Viết các biểu thức sau dưới dạng bình phương của một tổng hoặc một hiệu
a, 1-2x+X^2
b, 4y+4+y^2
c, 1/16+1/2x+x^2
d, 36x^2+12xy+y^2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, 1-2x+x^2 = x^2 - 2x.1 + 1^2= (x-1)^2
b, 4y+4+y^2 = y^2 + 2y.2+ 2^2 = (y+2)^2
c, 1/16+1/2x+x^2 = x^2 + 2.x.\(\frac{1}{4}\)+ (1/4)^2 = (x+1/4)^2
d, 36x^2+12xy+y^2 = (6x)^2 + 2.6x.y + y^2 = (6x+y)^2
a) \(1-2x+x^2=\left(1-x\right)^2=\left(x-1\right)^2\)
b) \(4y+4+y^2=y^2+4y+4=\left(y+2\right)^2\)
c) \(\frac{1}{16}+\frac{1}{2}x+x^2=\left(x+\frac{1}{4}\right)^2\)
d) \(36x^2+12xy+y^2=\left(6x+y\right)^2\)
\(a,=\left(x^2y+3\right)^2\\ b,=\left(2x+y\right)^2\\ c,=\left(5y^2-1\right)^2\)
\(a,=\left(x-4\right)^2\\ b,=\left(\dfrac{1}{2}xy^2+1\right)^2\)
a) Ta có: \(\left(x^2+9x+18\right)^2+2\left(x^2+9x\right)+37\)
\(=\left(x^2+9x+18\right)^2+2\cdot\left(x^2+9x+18\right)-36+37\)
\(=\left(x^2+9x+19\right)^2\)
b) Ta có: \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)
\(=\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+2\left(x+1\right)\left(y+1\right)\)
\(=\left(x^2+2x+2+y^2+2y\right)^2\)
Lời giải:
a. $-8x+16+x^2=x^2-2.x.4+4^2=(x-4)^2$
b. $xy^2+\frac{1}{4}x^2y^4+1=(\frac{1}{2}xy^2)^2+2.\frac{1}{2}xy^2.1+1^2$
$=(\frac{1}{2}xy^2+1)^2$
a: \(x^2-8x+16=\left(x-4\right)^2\)
b: \(\dfrac{1}{4}x^2y^4+xy^2+1=\left(\dfrac{1}{2}xy^2+1\right)^2\)
này mình có vài câu không làm được, xin lỗi bạn nha
\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)
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, \(\left(2x-y\right)^2+4\left(x+y\right)^2-4\left(2x-y\right)\left(x+y\right)\)
\(=\left(2x-y\right)^2-2\left(2x-y\right)\left(2x+2y\right)+\left(2x+2y\right)^2\)
\(=\left(2x-y+2x+2y\right)^2=\left(4x+y\right)^2\)
a. x2 + 6x + 9 = (x + 3)2
b. 25 + 10x + x2 = (5 + x)2
c. x2 + 8x + 16 = (x + 4)2
d. x2 + 14x + 49 = (x + 7)2
e. 4x2 + 12x + 9 = (2x + 3)2
f. 9x2 + 12x + 4 = (3x + 2)2
h. 16x2 + 8 + 1 = (4x + 1)2
i. 4x2 + 12xy + 9y2 = (2x + 3y)2
k. 25x2 + 20xy + 4y2 = (5x + 2y)2
a) \(=\left(x+3\right)^2\)
b) \(=\left(x+5\right)^2\)
c) \(=\left(x+4\right)^2\)
d) \(=\left(x+7\right)^2\)
e) \(=\left(2x+3\right)^2\)
f) \(=\left(3x+2\right)^2\)
h) \(=\left(4x+1\right)^2\)
i) \(=\left(2x+3y\right)^2\)
k) \(=\left(5x+2y\right)^2\)
A)\(1-2x+x^2\)
\(=\left(1-x\right)^2\)
B)\(4y+4+y^2\)
\(=2^2+4y+y^2\)
\(=\left(2+y\right)^2\)
C)\(\frac{1}{16}+\frac{1}{2}x+x^2\)
\(=\left(\frac{1}{4}\right)^2+\frac{1}{2}x+x^2\)
\(=\left(\frac{1}{4}+x\right)\)
D)\(36x^2+12xy+y^2\)
\(=\left(6x+y\right)^2\)