\(a^2-2a+b^2+4b+4c^2-4c+6=0\)

\(=>\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\)

\(=>\left(a^2-2.a.1+1^2\right)+\left(b^2+2.b.2+2^2\right)+\left[\left(2c\right)^2-2.2c.1+1^2\right]=0\)

\(=>\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\left(1\right)\)

Vì : \(\left(a-1\right)^2\ge0\) với mọi a

\(\left(b+2\right)^2\ge0\) với mọi b

\(\left(2c-1\right)^2\ge0\) với mọi c

=>\(\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\ge0\) với mọi a,b,c

Để (1) thì \(\left(a-1\right)^2=\left(b+2\right)^2=\left(2c-1\right)^2=0=>a=1;b=-2;c=\frac{1}{2}\)

Vậy........

(x+y+z)(x²+y²+z²-xy-yz-xz)

Có thể xem trên wikipedia với từ khóa 7 HDT đáng nhớ mục liên quan nhé bạn​

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\(x^3+y^3+z^3-3xyz\)

\(=x^3+y^3+z^3-xyz-xyz-xyz\)

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

Bạn áp dụng hằng đẳng thức \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)

Ta có : \(\left(20^2+18^2+16^2+...+4^2+2^2\right)-\left(19^2+17^2+15^2+...+3^2+1^2\right)\)

\(=\left(20^2-19^2\right)+\left(18^2-17^2\right)+\left(16^2-15^2\right)+...+\left(4^2-3^2\right)+\left(2^2-1^2\right)\)

\(=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+\left(16-15\right)\left(16+15\right)+...+\left(4-3\right)\left(4+3\right)+\left(2-1\right)\left(2+1\right)\)

\(=1+2+3+...+20=\frac{20.21}{2}=210\)

\(\left(x-a\right)\left(x+a\right)\left(x+2a\right)+a^4\)

\(=\left(x^2-a^2\right)\left(x+2a\right)+a^4\)

\(=\left(x^3+2ax^2-a^2x+2a^3\right)+a^4\)

Ta có:\(x^2-4\)

\(=x^2-2^2\)

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

\(x^2-4\)

\(=x^2-2^2\)

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

1) A=1995²-1994.1996

      =1995²-(1995-1)(1995+1)

      =1995²-(1995²-1)

      =1

2) B=9⁸.2⁸-(18⁴-1)(18⁴+1)

      =(9.2)⁸-[(18⁴)²-1]

      = 18⁸-18⁸+1

      =1

3) C=163²+74.163+37²

        =163²+2.37.163+37²

      =163²+2.163.37+37²

      =(163+37)².2

      =80000