng minh \(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)=x^4-y^4\)
Ta có: \(VT=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\)
\(=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4\)
\(=x^4-y^4\)
=VP(đpcm)
d) Ta có: \(VT=\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)
\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5\)
\(=x^5+y^5\)
=VP(đpcm)
Tìm x, y, z thỏa mãn đẳng thức sau: (x - z)2 + (y - z)2 + y2 + z2 = 2xy - 2yz + 6z + 9
Chứng minh các đẳng thức sau:
a) (x-1) (x^2 + x+ 1) = x^3 -1
b) (x^3+x^2y + xy^2 + y^3) (x-y) = x^4 - y^4
c) (x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2 yz + 2zx
a) \(VT=\left(x-1\right)\left(x^2+x+1\right)\)
\(=x^3+x^2+x-x^2-x-1\)
\(=x^3-1=VP\)
b) \(VT=\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)\)
\(=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4\)
\(=x^4-y^4=VP\)
c) \(VT=\left(x+y+z\right)^2\)
\(=\left(x+y\right)^2+2\left(x+y\right)z+z^2\)
\(=x^2+2xy+y^2+2xz+2yz+z^2\)
\(=x^2+y^2+z^2+2xy+2yz+2zx=VP\)
Chúc bạn học tốt.
Tìm x,y,z thỏa mãn đẳng thức sau :
(x-z)2 + (y-z)2 +y2 +z2 = 2xy -2yz + 6z - 9
Chứng minh đẳng thức:
1) (xy+z)2 -x2y2 = z(2xy+z)
2) (x2+y2)2 -4x2y2 = (x+y)2 (x-y)2
1, \(\left(xy+z\right)^2-x^2y^2=z\left(2xy+z\right)\)
Biến đổi VT :\(\left(xy+z\right)^2-x^2y^2\)
\(=x^2y^2+2xyz+z^2-x^2y^2\)
\(=2xyz+z^2\)
\(=z\left(2xy+z\right)\) = VP
Vậy \(\left(xy+z\right)^2-x^2y^2=z\left(2xy+z\right)\)
2, \(\left(x^2+y^2\right)^2-4x^2y^2=\left(x+y\right)^2\left(x-y\right)^2\)
Biến đổi VT: \(\left(x^2+y^2\right)^2-4x^2y^2\)
\(=x^4+2x^2y^2+y^4-4x^2y^2\)
\(=x^4-2x^2y^2+y^4\)
Biến đổi VP: \(\left(x+y\right)^2\left(x-y\right)^2\)
\(=\left(x^2+2xy+y^2\right)\left(x^2-2xy+y^2\right)\)
\(=x^4-2x^3y+x^2y^2+2x^3y-4x^2y^2+2xy^3+x^2y^2-2xy^3+y^4\)\(=x^4-2x^2y^2+y^4\)
Ta có VT = VP
Vậy \(\left(x^2+y^2\right)^2-4x^2y^2=\left(x+y\right)^2\left(x-y\right)^2\)
1 ) \(VT=\left(xy+z\right)^2-x^2y^2\)
\(=x^2y^2+2xyz+z^2-x^2y^2\)
\(=2xyz+z^2\)
\(=z\left(2xy+z\right)=VP\left(đpcm\right)\)
2 ) \(VT=\left(x^2+y^2\right)^2-4x^2y^2\)
\(=x^4+2x^2y^2+y^4-4x^2y^2\)
\(=x^4+y^4-2x^2y^2\)
\(=\left(x^2-y^2\right)^2\)
\(=\left[\left(x-y\right)\left(x+y\right)\right]^2\)
\(=\left(x-y\right)^2\left(x+y\right)^2=VP\left(đpcm\right)\)
Giải chi tiết nha
Thu gọn biểu thức sau:
2
x
y
−
2
y
z
⋅
z
+
x
y
+
1
2
z
2
y
+
2
z
y
⋅
z
2xy−2yz⋅z+xy+12z2y+2zy⋅z
Chứng minh đẳng thức:
a) (x-y-z)2 = x2 + y2 + z2 - 2xy + 2yz - 2zx
b) (x+y-z)2 = x2 + y2 + z2 + 2xy - 2yz - 2zx
c) (x-y)(x3 + x2y + xy2 + y3 = x4 - y4
d) (x+y)(x4 - x3y + x2y2 - xy3 + y4) = x5 + y5
a) Ta có: \(VP=x^2+y^2+z^2-2xy+2yz-2zx\)
\(=\left(x^2-xy-xz\right)+\left(y^2-xy+yz\right)+\left(z^2-yz-zx\right)\)
\(=x\left(x-y-z\right)+y\left(y-x+z\right)+z\left(z-y-x\right)\)
\(=x\left(x-y-z\right)-y\left(x-y-z\right)-z\left(x-y-z\right)\)
\(=\left(x-y-z\right)\left(x-y-z\right)\)
\(=\left(x-y-z\right)^2=VT\)(đpcm)
b) Ta có: \(VP=x^2+y^2+z^2+2xy-2yz-2zx\)
\(=\left(x^2+xy-zx\right)+\left(y^2+xy-2yz\right)+\left(z^2-yz-zx\right)\)
\(=x\left(x+y-z\right)+y\left(x+y-z\right)+z\left(z-y-x\right)\)
\(=\left(x+y-z\right)\left(x+y\right)-z\left(x+y-z\right)\)
\(=\left(x+y-z\right)\left(x+y-z\right)\)
\(=\left(x+y-z\right)^2=VT\)(đpcm)
c) Ta có: \(VP=x^4-y^4\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x^3+xy^2+x^2y+y^3\right)=VT\)(đpcm)
d) Ta có: \(VT=\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)
\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5\)
\(=x^5+y^5=VP\)(đpcm)
c/m: a) x^2 +y^2+z^2>=2xy-2xz+2yz
b) x^4+y^4+z^2+1 >=2x.(xy^2-x+z+1)