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\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(2\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}\right)=2.\frac{15}{93}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(2x=90\)
\(x=\frac{90}{2}=45\)
Vậy \(x=45\)
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Ta có : \(\left(3-x\right)\left|x+5\right|=0\)
\(\Leftrightarrow\orbr{\begin{cases}3-x=0\\\left|x+5\right|=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
Ta có :
\(\left(3-x\right)\left|x+5\right|=0\)
\(\Rightarrow\orbr{\begin{cases}3-x=0\\\left|x+5\right|=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x+5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
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Đặt A= 1.2+2.3 +.......+99.100
3A= 1.2.3+2.3.4+3.4.3 +......+ 99.100.3
3A= 1.2. (3 - 0) + 2.3.(4 - 1) +3.4. (5 - 2)....... . 99.100. (101 - 98)
3A = (1.2.3 + 2.3.4 + 3.4.5 +...... + 99.100.101) - (0.1.2 + 1.2.3 + 2.3.4 +.......+ 98.99.100)
3A = 99.100.101 - 0.1.2
3A = 999900 - 0
3A= 999900
A= 999900 : 3
A = 333300
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a)Ta có: \(\frac{1}{3}+\frac{2}{3}:x=-7\)
\(\Rightarrow\frac{2}{3}:x=\left(-7\right)-\frac{1}{3}=\frac{-22}{3}\)
\(\Rightarrow x=\frac{2}{3}:\frac{-22}{3}=\frac{-1}{11}\)
c) Ta có:(2x-3).(6-2x) = 0
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\6-2x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=3\\2x=6\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}}\)
Vậy x = 2 hoặc x = \(\frac{3}{2}\)
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\(\frac{1}{2x-1}=\frac{y}{5}+\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2x-1}=\frac{3y+5}{15}\)
\(\Leftrightarrow\left(2x-1\right).\left(3y+5\right)=15\)
\(\Rightarrow2x-1;3y+5\inƯ\left(15\right)=\left\{1;3;5;15\right\}\)(mình lười nên lấy x ,y \(\ge0\)nếu âm thì bạn lấy thêm)
5
vậy các cặp giá trị (x,y)={(1;10/3),(2;0),(8;-4/3),(3;-2/3)}