Giải

Ta có: \(101^2-1=\left(101-1\right)\left(101+1\right)\)(Áp dụng hằng đẳng thức: \(A^2-B^2=\left(A-B\right)\left(A+B\right)\), ở đây, 1 cũng chính là 1 mũ 2)

\(=100.102\)

\(=10200\)

Vậy biểu thức \(101^2-1=10200\)

A) x² -3x+xy-3y=x²+xy-3x-3y=x(x+y)-3(x+y)=(x+y)(x-3)

\(x^2-3x+xy-3y\)

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

\(=x.\left(x+y\right)-3.\left(x+y\right)\)

\(=\left(x-3\right).\left(x+y\right)\)

\(2x^2-x+2xy-y\)

\(=2x^2-\left(x-2xy+y\right)\)

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

\(=\left(\sqrt{2}x\right)^2-\left(x-y\right)^2\)

\(=\left(\sqrt{2}x-x+y\right).\left(\sqrt{2}x+x-y\right)\)

\(x^4+x^3+2x^2+x+1\)

\(=\left(x^4+2x^2+1\right)+\left(x^3+x\right)\)

\(=\left(x^2+1\right)^2+x.\left(x^2+1\right)\)

\(=\left(x^2+1\right).\left(x^2+1+x\right)\)

\(16+2xy-x^2-y^2\)

\(=16-x^2+2xy-y^2\)

\(=16-\left(x^2-2xy+y^2\right)\)

\(=4^2-\left(x-y\right)^2\)

\(=[4-\left(x-y\right)].[4+\left(x-y\right)]\)

\(=\left(4-x+y\right).\left(4+x-y\right)\)

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

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

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

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

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

\(5.\left(2x-3\right)+6x=10x-15+6x=16x-15\)

\(\left(x+5\right).\left(x+1\right)-x^2=x^2+x+5x+5-x^2=6x+5\)

\(\left(x-3\right).\left(x+3\right)+\left(2-x\right)^2=x^2-9+4-4x+x^2=2x^2-4x-5\)

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

\(\left(x+2\right)^2-3x.\left(x+1\right)=x^2+4x+4-3x^2-3x=-2x^2+x+4\)

\(\left(x+4\right).\left(x^2-4x+16\right)=x^3+4^3=x^3+64\)

\(\left(3x-2\right).\left(9x^2+6x+4\right)=\left(3x\right)^3-2^3=27x^3-8\)

a=130 và d=50

=>a+d=180

mà a và d là hai góc tcp

=>ab//cd

=>là hình thang

mà hai đg chéo ac=bd

=>HBH

Gửi bạn nè. Chúc bạn học tốt !