Ta có : ( a - b )²  + 4ab

= a² - 2ab + b² + 4ab

= a² + 2ab + b²

= ( a + b )² ( Vế trái )

Do đó : ( a + b )² = ( a - b )² + 4ab 

+) Biến đổi vế phải ta có :

\(\left(A-B\right)^2+4AB\)

\(=A^2-2AB+B^2+4AB\)

\(=A^2+2AB+B^2=\left(A+B\right)^2=VT\left(đpcm\right)\)

(x + 3)^2 = 9(2x - 1)^2

=> x^2 + 6x + 9 = 9(4x^2 - 4x + 1) 

=> x^2 + 6x + 9 = 36x^2 - 36x + 9

=> x^2 - 36x^2 + 6x + 36x + 9 - 9 = 0

=> -35x^2 + 42x = 0

=> x(-35x + 42) = 0

=> x = 0 hoặc -35x + 42 = 0

=> x = 0 hoặc x = 42/35

\(\left(x+3\right)^2=9\left(2x-1\right)^2\)

\(\left(x+3\right)^2=3^2\cdot\left(2x-1\right)^2\)

\(\left(x+3\right)^2=\left(6x-3\right)^2\)

\(\left(x+3\right)^2-\left(6x-3\right)^2=0\)

\(\left(x+3-6x+3\right)\cdot\left(x+3+6x-3\right)=0\)

\(\left(6-5x\right)\cdot7x=0\)

\(\orbr{\begin{cases}6-5x=0\\7x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=0\end{cases}}}\)

\(a^3+b^3+2019ab\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+2019ab\)

\(=673^3-2019ab+2019ab\)

\(=673^3=47832147\)

\(=x\left(\frac{x^2}{4}+x+1\right)=x\left(\frac{x}{2}+1\right)^2\)

\(\left(2x-1\right)\left(2x+1\right)\left(4x^2+1\right)\)

\(=\left(4x^2-1\right)\left(4x^2+1\right)\)

\(=16x^4-1\)

cái này là tính mà bạn

\(12x^2+7x-10\)

\(=12x^2-8x+15x-10\)

\(=4x\left(3x-2\right)+5\left(3x-2\right)\)

\(=\left(4x+5\right)\left(3x-2\right)\)

mà sao bạn tách đc  như v

\(a^2+b^2=\left(a+b\right)^2-2ab=7^2-24=25\)

\(\left(a-b\right)^2=\left(a+b\right)^2-4ab=7^2-4.12=1\)

\(\Rightarrow a-b=-1\)

\(\Rightarrow A=\left(-1\right)^5=?\)

\(B=\left(a^2+b^2\right)^2-2\left(ab\right)^2=25^2-2.12^2=?\)