Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(4x^2-4x+1=\left(2x-1\right)^2\)
b, \(x^2+4xy+4y^2=\left(x+2y\right)^2\)
c, \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
d, \(x^2+12xy+36y^2=\left(x+6y\right)^2\)
e, \(x^2-12xy+36y^2=\left(x-6y\right)^2\)
a, \(4x^2-4x+1\)
\(=4x^2-2x-2x+1=2x.\left(2x-1\right)-\left(2x-1\right)\)
\(=\left(2x-1\right)^2\)
b, \(x^2+4xy+4y^2\)
\(=x^2+2xy+2xy+4y^2\)
\(=x.\left(x+2y\right)+2y.\left(x+2y\right)\)
\(=\left(x+2y\right)^2\)
Chúc bạn học tốt!!! (bạn nhờ mình giải chi tiết bài này á)
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
a, \(25x^2+5xy+\frac{1}{4}y^2=\left(5x\right)^2+2.5x.\frac{1}{2}y+\left(\frac{1}{2}y\right)^2\)
\(=\left(5x+\frac{1}{2}y\right)^2\)
b, \(9x^2+12x+4=\left(3x\right)^2+2.3x.2+2^2=\left(3x+2\right)^2\)
c, \(x^2-6x+5-y^2-4y=\left(x^2-6x+9\right)-\left(y^2+4y+4\right)\)
\(=\left(x-3\right)^2-\left(y+2\right)^2=\left(x-y-5\right)\left(x+y-1\right)\)
d, \(\left(2x-y\right)^2+4\left(x+y\right)^2-4\left(2x-y\right)\left(x+y\right)\)
\(=\left(2x-y\right)^2-2\left(2x-y\right)\left(2x+2y\right)+\left(2x+2y\right)^2\)
\(=\left(2x-y+2x+2y\right)^2=\left(4x+y\right)^2\)
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
phân tích các đa thức sau thành nhân tử
a) 4x^2 - 4xy + 4y^2
\(=\) \(\left(2x\right)^2-4xy+\left(2y\right)^2\)
\(=\left(2x-2y\right)^2\)
b) x^2 - 4xy +4y^2
\(=x^2-4xy+\left(2y\right)^2\)
\(=\left(x-2y\right)^2\)
c) x^2 + 10x + 25
\(=x^2+2.x.5+5^2\)
\(=\left(x+5\right)^2\)
d)x^2 - 10x + 25
\(=x^2-2.x.5+5^2\)
\(=\left(x-5\right)^2\)
e) 81 - (x+1)^2
\(=9^2-\left(x+1\right)^2\)
\(=\left(9-x-1\right)\left(9+x+1\right)\)
f) 16x^2 - 64 (y + 1)^2
\(=16x^2-8^2\left(y+1\right)^2\)
\(=16x^2-\left(8y+8\right)^2\)
\(=\left(16-8y-8\right)\left(16+8y+8\right)\)
p/s: ko chắc câu cuối đâu :v
a) \(A=x^2+2y^2+2xy+4x+6y+19\)
\(=\left[\left(x^2+2xy+y^2\right)+2.\left(x+y\right).2+4\right]+\left(y^2+2y+1\right)+14\)
\(=\left[\left(x+y\right)^2+2\left(x+y\right).2+2^2\right]+\left(y+1\right)^2+14\)
\(=\left(x+y+2\right)^2+\left(y+1\right)^2+14\ge14\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x+y+2=0\\y=-1\end{cases}}\Leftrightarrow x=y=-1\)
b)Đề có gì đó sai sai...
c) Tương tự câu b,em cũng thấy sai sai...HÓng cao nhân giải ạ!
b) \(P=2x^2+y^2+2xy-2y-4\)
\(\Leftrightarrow2P=4x^2+2y^2+4xy-4y-8\)
\(\Leftrightarrow2P=\left(4x^2+4xy+y^2\right)+\left(y^2-4y+4\right)-12\)
\(\Leftrightarrow2P=\left(2x+y\right)^2+\left(y-2\right)^2-12\ge-12\forall x;y\)
Có \(2P\ge-12\Leftrightarrow P\ge-6\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2x+y=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}}\)
A
A