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a: \(=4x^2+20x+25+4x^2-20x+25-\left(4x^2-1\right)\)
\(=8x^2+50-4x^2+1=4x^2+51\)
b: \(=8a^3+12a^2b+6ab^2+b^3+8a^3-12a^2b+6ab^2-b^3-16a^3\)
\(=12ab^2\)
c: \(\left(2x-1\right)^3-\left(x-2\right)\left(x^2+2x+4\right)-7x^3-2x\)
\(=\left(2x-1\right)^3-x^3+8-7x^3-2x\)
\(=8x^3-12x^2+6x-1-8x^3-2x+8\)
\(=-12x^2+4x+7\)
d: \(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+3\)
\(=-3x^2+4x+3\)

a)(3x-1)2+2(3x-1)(2x+1)2(2x+1)=48x^4+56x^3+21x^2-12x-1 cái này tra google
b)(x2+1)(x-3)-(x-3)(x2+3x+9)=(x2+1)(x-3)-(x-3)(x+3)2=(x-3)[(x2+1)-(x+3)2 ]
c)(2x+3)2+(2x+5)2-2(2x+3)(2x+5)=(2x+3)2+(2x+5)2-(2x+3)(2x+5)-(2x+3)(2x+5)=(2x+3)(2x+3-2x+5)+(2x+5)(2x+5-2x+3)
=8(2x+3)+8(2x+5)=8(2x+3+2x+5)
=8(4x+8)
d)(x-3)(x+3)-(x-3)2 =(x-3)(x+3)-(x-3)(x-3)=(x-3)(x+3-x-3)=0
e)(2x+1)2+2(4x2-1)+(2x-1)2 =(2x+1)2+2[(2x)2 -1]+(2x-1)2 =(2x+1)(2x+1+2x-1)+(2x-1)(2x+1+2x-1)=4x(2x+1)+4x(2x-1)
=4x(2x+1+2x-1)=16x2
f)(x2-1)(x+2)-(x-2)(x2+2x+4)= (x2-1)(x+2)-(x-2)(x+2)2 =(x2-1)(x+2)-(x2-22)(x+2)=(x+2)(x2-1-x2-22) mình đoán câu f khai triển ra thế này nhưng kq không giống nhau nên chắc bạn phải tự làm rồi

f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)
\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)
\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)
\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)
\(-x^3=27\)
\(x=-3\)

\(12\left(x-2\right)\left(x+2\right)-3\left(2x+3\right)^2\)=52\(\Leftrightarrow12\left(x^2-2^2\right)-3\left(4x^2+12x+9\right)=52\)
\(\Leftrightarrow12x^2-48-12x^2-36x-27-52=0\)
\(\Leftrightarrow-36x-127=0\)
\(\Leftrightarrow x=-3.52\)
Bạn học hằng đẳng thức chưa bạn , bạn chỉ cần nắp chúng vào là làm đc thôi

a ) \(\left(2x-3\right)^3-x\left(2x-1\right)^2\)
\(=\left(2x\right)^3-3.\left(2x\right)^2.3+3.2x.3^2-3^3-x.\left(2x\right)^2-2.2x.1+1^2\)
\(=8x^3-3.4x^2.3+3.2x.9-27-x.4x^2-2.2x.x+1\)
\(=8x^3-36x^2+54x-27-4x^3-4x^2+1\)
\(=4x^3-40x^2+54x-26\)
a ) (2x−3)^3−x(2x−1)^2
=(2x)^3−3.(2x)^2.3+3.2x.3^2−33^−x.(2x)^2−2.2x.1+1^2
=8x^3−3.4x^2.3+3.2x.9−27−x.4x^2−2.2x.x+1
=8x^3−36x^2+54x−27−4x^3−4x2+1
=4x^3−40x^2+54x−26

Bài1:
\(a,\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\\ \Leftrightarrow\left(2x+3\right)^2-4x^2+1=22\\ \Leftrightarrow\left(2x+3-2x\right)\left(2x+3+2x\right)=21\\ \Leftrightarrow3\left(4x+3\right)=21\\ \Leftrightarrow4x+3=7\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\\ Vậy....\\ b,\left(2x-1\right)^3-4x^2\left(2x-3\right)=5\\ \Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\\ \Leftrightarrow6x=6\\ \Leftrightarrow x=1\\ Vậy...\)
Các câu sau cũng như thế
Bài2:
\(A=x^2+20x+9\\ =\left(x^2+20x+100\right)-91\\ =\left(x+10\right)^2-91\)
Với mọi x thì \(\left(x+10\right)^2\ge0\\ \Rightarrow\left(x+10\right)^2-91\ge-91\)
Hay \(A\ge-91\)
Để A=-91 thì
\(\left(x+10\right)^2=0\\ \Leftrightarrow x+10=0\\ \Leftrightarrow x=-10\)
Vậy...
\(B=4x^2+5x+7\\ =\left(4x^2+5x+\dfrac{25}{16}\right)+5,4375\\ =\left(2x+\dfrac{5}{4}\right)^2+5,4375\)
Với mọi x;y thì \(\left(2x+\dfrac{5}{4}\right)^2+5,4375\ge5,4375\)
Hay \(A\ge5,4375\)
Để \(A=5,4375\) thì \(\left(2x+\dfrac{5}{4}\right)^2=0\\ \Leftrightarrow2x+\dfrac{5}{4}=0\\ \Leftrightarrow x=\dfrac{-5}{8}\)
Vậy....
\(\left(2x-3\right)^2-\left(2x+1\right)\left(x-1\right)+3\left(2x-3\right)\)
\(=\left(2x-3\right)\left(2x-3+3\right)+\left(2x+1\right)\left(x-1\right)\)
\(=\left(2x-3\right).2x+2x^2-x-1\)
\(=4x^2-6x+2x^2-x-1\)
\(=6x^2-7x-1\)