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Lời giải:
a. $x^2+y^2+4y+13-6x$
$=(x^2-6x+9)+(y^2+4y+4)$
$=(x-3)^2+(y+2)^2$
b.
$4x^2-4xy+1+2y^2-2y$
$=(4x^2-4xy+y^2)+(y^2-2y+1)$
$=(2x-y)^2+(y-1)^2$
c.
$x^2-2xy+2y^2+2y+1$
$=(x^2-2xy+y^2)+(y^2+2y+1)$
$=(x-y)^2+(y+1)^2$
a. \(x^2+y^2+4y+12-6x=\left(x^2-6x+9\right)+\left(y^2+4y+4\right)=\left(x-3\right)^2+\left(y+2\right)^2\)b. \(4x^2-4xy+1+2y^2-2y=\left(4x^2-4xy+y^2\right)+\left(y^2-2y+1\right)=\left(2x-y\right)^2+\left(y-1\right)^2\)c. \(x^2-2xy+2y^2+2y+1=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x-y\right)^2+\left(y+1\right)^2\)
b:=y^2+2y+1+9x^2-12x+4
=(y+1)^2+(3x-2)^2
a:
SỬa đề: 5y^2
=y^2-10y+25+9x^2+4y^2-12xy
=(y-5)^2+(3x-2y)^2
\(a.\)
\(z^2-6z+5-t^2-4t\)
\(=z^2-6z+9-\left(t^2+4t+4\right)\)
\(=\left(z-3\right)^2-\left(t+2\right)^2\)
\(b.\)
\(4x^2-12x-y^2+2y+1\)
Câu này đề sai sao ấy em !
b, mik nghĩ đề sửa thành: \(4x^2-12x-y^2+2y+8\)
\(=4x^2-12x+9-y^2+2y-1\)
\(=\left(2x\right)^2-2.2.3.x+3^2-\left(y^2-2y+1\right)\)
\(=\left(2x-3\right)^2-\left(y-1\right)^2\)
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
4x²y⁴ - 4xy³ + y²
= (2xy²)² - 2.2xy².y + y²
= (2xy² - y)²
------------
Sửa đề:
(x - 2y)² - 4(x - 2y) + 4
= (x - 2y)² - 2.(x - 2y).2 + 2²
= (x - 2y - 2)²
------------
25x² - 5xy + 1/4 y²
= (5x)² - 2.5xy.y/2 + (y/2)²
= (5x - y/2)²
\(4x^2y^4-4xy^3+y^2\)
\(=\left(2xy^2\right)^2-2\cdot2xy^2\cdot y+y^2\)
\(=\left(2xy^2-y\right)^2\)
_____
\(\left(x-2y\right)^2-4\left(x-2y\right)+4\)
\(=\left(x-2y\right)^2-2\cdot\left(x-2y\right)\cdot2+2^2\)
\(=\left[\left(x-2y\right)-2\right]^2\)
\(=\left(x-2y-2\right)^2\)
____
\(25x^2-5xy+\dfrac{1}{4}y^2\)
\(=\left(5x\right)^2-2\cdot\dfrac{5}{2}xy+\left(\dfrac{1}{2}y\right)^2\)
\(=\left(5x\right)^2-2\cdot\dfrac{1}{2}y\cdot5x+\left(\dfrac{1}{2}y\right)^2\)
\(=\left(5x-\dfrac{1}{2}y\right)^2\)
A= 2x^2 + y^2 - 2xy -2x+3
A= x^2-2xy + y^2 + x^2 - 2x+ 1 +2
A= (x-y)^2 + (x-1)^2 + 2
(x-y)^2> hoặc = 0 với mọi giá trị của x
(x-1)^2 > hoặc =0 với mọi giá trị của x
=> (x-y)^2 + (x-1)^2 > hoặc =0 với mọi giá trị của x
=> (x-y)^2 + (x-1)^2 + 2 > hoặc =2
=> A lớn hơn hoặc bằng 2
=> GTNN của A=2 tại x=y=1
a) \(x^2+10x+26+y^2+2y\)
= \(x^2+10x+25+y^2+2y+1\)
= \(\left(x+5\right)^2+\left(y+1\right)^2\)
b) \(x^2-2xy+2y^2+2y+1\)
= \(x^2-2xy+y^2+y^2+2y+1\)
= \(\left(x-y\right)^2+\left(y+1\right)^2\)
c) \(z^2-6z+5-t^2-4t\)
= \(z^2-6z+9-\left(t^2+4t+4\right)\)
= \(\left(z-3\right)^2-\left(t+2\right)^2\)
d) \(4x^2-12x-y^2+2y+1\)
Hình như câu này sai đề -_-
a, \(x^2+10x+26+y^2+2y\)
\(=\left(x^2+2.x.5+5^2\right)+\left(1^2+2.1.y+y^2\right)\)
\(=\left(x+5\right)^2+\left(y+1\right)^2\)
b, \(x^2-2xy+2y^2+2y+1\)
\(=x^2-2xy+y^2+y^2+2y+1\)
\(=\left(x^2-2.x.y+y^2\right)+\left(y^2+2.y.1+1^2\right)\)
\(=\left(x-y\right)^2+\left(y+1\right)^2\)
c,\(z^2 -6z+5-t^2-4t\)
\(=-\left(t^2+4t-z^2+6z-5\right)\)
\(=-\left(t^2+2.t.2+2^2-z^2+2.z.3-3^2\right)\)
\(=-\left(\left(t^2+2.t.2+2^2\right)-\left(z^2-2.z.3+3^2\right)\right)\)
\(=-\left(\left(t+2\right)^2-\left(z-3\right)^2\right)\)
\(=\left(z-3\right)^2-\left(t+2\right)^2\)
d, Không biết làm hihi :)
\(x^2+3y^2-4x+6y+7=0\\ \Leftrightarrow\left(x^2-4x+4\right)+\left(3y^2+6y+3\right)=0\\ \Leftrightarrow\left(x-2\right)^2+3\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
\(3x^2+y^2+10x-2xy+26=0\\ \Leftrightarrow\left(x^2-2xy+y^2\right)+\left(2x^2+10x+\dfrac{25}{8}\right)+\dfrac{183}{8}=0\\ \Leftrightarrow\left(x-y\right)^2+2\left(x^2+2\cdot\dfrac{5}{2}x+\dfrac{25}{4}\right)+\dfrac{183}{8}=0\\ \Leftrightarrow\left(x-y\right)^2+2\left(x+\dfrac{5}{2}\right)^2+\dfrac{183}{8}=0\\ \Leftrightarrow x,y\in\varnothing\)
Sửa đề: \(3x^2+6y^2-12x-20y+40=0\)
\(\Leftrightarrow\left(3x^2-12x+12\right)+\left(6y^2-20y+\dfrac{50}{3}\right)+\dfrac{34}{3}=0\\ \Leftrightarrow3\left(x-2\right)^2+6\left(y^2-2\cdot\dfrac{5}{3}y+\dfrac{25}{9}\right)+\dfrac{34}{3}=0\\ \Leftrightarrow3\left(x-2\right)^2+6\left(y-\dfrac{5}{3}\right)^2+\dfrac{34}{3}=0\\ \Leftrightarrow x,y\in\varnothing\)
\(2\left(x^2+y^2\right)=\left(x+y\right)^2\\ \Leftrightarrow2x^2+2y^2=x^2+2xy+y^2\\ \Leftrightarrow x^2-2xy+y^2=0\\ \Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x-y=0\Leftrightarrow x=y\)
a)\(x^2+10x+26+y^2+2y\)
\(=x^2+10x+25+y^2+2y+1\)
\(=\left(x+5\right)^2+\left(y+1\right)^2\)
b)\(x^2-2xy+2y^2+1\)
\(=x^2-2xy+y^2+y^2+1\)
\(=\left(x-y\right)^2+y^2+1\)
c có lẽ sai ?