$\\$
Bài `2.`
`a,`
`2 (a^2 + b^2) = (a+b)^2`
`-> 2a^2 + 2b^2 = a^2 + 2ab + b^2`
`-> 2a^2 + 2b^2 - a^2 - 2ab - b^2=0`
`-> (2a^2 - a^2) - 2ab +(2b^2 -b^2)=0`
`-> a^2 - 2ab + b^2=0`
`-> (a-b)^2=0`
`->a-b=0`
`->a=b` (đpcm)
`b,`
`(a-b)^2 + (b-c)^2 + (c-a)^2 = 4 (a^2+ b^2 +c^2- ab - bc - ca)`
`-> a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ac + a^2 = 4a^2 + 4b^2 + 4c^2 - 4ab - 4bc - 4ca`
`-> a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ac + a^2 - 4a^2 - 4b^2 - 4c^2 + 4ab + 4bc + 4ca =0`
`-> (a^2 + a^2 - 4a^2) + (b^2 + b^2 - 4b^2) + (c^2 + c^2 - 4c^2) + (-2ab + 4ab) + (-2bc + 4bc) + (-2ac + 4ac)=0`
`-> -2a^2 - 2b^2 - 2c^2 + 2ab + 2bc + 2ac=0`
`-> - (2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac)=0`
`-> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac =0`
`-> (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (a^2 - 2ac + c^2)=0`
`-> (a-b)^2 + (b-c)^2 + (a-c)^2=0`
Với mọi `a,b,c` có : $\begin{cases} (a-b)^2 ≥0\\(b-c)^2≥0\\(a-c)^2 ≥0 \end{cases}$
`-> (a-b)^2 + (b-c)^2 + (a-c)^2 ≥0∀a,b,c`
Dấu "`=`" xảy ra khi :
`↔` $\begin{cases} (a-b)^2 =0\\(b-c)^2=0\\(a-c)^2 =0 \end{cases}$
`↔` $\begin{cases} a-b=0\\b-c=0\\a-c=0 \end{cases}$
`↔` $\begin{cases} a=b\\b=c\\a=c \end{cases}$
`↔ a=b=c` (đpcm)
$\\$
Bài `3.`
`a,`
`A = (x-y-z)^2 + 2 (x-y-z) (y+z) + (y+z)^2`
`-> A = (x-y-z + y+z)^2`
`-> A = x^2`
Vậy `A=x^2`
`b,`
`B = (x-y+z)^2 - (2x-1)^2 - (1-3y)^2 + 12`
`-> B = [(x-y) + z]^2 - (4x^2 - 4x + 1) - (1 - 6y + 9y^2) + 12`
`-> B = [(x-y)^2 + 2 (x-y) . z + z^2] - 4x^2 + 4x-1 - 1 + 6y - 9y^2 + 12`
`-> B = [x^2 - 2xy + y^2 + 2xz - 2yz + z^2] - 4x^2 + 4x-1 - 1 + 6y - 9y^2 + 12`
`-> B = x^2 + y^2 + z^2 - 2xy - 2yz + 2xz - 4x^2 + 4x-1 - 1 + 6y - 9y^2 + 12`
`-> B = -3x^2 - 8y^2 + z^2 - 2xy - 2yz + 2xz + 4x + 10 +6y`
Vậy `B = -3x^2 - 8y^2 + z^2 - 2xy - 2yz + 2xz + 4x + 10 +6y`
