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

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

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

\(=4x+3\)

b) \(5\left(x+2\right)\left(x-2\right)-\frac{1}{2}\left(6-8x\right)+17\)

\(=5\left(x^2-4\right)-3+4x+17\)

\(=5x^2-20-3+4x+17\)

\(=5x^2-6+4x\)

uầy ,bn cứ nhân vào là đc

Thôi giúp luôn  =.=

\(\left(x+3\right)^2+\left(2x+1\right)\left(3x-5\right)-2x\left(3-x\right)+4x+25\)

\(=x^2+6x+9+6x^2-10x+3x-5-6x+2x^2+4x+25\)

\(=9x^2-3x+29\)

\(\left(x+3\right)^2+\left(2x+1\right)\left(3x-5\right)-2x\left(3-x\right)+4x+25\)

\(=x^2+6x+9+6x^2-10x+3x-5-6x+2x^2+4x+25\)

\(=9x^2-3x+29\)

hok tốt!

8/3 - 2X

\(\frac{8}{3}\)

\(a,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\\ =\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|\\ =\sqrt{3}+\sqrt{2}-\left(\sqrt{3}-\sqrt{2}\right)\\ =\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\\=2\sqrt{2} \)

\(b,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1}+\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}+1+\sqrt{3}-1\\ =2\sqrt{3}\)

\(c,=x-4+\sqrt{\left(4^2-2.4.x+x^2\right)}\\ =x-4+\sqrt{\left(4-x\right)^2}\\ =x-4+\left|4-x\right|\\ =x-4+x-4=2x-8\)    (vì \(x>4\) )

@seven 

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

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

\(=x^4+4x^2+4-x^4+16=4x^2+20=4\left(x^2+5\right)\)