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7 tháng 9 2020

\(\left(x+y-z\right)^2+2\left(z-x-y\right)\left(x+y\right)+\left(x+y\right)^2\)

\(=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2\)

\(\left[\left(x+y-z\right)-\left(x+y\right)\right]^2=z^2\)

7 tháng 9 2020

\(\left(x+y-z\right)^2+2\left(z-x-y\right)\left(x+y\right)+\left(x+y\right)^2\)

\(=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2\)

\(=\left(x+y-z-x+y\right)^2\)

\(=-z^2\)

27 tháng 7 2016

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27 tháng 7 2016

\(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)

\(=x^2+y^2+z^2-2xy-2yz+2xz+z^2-2yz+y^2+\left(2y-2z\right)\left(x-y+z\right)\)

\(=x^2+y^2+z^2-2xy-2yz+2xz+z^2-2yz+y^2+2xy-2y^2+2yz-2xz+2yz-2z^2\)

\(=x^2\)

6 tháng 6 2017

\(a,\left(x+y\right)^2+\left(x-y\right)^2=x^2+2xy+y^2+x^2-2xy+y^2=2\left(x^2+y^2\right)\)\(b,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2=2x^2-2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3x^2\)\(c,\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)=\left[\left(x-y+z\right)-\left(z-y\right)\right]^2=\left(x-2y\right)^2\)

17 tháng 6 2017

a) \(\left(x+y\right)^2+\left(x-y\right)^2\)

=\(\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)\)

=\(x^2+2xy+y^2+x^2-2xy+y^2\)

\(2x^2+2y^2=2\left(x^2+y^2\right)\)

b) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)

=\(\left[\left(x-y\right)+\left(x+y\right)\right]^2\)

= \(\left(x-y+x+y\right)^2\)

\(=2x^2\)

c) \(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)

\(=\left(x-y+z\right)^2-2\left(x-y+z\right)\left(z-y\right)+\left(z-y\right)^2\)

\(=\left[\left(x-y+z\right)-\left(z-y\right)\right]^2\)

= \(\left(x-y+z-z+y\right)^2=x^2\)

28 tháng 6 2016

1)  2xy2+x2y4+1=(xy2)2+2xy2.1+12=(xy2+1)2

2)

a)2(x-y)(x+y)+(x+y)2+(x-y)2=(x+y+x-y)2=(2x)2=4x2

b)(x-y+z)2+(z-y)2+2(x-y+z)(y-z)

=(x-y+z)2+(y-z)2+2(x-y+z)(y-z)

=(x-y+z+y-z)2

=x2

20 tháng 4 2017

Bài giải:

a) (a + b)2 – (a – b)2 = (a2 + 2ab + b2) – (a2 – 2ab + b2)

= a2 + 2ab + b2 – a2 + 2ab - b2 = 4ab

Hoặc (a + b)2 – (a – b)2 = [(a + b) + (a – b)][(a + b) – (a – b)]

= (a + b + a – b)(a + b – a + b)

= 2a . 2b = 4ab

b) (a + b)3 – (a – b)3 – 2b3

= (a3 + 3a2b + 3ab2 + b3) – (a3 – 3a2b + 3ab2 – b3) – 2b3

= a3 + 3a2b + 3ab2 + b3 – a3 + 3a2b - 3ab2 + b3 – 2b3

= 6a2b

Hoặc (a + b)3 – (a – b)3 – 2b3 = [(a + b)3 – (a – b)3] – 2b3

= [(a + b) – (a – b)][(a + b)2 + (a + b)(a – b) + (a – b)2] – 2b3

= (a + b – a + b)(a2 + 2ab + b2 + a2 – b2 + a2 – 2ab + b2) – 2b3

= 2b . (3a2 + b2) – 2b3 = 6a2b + 2b3 – 2b3 = 6a2b

c) (x + y + z)2 – 2(x + y + z)(x + y) + (x + y)2

= x2 + y2 + z2+ 2xy + 2yz + 2xz – 2(x2 + xy + yx + y2 + zx + zy) + x2 + 2xy + y2

= 2x2 + 2y2 + z2 + 4xy + 2yz + 2xz – 2x2 – 4xy – 2y2 – 2xz – 2yz = z2

AH
Akai Haruma
Giáo viên
13 tháng 11 2023

Lời giải:

$xy+yz+xz=1$
$\Rightarrow x^2+1=x^2+xy+yz+xz=(x+y)(x+z)$

Tương tự: $y^2+1=(y+z)(y+x); z^2+1=(z+x)(z+y)$

Khi đó:

\(\sum \sqrt{\frac{(x^2+1)(y^2+1)}{z^2+1}}=\sum \sqrt{\frac{(x+y)(x+z)(y+x)(y+z)}{(z+x)(z+y)}}=\sum \sqrt{(x+y)^2}\)

$=\sum (x+y)=2(x+y+z)$