Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: (a+b+c)^2 + a^2 + b^2 + c^2
= a^2 +b^2 +c^2 + 2ab + 2ac + 2bc + a^2 + b^2 + c^2
= (a^2 +2ab+ b^2) + (b^2 +2bc+ c^2) +(c^2 +2ac+ a^2 )
= (a+b)^2 +(b+c)^2 +(c+a)^2
\(\left(a^2+b^2+c^2\right)+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
Bài 2 :
a ) \(A=\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(A=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)
\(A=\left(a^2+2ab+b^2\right)+\left(a^2+2ac+c^2\right)+\left(b^2+2bc+c^2\right)\)
\(A=\left(a+b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2\)
a. ( a + b + c)2 + a2 + b2 + c2
= a2 + b2 + c2 + 2ab + 2ac + 2bc + a2 + b2 + c2
= (a+b)2 + (b+c)2 + (a+c)2
b. 2.(a-b).(c-b) + 2.(b-a).(c-a) + 2.(b-c).(a-c)
đặt a - b = x; b-c = y; c-a = z => x + y + z = 0 (1)
ta có: 2.x.(-y) + 2.(-x).z + 2.y.(-z)
= -2xy - 2xz - 2yz = -2.(xy+xz+yz)
ta có: (x+y+z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz
02 = x2 + y2 + z2 + 2.(xy+yz+xz)
=> x2 + y2 + z2 = -2.(xy+yz+xz) (2)
Từ (2) => 2.(a-b).(c-b) + 2.(b-a) .(c-a) + 2.(b-c).(a-c) = x2 + y2 + z2
= (a-b)2 + (b-c)2 + (c-a)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 + 2.x.1+ 12 = ( x + 1)2
b) 9x2 + y2 + 6xy = (3x)2 + 2.3.x.y + y2 = (3x + y)2
c) 25a2 + 4b2 – 20ab = (5a)2 – 2.5.a.2b. + (2b)2 = (5a – 2b)2
Hoặc 25a2 + 4b2 – 20ab = (2b)2 – 2.2b.5a. + (5a)2 = (2b – 5a)2
d) x2 – x + \(\dfrac{1}{4}\) = x2 – 2.x. \(\dfrac{1}{2}\) + ( \(\dfrac{1}{2}\))22 = ( x - \(\dfrac{1}{2}\) )2
Hoặc x2 – x + \(\dfrac{1}{4}\) = \(\dfrac{1}{4}\) - x + x2 = (\(\dfrac{1}{2}\))2 – 2. \(\dfrac{1}{2}\).x + x2 = (\(\dfrac{1}{2}\) - x)2
a) x2 + 2x + 1 = x2+ 2 . x . 1 + 12
= (x + 1)2
b) 9x2 + y2+ 6xy = (3x)2 + 2 . 3 . x . y + y2 = (3x + y)2
c) 25a2 + 4b2– 20ab = (5a)2 – 2 . 5a . 2b + (2b)2 = (5a – 2b)2
Hoặc 25a2 + 4b2 – 20ab = (2b)2 – 2 . 2b . 5a + (5a)2 = (2b – 5a)2
d) x2 – x + 1414 = x2 – 2 . x . 1212 + (12)2(12)2= (x−12)2(x−12)2
Hoặc x2 – x + 1414 = 1414 - x + x2 = (12)2(12)2 - 2 . 1212 . x + x2 = (12−x)2
oyfngg
em mới hc lớp 7 à