Số chia bé nhất có thể có là 1

Suy ra số dư = thương = 1

Vậy số đó \(=1\times68+1=69\)

Người 1 bắt tay 119 người còn lại

Người 2 bắt tay 118 người còn lại

.....

Người 119 bắt tay 1 người còn lại

Vậy có \(1+2+...+119=\dfrac{\left(119+1\right)\cdot119}{2}=7140\) cái bắt tay

Có gì để thu gọn hả

Hệ số: `2-\sqrt{5}`

Biến: `x`

Bậc `1`

\(=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+...+\left(2003-2004\right)\left(2003+2004\right)+2005^2\\ =-\left(1+2\right)-\left(3+4\right)-...-\left(2003+2004\right)+2005^2\\ =-\left(1+2+3+...+2003+2004\right)+2005^2\\ =-\dfrac{\left(2004+1\right)\cdot2004}{2}+2005^2\\ =2011015\)

\(a,\\ \text{ĐL đường p/g: }\dfrac{AB}{AC}=\dfrac{BD}{CD}=\dfrac{36}{60}=\dfrac{3}{5}\\ \text{Hệ thức lượng: }\dfrac{HB}{HC}=\dfrac{\dfrac{AB^2}{BC}}{\dfrac{AC^2}{BC}}=\dfrac{AB^2}{AC^2}=\left(\dfrac{3}{5}\right)^2=\dfrac{9}{25}\\ b,BC=BD+CD=HB+HC=96\left(cm\right)\\ \to\dfrac{9}{25}HC+HC=96\\ \to HC=\dfrac{1200}{17}\to HB=\dfrac{432}{17}\\ \to AH=\sqrt{HC\cdot HB}=\dfrac{720}{17}\left(cm\right)\)

\(14,\\ a,B=\left(2x+1\right)^3=\left(2\cdot\dfrac{1}{2}+1\right)^3=2^3=8\\ b,C=\left(x+2y-2\right)^3=\left(20+2\cdot1-2\right)^3=20^3=8000\\ 15,\\ a,11^3-1=\left(11-1\right)\left(11^2+11+1\right)=10\cdot133=1330\\ b,x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=6^3+3\cdot9\cdot6=378\\ 16,\\ a,M=x^3+27-\left(27-8x^3\right)=9x^3=9\cdot20^3=72000\\ b,N=x^3-8y^3+16y^3=x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)=0\)

Đặt \(A=\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\ge0\)

\(\to A^2=\dfrac{2+\sqrt{3}}{2-\sqrt{3}}+\dfrac{2-\sqrt{3}}{2+\sqrt{3}}+2\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\cdot\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\\ \to A^2=\dfrac{\left(2+\sqrt{3}\right)^2+\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2\\ \to A^2=\dfrac{7+4\sqrt{3}+7-4\sqrt{3}}{4-3}+2\\ \to A^2=14+2=16\\ \to A=4\)

`n_{Na_2O}={4,6}/{62}\approx 0,074(mol)`

`Na_2O+H_2O->2NaOH`

`0,074->0,074->0,148(mol)`

`a)\ C_{M\ NaOH}={0,148}/{0,2}=0,74M`

`b)\ C\%_{NaOH}={0,148.40}/{200.1,12}.100\%\approx 2,64\%`

\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\\ \to\dfrac{1}{23,04}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\\ \to\dfrac{1}{23,04}=\dfrac{1}{AB^2}+\dfrac{1}{\dfrac{3}{4}AB^2}\\ \to\dfrac{1}{AB^2}+\dfrac{4}{3AB^2}=\dfrac{1}{23,04}\\ \to\dfrac{7}{3AB^2}=\dfrac{1}{23,04}\\ \to AB^2=53,76\\ \to AB=\dfrac{8\sqrt{21}}{5}\left(cm\right)\\ \to AC=\dfrac{32\sqrt{21}}{15}\left(cm\right)\\ \to BC=\sqrt{AB^2+AC^2}=\dfrac{8\sqrt{21}}{3}\left(cm\right)\)

Hệ thức lượng:

\(HB=\dfrac{AB^2}{BC}=\dfrac{24\sqrt{21}}{25}\left(cm\right)\\ HC=\dfrac{AC^2}{BC}=\dfrac{7168-200\sqrt{21}}{75}\left(cm\right)\)