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a) Sửa đề: \(1-4x+4x^2\)
Ta có: \(1-4x+4x^2\)
\(=1^2-2\cdot1\cdot2x+\left(2x\right)^2\)
\(=\left(1-2x\right)^2\)
b) Ta có: \(x^2-x+\frac{1}{4}\)
\(=x^2-2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2\)
\(=\left(x-\frac{1}{2}\right)^2\)
c) Ta có: \(\left(x-y\right)^2-2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x-y-x-y\right)^2\)
\(=\left(-2y\right)^2=4y^2\)
a) \(9x^2+6x+1=\left(3x+1\right)^2\)
b)\(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)
c)\(x^2y^4-2xy^2+1=\left(xy^2-1\right)^2\)
d) \(x^2+\frac{2}{3}x+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2\)
a) 9x2 + 6x + 1 = ( 3x + 1 )2
b) x2 - x + 1/4 = ( x - 1/2)2
c) x2 . y4 - 2xy2 + 1 = ( xy2 - 1 ) 2
d) x2 + 2/3x + 1/9 = (x+1/3)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
a)x2+2x+1=x2+2x.1+12=(x+1)2
b)x2-x+\(\frac{1}{4}\)=x2-2.x.\(\frac{1}{2}\)+\(\left(\frac{1}{2}\right)^2\)=\(\left(x-\frac{1}{2}\right)^2\)
a, (x+2)^2
b, (x-3)^2
c, (2x+3)^2
d, (3x-1)^2
e, (x+5)^2
g, (4x-1)^2
a) x2 + 4x + 4 = ( x + 2 )2
b) x2 - 6x + 9 = (x-3)2
c) 4x2 + 12x + 9 = (2x)2 + 2.2x.3 + 3^2 = (2x + 3)2
d) 9x2 - 6x + 1 = (3x)2 - 2.3x.1 + 1^2 = (3x-1)2
e) x2 + 25 +10x = x2 + 2.x.5 + 52 = (x+5)2
g) 16x2 +1 - 8x = (4x)2 - 2.4x.1 + 1^2 = (4x-1)2
*\(x^2-x+\frac{1}{4}=x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=\left(x-\frac{1}{2}\right)^2\)
*\(25a^2+4b^2-20ab=\left(5a\right)^2-2.5a.2b+\left(2b\right)^2=\left(5a-2b\right)^2\)
*\(9x^2+y^2+6xy=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)
1) viết biểu thức dưới dạng bình phương của 1 tổng hoặc 1 hiệu
a) x2 - 2x +1
= x2 - 2 . x . 1 + 12
= ( x + 1 )2
b) 81x2 + 1 +18x
= 81x2 + 18x + 1
= (9x)2 + 2 . 9x . 1 + 12
= (9x + 1)2
c)1/4 + x2 + x
= x2 + 2 . x . 1/2 + (1/2)2
= (x + 1/2)2
~.~
Ta có: \(x^4+x^3+\frac{1}{4}x^2\)
\(=x^2\left(x^2+x+\frac{1}{4}\right)\)
\(=x^2\left[x^2+2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]\)
\(=x^2\cdot\left(x+\frac{1}{2}\right)^2\)
\(=\left(x^2+\frac{1}{2}x\right)^2\)