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\(\dfrac{x}{6}+\dfrac{x}{10}+\dfrac{x}{15}+........+\dfrac{x}{78}=\dfrac{220}{39}\)
\(\Leftrightarrow\dfrac{2x}{12}+\dfrac{2x}{20}+........+\dfrac{2x}{156}=\dfrac{220}{39}\)
\(\Leftrightarrow2x\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+..........+\dfrac{1}{12.13}\right)=\dfrac{220}{39}\)
\(\Leftrightarrow2x\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=\dfrac{220}{39}\)
\(\Leftrightarrow2x.\dfrac{10}{39}=\dfrac{220}{39}\)
\(\Leftrightarrow x.\dfrac{20}{39}=\dfrac{220}{39}\)
\(\Leftrightarrow x=11\)
Vậy ...
\(x.\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{78}\right)=\dfrac{220}{39}\)
\(x.\dfrac{20}{39}=\dfrac{220}{39}\)
\(x=\dfrac{220}{39}:\dfrac{20}{39}\)
x\(=11\)
A =\(2.\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+......+\dfrac{1}{156}\right)\)
A =\(2.\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+..........+\dfrac{1}{12.13}\right)\)
A =2.\(\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)
A=\(2.\dfrac{10}{39}=\dfrac{20}{39}\)
Câu 2:
\(\Leftrightarrow x\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{78}\right)=\dfrac{220}{39}\)
\(\Leftrightarrow2x\left(\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{156}\right)=\dfrac{220}{39}\)
\(\Leftrightarrow x\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{12}-\dfrac{1}{13}\right)=\dfrac{110}{39}\)
\(\Leftrightarrow x\cdot\dfrac{10}{39}=\dfrac{110}{39}\)
=>x=11
Tìm số nguyên x, biết:
a) \(\dfrac{-28}{35}=\dfrac{16}{x};\) b) \(\dfrac{x+7}{15}=\dfrac{-24}{36}.\)
\(a.\)
\(\dfrac{-28}{35}=\dfrac{16}{x}\)
\(\Rightarrow x=\dfrac{35\cdot16}{-28}=\dfrac{5\cdot7\cdot4\cdot4}{-7\cdot4}=-20\)
\(b.\)
\(\dfrac{x+7}{15}=\dfrac{-24}{36}\)
\(\Rightarrow x+7=\dfrac{15\cdot-24}{36}=\dfrac{5\cdot3\cdot-12\cdot2}{12\cdot3}=-10\)
\(\Leftrightarrow x=-17\)
Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.
b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
=>\(\dfrac{-3}{5}.x=1\)
=>\(x=1:\dfrac{-3}{5}\)
=>\(x=\dfrac{-5}{3}\)
Vậy \(x=\dfrac{-5}{3}\)
\(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ \dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ 2\cdot\left[\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+...+\dfrac{1}{x\left(x+1\right)}\right]=\dfrac{2}{9}\\ \dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2}{9}:2\\ \dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{9}\\ \dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{4}{9}\\ \dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\\ \dfrac{1}{x+1}=\dfrac{1}{18}\\ x+1=18\\ x=17\)
Vậy x = 17
1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)
\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)
\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)
\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)
Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)
Gọi biểu thức là A
\(A=\dfrac{2x}{12}+\dfrac{2x}{20}+\dfrac{2x}{30}+....+\dfrac{2x}{156}=\dfrac{200}{39}\)
Ta có công thức :
\(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)
Áp dụng công thức trên, ta có :
\(A=\dfrac{2x}{3.4}+\dfrac{2x}{4.5}+\dfrac{2x}{5.6}+....+\dfrac{2x}{12.13}\)
\(A=2x.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{12}-\dfrac{1}{13}\right)\)
\(A=2x.\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)
\(A=2x.\left(\dfrac{10}{39}\right)=\dfrac{200}{39}\)
\(A=2x=\dfrac{200}{39}:\dfrac{10}{39}\)
\(2x=20\)
\(\Rightarrow x=10\)
mink nghĩ vậy bạn ạ