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NN
17 tháng 2 2022
Tham Khảo:
Every year, on January 13, Lim festival is held in Tien Du, Bac Ninh. While the festival is going on, there's a lot of activity. Like other festivals, the Lim festival is divided into ceremonies and festivals. The ceremony is organized with traditional rituals such as worshiping and rituals.
Z
17 tháng 2 2022
refer
Every year, on January 13, Lim festival is held in Tien Du, Bac Ninh. While the festival is going on, there's a lot of activity. Like other festivals, the Lim festival is divided into ceremonies and festivals. The ceremony is organized with traditional rituals such as worshiping and rituals.
NM
Nguyễn Minh Quang
Giáo viên
10 tháng 8 2021
ta có \(\left(x-1\right)\left(3-x\right)\le\left(\frac{x-1+3-x}{2}\right)^2=1\le\left|y-2\right|+1\)
Dấu bằng xart ra khi:
\(\hept{\begin{cases}x-1=3-x\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=2\end{cases}}\)Vậy phương trình có nghiệm duy nhất (2,2)
15 tháng 8 2021
\(B=\left|x+1\right|+\left|x-4\right|+\left|2x-5\right|\ge\left|2x-3\right|+\left|2x-5\right|=\left|2x-3\right|+\left|5-2x\right|\)
\(\ge\left|2x-3+5-2x\right|=\left|2\right|=2\)
Dấu ''='' xảy ra khi \(\left(x+1\right)\left(4-x\right)\ge0;\left(2x-3\right)\left(5-2x\right)\ge0\)
\(-1\le x\le4;\frac{3}{2}\le x\le\frac{5}{2}\Rightarrow-1\le x\le4\)
Vậy GTNN của B bằng 2 tại -1 =< x =< 4
31 tháng 3 2022
\(a;\left(\cos a-\sin a\right)\left(cosa+sina\right)=cos^2a-sin^2a=1-sin^2a-sin^2a=1-2sin^2a\)
\(b;VP=\left(2cosa-1\right)\left(2cosa+1\right)=4cos^2a-1=4\left(1-sin^2a\right)-1=3-4sin^2a=VT\)
e;\(\dfrac{1}{1+tana}+\dfrac{1}{1+cota}=1\Leftrightarrow cota+tana+2=\left(cota+1\right)\left(tana+1\right)\Leftrightarrow cota+tana+2=cota.tana+cota+tana+1\Leftrightarrow cota+tana+2=1+cota+tana+1\Leftrightarrow0=0\left(đúng\right)\Rightarrow VT=VP\)
\(d;sin^3a+cos^3a=\left(sina+cosa\right)\left(sin^2a-sina.cosa+cos^2a\right)=\left(sina+cosa\right)\left(1-sina.cosa\right)\left(đpcm\right)\left(hđt:a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\right)\)
\(c;sin^2a.cosa+sina.cos^2a=\left(sina.cosa\right)\left(sin^2+cos^2\right)=sina.cosa\)
\(f;;tana+\dfrac{cosa}{1+sina}=\dfrac{sina}{cosa}+\dfrac{cosa}{1+sina}=\dfrac{sina+sin^2a+cos^2a}{cosa\left(1+sina\right)}=\dfrac{1+sina}{cosa\left(1+sina\right)}=\dfrac{1}{cosa}\)
\(g;1+cot^2a=\dfrac{1}{sin^2a}=\dfrac{1}{1-cos^2a}=\dfrac{1}{\left(1-cosa\right)\left(1+cosa\right)}\left(đpcm\right)\)
\(h;\dfrac{1+cosa}{1-cosa}-\dfrac{1-cosa}{1+cosa}=\dfrac{\left(cosa+1\right)^2-\left(cosa-1\right)^2}{1-cosa^2}=\dfrac{\left(cosa+1-cosa+1\right)\left(cosa+1+cosa-1\right)}{1-cos^2a}=\dfrac{4cosa}{sin^2a}\left(đpcm\right)\)
\(k;\dfrac{1+cosa}{sina}-\dfrac{sina}{1+cosa}=\dfrac{\left(cosa+1\right)^2-sin^2a}{sina\left(1+cosa\right)}=\dfrac{cos^2a+2cosa+1-sin^2a}{sina\left(1+cosa\right)}=\dfrac{2cos^2a+2cosa}{sina\left(1+cosa\right)}=\dfrac{2cosa\left(1+cosa\right)}{sina\left(1+cosa\right)}=\dfrac{2cosa}{sina}=2cota\left(đpcm\right)\)
\(m;;;\Leftrightarrow sin^3a=cosa\left(1+cosa\right)\left(tana-sina\right)=\left(cosa+cos^2a\right)\left(tana-sina\right)\Leftrightarrow sin^3a=\left(cosa+cos^2a\right)\left(\dfrac{sina}{cosa}-sina\right)=sina-sina.cosa+cosa.sina-cos^2a.sina\Leftrightarrow sin^3a=sina-cos^2a.sina\Leftrightarrow sin^3a-sina\left(1-cos^2a\right)=0\Leftrightarrow sin^3a-sina.sin^2a=0\Leftrightarrow0=0\left(đúng\right)\Rightarrowđpcm\)
29 tháng 11 2023
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\\ n_{Al}=a;n_{Mg}=b\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}1,5a+b=0,4\\27a+24b=7,8\end{matrix}\right.\\ \Rightarrow a=0,2;b=0,1\\ \%m_{Al}=\dfrac{0,2.27}{7,8}\cdot100=69,23\%\\ \%m_{Mg}=100-69,23=30,77\%\)
Q9
1
2 tháng 1 2022
Chữ khó nhìn quá bạn!
1, I suggest collecting boys
2, How about going to the beach
3, Because she was very tired, she went to bed
Q9
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