1 + 2 = 3

ĐỔI TK NHA

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kb thỏai mái

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刚发少吃醋把做下 >:33

mik ko hiu

<=>2x^2+2x+x+1=0

<=>2x(x+1)+(x+1)=0

<=>(2x+1)(x+1)=0

<=>\(\orbr{\begin{cases}2x+1=0\\x+1=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=-\frac{1}{2}\\x=-1\end{cases}}\)

vậy..........

\(2x^2+3x+1=0.\)

\(\Rightarrow2x^2+2x+x+1=0\)

\(\Rightarrow\left(2x^2+2x\right)+\left(x+1\right)=0\)

\(\Rightarrow2x\left(x+1\right)+\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(2x+1\right)=0\)

\(\orbr{\begin{cases}x+1=0\Rightarrow x=-1\\2x+1=0\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}\end{cases}}\)

\(\Rightarrow S=\left\{-1;\frac{1}{2}\right\}\)

Có 2 trường hợp xảy ra

TH1: x=3

TH2: x=-3

#Hk_tốt

#Ngọc's_Ken'z

\(\left|x-3\right|+\left|3-x\right|=12\)

\(\Leftrightarrow\left|x-3\right|+\left|x-3\right|=12\)

\(\Leftrightarrow2\left|x-3\right|=12\)

\(\Leftrightarrow\left|x-3\right|=6\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=6\\x-3=-6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=9\\x=-3\end{cases}}\)

Vậy....

\(\frac{1}{2\cdot4}+\frac{1}{6\cdot8}+...+\frac{1}{96\cdot98}+\frac{1}{98\cdot100}\)

\(=\frac{1}{2}\left[\frac{2}{2\cdot4}+\frac{2}{6\cdot8}+...+\frac{2}{96\cdot98}+\frac{2}{98\cdot100}\right]\)

\(=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right]\)

\(=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{100}\right]=\frac{1}{2}\left[\frac{50}{100}-\frac{1}{100}\right]=\frac{1}{2}\cdot\frac{49}{100}=\frac{49}{200}\)

\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}...+\frac{1}{96.98}+\frac{1}{98.100}\)

\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{96.98}+\frac{2}{98.100}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=\frac{1}{2}.\frac{49}{100}\)

\(=\frac{49}{200}\)

~Học tốt~

Đặt \(B=\frac{C}{D}\)

Biến đổi D : \(D=\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}\)

                         \(=\left(99+1\right)+\left(\frac{98}{2}+1\right)+...+\left(\frac{1}{99}+1\right)-99\)

                          \(=100+\frac{100}{2}+...+\frac{100}{99}+\frac{100}{100}-100\)

                           \(=100.\left(\frac{1}{2}+...+\frac{1}{100}\right)\)

\(\Rightarrow B=\frac{\frac{1}{2}+...+\frac{1}{100}}{100.\left(\frac{1}{2}+...+\frac{1}{100}\right)}=\frac{1}{100}\)