Câu a :
\(\left(2x+1\right)^2-4x\left(x-5\right)\)
\(=4x^2+4x+1-4x^2+20\)
\(=4x+19\)
Câu b :
\(\left(x+3\right)^2-\left(x+1\right)\left(x-1\right)\)
\(=x^2+6x+9-x^2-1\)
\(=6x-8\)
Câu c :
\(\left(x-5\right)^2-\left(x+2\right)^2\)
\(=\left(x-5-x-2\right)\left(x-5+x+2\right)\)
\(=-7\left(2x-3\right)\)
Bạn trình bày cho rõ ràng xem nào.
\(\text{b) }\left(x+3\right)^2-\left(x+1\right)\left(x-1\right)\\ =\left(x+3\right)^2-\left(x^2-1^2\right)\\ =x^2+2\cdot x\cdot3+3^2-x^2+1\\ =\left(x^2-x^2\right)+6x+\left(9+1\right)\\ =6x+10\\ \)
\(\text{c) }\left(x-5\right)^2-\left(x+2\right)^2\\ =\left(x^2-2\cdot x\cdot5+5^2\right)-\left(x^2+2\cdot x\cdot2+2^2\right)\\ =x^2-10x+25-x^2-4x-4\\ =\left(x^2-x^2\right)-\left(10x+4x\right)+\left(25-4\right)\\ =-14x+21\\ \)
\(\text{d) }\left(x+3\right)^2-\left(x-3\right)^2\\ =\left(x^2+2\cdot x\cdot3+3^2\right)-\left(x^2-2\cdot x\cdot3+3^2\right)\\ =x^2+6x+9-x^2+6x-9\\ =\left(x^2-x^2\right)+\left(6x+6x\right)+\left(9-9\right)\\ =12x\\ \)
\(\text{e) }2x\left(x+1\right)-\left(x+3\right)^2-x^2\\ =2x^2+2x-\left(x^2+2\cdot x\cdot3+3^2\right)-x^2\\ =2x^2+2x-x^2-6x-9-x^2\\ =\left(2x^2-x^2-x^2\right)+\left(2x-6x\right)-9\\ =-4x-9\\ \)
\(\text{g) }\left(x+3\right)^2+\left(x+2\right)^2-2\left(x+3\right)\left(x+2\right)\\ =\left[\left(x+3\right)-\left(x+2\right)\right]^2\\ =\left(x+3-x-2\right)^2\\ =1^2\\ =1\\ \)
\(\text{a) }\left(2x+1\right)^2-4x\left(x+1\right)\\ =\left(2x\right)^2+2\cdot2x\cdot1+1^2-4x^2-4x\\ =4x^2+4x+1-4x^2-4x\\=\left(4x^2-4x^2\right)+\left(4x-4x\right)+1\\ =1 \)
a)\(\left(2x+1\right)^2-4x.5x=4x^2+4x+1-20x^2\)
=\(-16x^2+4x+1\)
d)\(\left(x+3\right)^2\)-\(\left(x+1\right)\left(x-1\right)\)=\(x^2+6x+9+x^2-1\)=\(2x^2+6x+8\)
e)\(2x\left(x+1\right)-\left(x+3\right)^2-x^2=2x^2+2x-x^2-6x-9-x^2\)
=-4x-9