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a) x2+10x+26+y2+2y
=x2+10x+25+y2+2y+1
=(x+5)2+(y+1)2
b) z2-6z+5-t2-4t
=z2-6z+9-t2-4t-4
=(z-3)2-(t2+4t+4)
=(z-3)2-(t+2)2
c)x2-2xy+2y2+2y+1
=x2-2xy+y2+y2+2y+1
=(x-y)2+(y+1)2
d) 4x2-12x-y2+2y+8
=4x2-12x+9-y2+2y-1
=(2x-3)2-(y2-2y+1)
=(2x-3)2-(y-1)2
1a/ z2 - 6z + 5 - t2 - 4t = z2 - 2 . 3z + 32 - 4 - t2 - 4t = (z2 - 2 . 3z + 32) - (22 + 2 . 2t + t2) = (z - 3)2 - (2 + t)2
b/ x2 - 2xy + 2y2 + 2y2 + 1 = x2 - 2xy + y2 + y2 + 2y + 1 = (x2 - 2xy + y2) + (y2 + 2y + 1) = (x - y)2 + (y + 1)2
c/ 4x2 - 12x - y2 + 2y + 8 = (2x)2 - 12x - y2 + 2y + 32 - 1 = [ (2x)2 - 2 . 3 . 2x + 32 ] - (y2 - 2y + 1) = (2x - 3)2 - (y - 1)2
2a/ (x + y + 4)(x + y - 4) = x2 + xy - 4x + xy + y2 - 4y + 4x + 4y + 16 = x2 + (xy + xy) + (-4x + 4x) + (-4y + 4y) + y2 + 16
= x2 + 2xy + y2 + 42 = (x + y)2 + 42
b/ (x - y + 6)(x + y - 6) = x2 + xy - 6x - xy - y2 + 6y + 6x + 6y - 36 = x2 + (xy - xy) + (-6x + 6x) + (6y + 6y) - y2 - 36
= x2 - y2 + 12y - 62 = x2 - (y2 - 12y + 62) = x2 - (y2 - 2 . 6y + 62) = x2 - (y - 6)2
c/ (y + 2z - 3)(y - 2z - 3) = y2 -2yz - 3y + 2yz - 4z2 - 6z - 3y + 6z + 9 = y2 + (-2yz + 2yz) + (-3y - 3y) + (-6z + 6z) - 4z2 + 9
= y2 - 6y - 4z2 + 9 = (y2 - 6y + 9) - 4z2 = (y - 3)2 - (2z)2
d/ (x + 2y + 3z)(2y + 3z - x) = 2xy + 3xz - x2 + 4y2 + 6yz - 2xy + 6yz + 9z2 - 3xz = (2xy - 2xy) + (3xz - 3xz) - x2 + (6yz + 6yz) + 9z2 + 4y2
= -x2 + 4y2 + 12yz + 9z2 = (4y2 + 12yz + 9z2) - x2 = [ (2y)2 + 2 . 2 . 3yz + (3z)2 ] - x2 = (2y + 3z)2 - x2
a) \(x^2+10x+26+y^2+2y\)
\(=x^2+2.5x+25+1+y^2+2y\)
\(=\left(x^2+2.5x+25\right)+\left(1+2y+y^2\right)\)
\(=\left(x+5\right)^2+\left(1+y\right)^2\)
b) \(x^2-2xy+2y^2+2y+1\)
\(=x^2-2xy+y^2+y^2+2y+1\)
\(=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)\)
\(=\left(x-y\right)^2+\left(y+1\right)^2\)
c) \(z^2-6z+13+t^2+4t\)
\(=z^2-2.3z+9+4+t^2+4t\)
\(=\left(z^2-2.3x+9\right)+\left(4+4t+t^2\right)\)
\(=\left(z-3\right)^2+\left(2+t\right)^2\)
d) \(4x^2+2z^2-4xz-2z+1\)
\(=4x^2+z^2+z^2-4xz-2z+1\)
\(=\left(4x^2-4xz+z^2\right)+\left(z^2-2z+1\right)\)
\(=\left(2x-z\right)^2+\left(z-1\right)^2\)
a) \(\Rightarrow\left(x^2+2\times5x+25\right)+\left(y^2+2y+1\right)\)
\(\Rightarrow\left(x+5\right)^2+\left(y+1\right)^2\)
Bạn làm bài kiểm tra hả sao nhiều bài tek. Mk làm mất khá nhiều tg luôn đó 
Có một số câu thì mình không làm được. Mong bạn thông cảm!!!
a) \(\left(5xy^3\right)^2-2.5xy^3.6yz^2+\left(6yz^2\right)^2\)=\(\left(5xy^3-6yz^2\right)^2\)
b) \(\left(\frac{1}{3}u^2v^3\right)^2-2.\frac{1}{3}u^2v^3.\frac{1}{2}u^3v+\left(\frac{1}{2}u^3v\right)^2\)=\(\left(\frac{1}{3}u^2v^3-\frac{1}{2}u^3v\right)^2\)
\(\text{1) }\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\\ =\dfrac{\left(x^7+x^6\right)+\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6+x^4+x^2+1}{x-1}\)
\(\text{3) }\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\\ =\dfrac{\left(x^2-2xy+y^2\right)+\left(2xz-2yz\right)+z^2}{\left(x^2-2xy+y^2\right)-z^2}\\ =\dfrac{\left(x-y\right)^2+2\left(x-y\right)z+z^2}{\left(x-y\right)^2-z^2}\\ =\dfrac{\left(x-y+z\right)^2}{\left(x-y+z\right)\left(x-y-z\right)}\\ =\dfrac{x-y+z}{x-y-z}\)
Bạn tự tách hđt nhé! Gõ mỏi tay :v~
\(\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2=\left(y+z-2x\right)^2+\left(z+x-2y\right)^2+\left(y+z-2z\right)^2\)
⇔ \(y^2-2yz+z^2+z^2-2xz+x^2+x^2-2xy+y^2=\)\(6(z^2-yz-xz+y^2-xy+x^2)\)
⇔ \(2\left(x^2+y^2+z^2-yz-xz-xy\right)\)=\(6(z^2-yz-xz+y^2-xy+x^2)\)
⇔ \(x^2+y^2+z^2-yz-xz-xy\) = \(3(z^2-yz-xz+y^2-xy+x^2)\)
⇔ \(2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)
⇔ \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
Mà \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\forall x;y;z\)
Do đó \(\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\)
⇒ \(x=y=z\)
j lắm thế :)))
Bài 2 : ~ bài 1 ngán quá =)))
a, Có
\(5x^2+10y^2-6xy-4x-2y+3\)
\(=\left(x^2-6xy+9y^2\right)+\left(4x^2-4x+1\right)+\left(y^2-2y+1\right)+1\)
\(=\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1>0\forall x;y\)
Do đó không tồn tại x , y tm \(5x^2+10y^2-6xy-4x-2y+3=0\)
b, \(x^2+4y^2+z^2-2x-6x+6y+15=0\)
Câu này đề sai :v bài ngta không cho 2 lần x vậy đâu bạn :)))
bài 1:
a) x2 + 10x + 26 + y2 + 2y
= (x2 + 10x + 25) + (y2 + 2y + 1)
= (x + 5)2 + (y + 1)2
b) z2 - 6z + 5 - t2 - 4t
= (z - 3)2 - (t + 2)2
c) x2 - 2xy + 2y2 + 2y + 1
= (x2 - 2xy + y2) + (y2 + 2y + 1)
= (x - y)2 + (y + 1)2
d) 4x2 - 12x - y2 + 2y + 1
= (4x2 - 12x ) - (y2 + 2y + 1)
= ......................................
ok mk nhé!! 4545454654654765765767587876968345232513546546575675767867876876877687975675