a, \(\left(a^2+b^2-2ab+2a-2b+1\right)+\left(b^2-2b+1\right)=0\)

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

Mà \(\left(a-b+1\right)^2\ge0,\left(b-1\right)^2\ge0\)

=> \(\hept{\begin{cases}a-b+1=0\\b=1\end{cases}\Rightarrow\hept{\begin{cases}a=0\\b=1\end{cases}}}\)

b,Tương tự 

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

=>\(\hept{\begin{cases}a=1\\b=1\end{cases}}\)

2a²b+4ab²-a²c+ac²-4b²c+2bc²-4abc

=2ab(a+2b)-ac(a+2b)+c²(a+2b)-2bc(a+2b)

=(a+2b)(2ab-ac+c²-2bc)

=(a+2b)\(\left[a\left(2b-c\right)-c\left(2b-c\right)\right]\)

=(a+2b)(2b-c)(a-c)

bạn vừa đăng câu này rồi mà

`4b^2+a^2+4ab`

`=(2b)^2+2.2b.a+a^2`

`=(a+2b)^2`

`-49-2a^4+14sqrt2a^2`

`=-(2a^4-14sqrt2a^2+49)`

`=-((sqrt2a^2)^2-2.sqrt2a^2.7+7^2)`

`=-(sqrt2a^2-7)^2`

\(a^2+4ab+4b^2=\left(a+2b\right)^2\)

\(-49-2a^4+14\sqrt{2a^2}=-\left(\sqrt{2a^2}-7\right)^2\)

\(A-B+C=2a^2-3ab+4b^2-3a^2-4ab+b^2+a^2+2ab+b^2\)

\(\Rightarrow A-B+C=-5ab+6b^2\)

A−B+C=−5ab+6b2

2a^2b + 4ab^2 -a^2c + ac^2 -4b^2c +2bc^2 - 4abc 
= (2a^2b - 4abc + 2bc^2) + (4ab^2 - 4b^2c) - (a^2c - ac^2) 
= 2b(a^2 - 2ac + c^2) + 4b^2(a - c) - ac(a - c) 
= 2b(a - c)^2 + 4b^2(a - c) - ac(a - c) 
= (a - c)  [ 2b(a - c) + 4b^2 - ac ] 
= (a - c) (2ab -2bc +4b^2 - ac) 
= (a - c)  [ (2ab - ac) + (4b^2 - 2bc) ] 
= (a - c) [a(2b - c) + 2b(2b - c)] 
= (a - c)(2b - c)(a + 2b)

TL:

=\(\left(2a^2b-4bc+2bc^2\right)+\left(4ab^2-4b^2c\right)-\left(a^2c-ac2\right)\) 

=\(2b\left(a^2-2c+c^2\right)+4b^2\left(a-c\right)-ac\left(a-c\right)\) 

=\(2b\left(a-c\right)+4b^2\left(a-c\right)-ac\left(a-c\right)\) 

=\(\left(a-c\right)\left(2b+4b^2-ac\right)\)

........................

Vậy......

\(2a^2b +4ab^2-a^2c +ac^2-4b^2c +2bc^2-4abc\)

\(=\left(2a^2b+4ab^2-2abc-4b^2c\right)-\left(a^2c+2abc-ac^2-2bc^2\right)\)

\(=2b\left(a^2-ac+2ab-2bc\right)-c\left(a^2-ac+2ab-2bc\right)\)

\(=\left(a^2-ac+2ab-2bc\right)\left(2b-c\right)\)

\(=\left[a\left(a-c\right)+2b\left(a-c\right)\right]\left(2b-c\right)\)

\(=\left(a+2b\right)\left(a-c\right)\left(2b-c\right)\)