\(\dfrac{2x^2-15x+25}{x-5}=\dfrac{x^2-10x+25}{x-5}+\dfrac{x\left(x-5\right)}{x-5}=\dfrac{\left(x-5\right)^2}{x-5}+x=x-5+x=2x-5\)

a) \(\left(3x-5\right)\left(3x+5\right)=9x^2-25\Leftrightarrow9x^2+15x-15x-25=9x^2-25\Leftrightarrow9x^2-25=9x^2-25\)(đúng)

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

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

\(\dfrac{3\sqrt{2}-6}{\sqrt{2}-1}=\dfrac{\left(3\sqrt{2}-6\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\dfrac{6-3\sqrt{2}-6}{1}=-3\sqrt{2}\)

a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2003.2004}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2003}-\dfrac{1}{2004}=1-\dfrac{1}{2004}=\dfrac{2003}{2004}\)b)Đặt  \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2003.2005}\)

\(\Rightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2003.2005}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2003}-\dfrac{1}{2005}=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)\(\Rightarrow A=\dfrac{1002}{2005}\)

\(1^3+2^3+3^3+4^3=1+8+27+64=100=10^2\)

\(1^3+2^3+3^3+4^3+5^3=1+8+27+64+125=225=15^2\)

d) \(\sqrt{x^2-6x+9}=2\Leftrightarrow\sqrt{\left(x-3\right)^2}=2\Leftrightarrow x-3=2\Leftrightarrow x=5\)

e) đk: \(x\ge2\)\(\sqrt{x^2-3x+2}=\sqrt{x-1}\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)f) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\Leftrightarrow2x-1=x-3\Leftrightarrow x=-2\)

Xét tam giác đều ABC có

G là trọng tâm của tam giác(gt)

=> 3 đường trung tuyến bằng nhau

=> \(GB=GC=AG=\dfrac{2}{3}AM=\dfrac{2}{3}.3=2\left(cm\right)\)

Bổ sung câu 3:

x=-1 không thỏa điều kiện

Vậy \(S=\varnothing\)

1) đk: \(x\ge5\)\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\Leftrightarrow2\sqrt[]{x-5}+\sqrt{x-5}-\dfrac{1}{3}.3\sqrt{x-5}=4\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\Leftrightarrow x-5=4\Leftrightarrow x=9\)3) đk: \(x\ge2\)\(\sqrt{x^2-4}=\sqrt{x-2}\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}=\sqrt{x-2}\Leftrightarrow\sqrt{x+2}=1\Leftrightarrow x+2=1\Leftrightarrow x=-1\)

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

f) \(\left(x+3\right)\left(x-4\right)=x^2-x-12\)

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