Tóm tắt:

\(45km/h\rightarrow1h\)

\(60km/h\rightarrow?h\)

             Thời gian đi từ a đến b với vận tốc 60km/h là:

                        \(\dfrac{45.1}{60}=\dfrac{3}{4}\left(h\right)\)

\(n\left(n-1\right)=240\Rightarrow n^2-n-240=0\Rightarrow\left(n-16\right)\left(n+15\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}n-16=0\\n+15=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}n=16\\n=-15\end{matrix}\right.\)

a) \(\sqrt{6-4\sqrt{2}}-\sqrt{6+4\sqrt{2}}=2-\sqrt{2}-2-\sqrt{2}=-2\sqrt{2}\)

b) \(\sqrt{14+6\sqrt{5}}-\sqrt{14-6\sqrt{5}}=3+\sqrt{5}-3+\sqrt{5}=2\sqrt{5}\)

c) \(\sqrt{11+4\sqrt{7}}-\sqrt{11-4\sqrt{7}}=\sqrt{7}+2-\sqrt{7}+2=4\)

d) \(\sqrt{19-8\sqrt{3}}+\sqrt{19+8\sqrt{3}}=4-\sqrt{3}+4+\sqrt{3}=8\)

a) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{3}\right)^2=\dfrac{x.y}{2.3}=\dfrac{54}{6}=9\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=36\\y^2=81\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm6\\y=\pm9\end{matrix}\right.\)

b) \(\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\left(\dfrac{x}{5}\right)^2=\left(\dfrac{y}{3}\right)^2=\dfrac{x^2-y^2}{5^2-3^2}=\dfrac{4}{16}=\dfrac{1}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{25}{4}\\y^2=\dfrac{9}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{5}{2}\\y=\pm\dfrac{3}{2}\end{matrix}\right.\)

\(x-3-\sqrt{x^2-6x+9}\left(1\right)=x-3-\sqrt{\left(x-3\right)^2}=x-3-\left|x-3\right|\)

TH1: \(x< 3\)

\(\left(1\right)=x-3+x-3=2x-6\)

TH2: \(x\ge3\)

\(\left(1\right)=x-3-x+3=0\)

a) \(x^4-25x^3=0\Leftrightarrow x^3\left(x-25\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^3=0\\x-25=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=25\end{matrix}\right.\)

b) \(\left(x-5\right)^2-\left(3x-2\right)^2=0\Leftrightarrow\left(x-5-3x+2\right)\left(x-5+3x-2\right)=0\Leftrightarrow-\left(2x+3\right)\left(4x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\4x-7=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{7}{4}\end{matrix}\right.\)

c) \(x^3-4x^2-9x+36=0\Leftrightarrow\left(x+3\right)\left(x-4\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\\x=3\end{matrix}\right.\)

a) \(40x^4-10x^2=10x^2\left(4x^2-1\right)=10x^2\left(2x-1\right)\left(2x+1\right)\)

b) \(16x^4-20x^2-y^2-5y=\left(4x^2-\dfrac{5}{2}\right)^2-\left(y-\dfrac{5}{2}\right)^2=\left(4x^2-\dfrac{5}{2}-y+\dfrac{5}{2}\right)\left(4x^2-\dfrac{5}{2}+y-\dfrac{5}{2}\right)=\left(4x^2-y\right)\left(4x^2+y-5\right)\)c)\(64a^2-9b^2-16a+1=\left(8a-1\right)^2-9b^2=\left(8a-1-3b\right)\left(8a-1+3b\right)\)d) \(5x^2+23x-10=5\left(x-\dfrac{2}{5}\right)\left(x+5\right)\)

\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}\)

(100 số số hạng)

\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{100}{\sqrt{100}}=\dfrac{100}{10}=10\)

Đặt \(a=\sqrt[3]{7+5\sqrt{2}},b=\sqrt[3]{7-5\sqrt{2}}\)

\(\Rightarrow\left\{{}\begin{matrix}a^3+b^3=14\\ab=-1\end{matrix}\right.\)

Ta có: \(x=a+b+2\Leftrightarrow x-2=a+b\Leftrightarrow\left(x-2\right)^3=\left(a+b\right)^3\Leftrightarrow x^3-6x^2+12x-8=a^3+b^3+3ab\left(a+b\right)\Leftrightarrow x^3-6x^2+12x-8=14+3\left(-1\right)\left(x-2\right)\Leftrightarrow x^3-6x^2+15x-28=0\Leftrightarrow\left(x-4\right)\left(x^2-2x+7\right)=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)

(do \(x^2-2x+7=\left(x-1\right)^2+6\ge6>0\))

Bài 3:

a) \(4x^2+4x+1=\left(2x+1\right)^2\)

b) \(9x^2-12x+4=\left(3x-2\right)^2\)

c) \(ab^2+\dfrac{1}{4}a^2b^4+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)