C

c

B, C và D

mấy cái đó là đúng hả bạn

a) \(\left(a^2+b+c\right)^2\)

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

\(=a^4+2a^2b+b^2+2a^2c+2bc+c^2\)

b) \(\left(a+b+c\right)^2\)

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

\(=a^2+2ab+b^2+2ca+2bc+c^2\)

a) (a^2+b+c)^2(a^2+b+c)^2

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

=a^4+2a2b+b^2+2a2c+2bc+c^2

b) (a+b+c)^2(a+b+c)^2

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

=a^2+2ab+b^2+2ca+2bc+c^2

a,hđt số 3 = \(\left(a^2+2a\right)^2-9\) 

b,hđt số 3=\(\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)(đổi dấu làm ngoặc khi trước nó là dấu trừ)=\(x^2-\left(y-6\right)^2\)

a) \(\left(a^2+2a+3\right)\left(a^2+2a-3\right)\)

\(=\left(a^2+2a\right)^2+3.\left(-3\right)\)

\(=\left(a^2+2a\right)^2-9\)

b) \(\left(x-y+6\right)\left(x+y-6\right)\)

\(=\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)

\(=x^2-\left(y-6\right)^2\)

(2a-b)² - 2 x ( 2a-b) x (a+b) + (a + b)²

= [(2a-b) - (a+b)]²

Ta có \(A\left(1\right)=B\left(-2\right)\Leftrightarrow12+2a+a^2=8-\left|2a+3\right|\left(-2\right)+a^2\)

\(\Leftrightarrow4+2a=2\left|2a+3\right|\)

đk a >= -2 

\(\left[{}\begin{matrix}4a+6=4+2a\\4a+6=-2a-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=-1\left(tm\right)\\a=-\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

(a²+2a+3)(a²-2a+3)=a⁴+2a²+9

Ta có: \(\left(\frac{a+b}{b}-\frac{2b}{b-a}\right)\cdot\frac{b-a}{a^2+b^2}+\left(\frac{a^2+1}{2a-1}-\frac{a}{2}\right):\frac{a+2}{1-2a}\)

\(=\frac{\left(a+b\right)\left(a-b\right)+2b^2}{b\left(a-b\right)}\cdot\frac{-\left(a-b\right)}{a^2+b^2}+\frac{2\left(a^2+1\right)-a\cdot\left(2a-1\right)}{2\left(2a-1\right)}\cdot\frac{-\left(2a-1\right)}{a+2}\)

\(=\frac{a^2-b^2+2b^2}{-b}\cdot\frac{1}{a^2+b^2}+\frac{2a^2+2-2a^2+a}{2}\cdot\frac{-1}{a+2}\)

\(=\frac{-1}{b}+\frac{-1}{2}=\frac{-2-b}{2b}\)