\(\left(a+b\right)^3-3ab.\left(a+b\right)=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]=\left(a+b\right)\left(a^2+b^2-ab\right)\)

Đặt \(A=1+2+2^2+2^3+...+2^{63}\)

\(\Rightarrow2A=2+2^2+2^3+2^4+2^5+...+2^{64}\)

\(\Rightarrow A=2A-A=\left(2+2^2+2^3+2^4+...+2^{64}\right)-\left(1+2+2^2+2^3+...+2^{63}\right)=2^{64}-1\left(đpcm\right)\)

Ta có: \(\left\{{}\begin{matrix}\sqrt{2008}+\sqrt{2005}< \sqrt{2015}+\sqrt{2009}\\\sqrt{2010}+\sqrt{2007}< \sqrt{2015}+\sqrt{2009}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{\sqrt{2008}+\sqrt{2005}}+\dfrac{1}{\sqrt{2010}+\sqrt{2007}}>\dfrac{2}{\sqrt{2015}+\sqrt{2009}}\)

\(\Leftrightarrow\dfrac{\sqrt{2008}-\sqrt{2005}}{3}+\dfrac{\sqrt{2010}-\sqrt{2007}}{3}>\dfrac{\sqrt{2015}-\sqrt{2009}}{3}\)

\(\Leftrightarrow\sqrt{2008}+\sqrt{2009}+\sqrt{2010}>\sqrt{2005}+\sqrt{2007}+\sqrt{2015}\)

Thời gian xe máy đi từ A đến B với vận tốc không thay đổi là:

              \(\dfrac{3}{4}.6=\dfrac{9}{2}=4,5\left(h\right)\)

\(B=\dfrac{\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}}{\dfrac{3}{2020}+\dfrac{3}{2021}-\dfrac{3}{2022}}-1=\dfrac{\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}}{3\left(\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}\right)}-1=\dfrac{1}{3}-1=-\dfrac{2}{3}\)

Bài 2:

a) \(\left(x+5\right)^2=x^2+10x+25\)

b) \(\left(\dfrac{5}{2}-t\right)^2=\dfrac{25}{4}-5t+t^2\)

c) \(\left(2u+3v\right)^2=4u^2+12uv+9v^2\)

d) \(\left(-\dfrac{1}{8}a+\dfrac{2}{3}bc\right)^2=\dfrac{1}{64}a^2-\dfrac{1}{6}abc+\dfrac{4}{9}b^2c^2\)

e) \(\left(\dfrac{x}{y}-\dfrac{1}{z}\right)^2=\dfrac{x^2}{y^2}-\dfrac{2x}{yz}+\dfrac{1}{z^2}\)

f) \(\left(\dfrac{mn}{4}-\dfrac{x}{6}\right)\left(\dfrac{mn}{4}+\dfrac{x}{6}\right)=\dfrac{m^2n^2}{16}-\dfrac{x^2}{36}\)

a) \(20a^4b^5c^2:\left(-5ab^2c\right)^2=20a^4b^5c^2:\left(25a^2b^4c^2\right)=\dfrac{4}{5}a^2b\)

b) \(\left(-15x^2y^3\right)^7:\left(15xy^3\right)^6-\left(32x^{18}y^5\right):\left(-4x^5y\right)^2=-15x^8y^3-2x^8y^3=-17x^8y^3\)

c) \(-13-13x^5y^2:\left(-2xy\right)-\left(x^2+2x+1\right):\left(x+1\right)=-13+\dfrac{13}{2}x^4y-\left(x+1\right)^2:\left(x+1\right)=-13+\dfrac{13}{2}x^4y-x-1=-14+\dfrac{13}{2}x^4y-x\)

\(\sqrt{4x^2-24x+45}=\sqrt{4\left(x-3\right)^2+9}\) 

Vì \(4\left(x-3\right)^2\ge0\Rightarrow4\left(x-3\right)^2+9\ge9\)

\(\Rightarrow\sqrt{4x^2-24x+45}\ge\sqrt{9}=3\)

\(ĐTXR\Leftrightarrow x=3\)