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\(x(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}=\) \(5\frac{1}{2}\)
\(x\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\right)=\frac{11}{2}\)
\(x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\right)=\frac{11}{2}\)
\(x\left(1-\frac{1}{12}\right)=\frac{11}{2}\)
\(x\cdot\frac{11}{12}=\frac{11}{2}\)
\(x=\frac{11}{2}:\frac{11}{12}\)
\(x=6\)
Vậy x = 6
\(\frac{x}{2}+\frac{x}{6}+\frac{x}{12}+\frac{x}{20}+...+\frac{x}{132}=5\frac{1}{2}\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{132}\right)=\frac{11}{2}\)
\(\Rightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\right)=\frac{11}{2}\)
\(\Rightarrow x\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}=\frac{11}{2}\right)\)
\(\Rightarrow x\left(1-\frac{1}{12}\right)=\frac{11}{2}\)
\(\Rightarrow x.\frac{11}{12}=\frac{11}{12}\)
\(\Rightarrow x=\frac{11}{12}:\frac{11}{12}=1\)
Vậy x = 1
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12/7:x+2/3=7/5
=> 12/7:x = 7/5-2/3
=> 12/7 :x = 11/15
=> x = 12/7 : 11/5
=> x = 180/77
bn k giúp mik nha
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Bài 1: Hơi thắc mắc một chút, ukm tìm x để phân số nguyên à bn:
\(a.\)\(\frac{6+x}{33}\)có giá trị nguyên
\(\Leftrightarrow6+x⋮33\)
\(\Leftrightarrow6+x\in B\left(33\right)=\left\{0;\pm33;\pm66;...\right\}\)
\(\Leftrightarrow x\in\left\{-6;27;-39;60;-72;...\right\}\)
Bài này sao sao ấy, nếu vậy thì sẽ có rất nhiều x thỏa mãn ( vô vàn luôn, ko giới hạn )
\(b.\)\(\frac{12+x}{43-x}\)có giá trị nguyên
\(\Leftrightarrow12+x⋮43-x\)
Ta thấy: \(43-x⋮43-x\forall x\in Z\)
\(\Rightarrow\left(12+x\right)+\left(43-x\right)⋮43-x\forall x\in Z\)
\(\Leftrightarrow12+x+43-x⋮43-x\forall x\in Z\)
\(\Leftrightarrow\left(12+43\right)+\left(x-x\right)⋮43-x\forall x\in Z\)
\(\Leftrightarrow55⋮43-x\forall x\in Z\)
\(\Leftrightarrow43-x\inƯ\left(55\right)=\left\{\pm1;\pm5;\pm11;\pm55\right\}\)
Sau đó bn lập bẳng kết quả và xét là đc nha, mk ko bt lập bảng kết quả trong OLM nên ko giúp bn đc, thứ lỗi nha.
Bài 2:
Câu hỏi của Sarimi chan - Toán lớp 5 - Học toán với OnlineMath
Câu hỏi của Phạm Huyền My - Toán lớp 5 - Học toán với OnlineMath
Vào link này nhé, bài của mk ở đây
Rất vui vì giúp đc bn !!!
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Ta có : \(\frac{1}{4}+\frac{1}{3}:\frac{1}{x}=\frac{11}{12}\)
\(\Rightarrow\frac{1}{3}:\frac{1}{x}=\frac{11}{12}-\frac{1}{4}\)
\(\frac{1}{3}:\frac{1}{x}=\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}:\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}\times\frac{3}{2}\)
\(\frac{1}{x}=\frac{1}{2}\)
=> x = 2
a) \(\frac{x\div3-16}{2}+21=38\)
\(\frac{x\div3-16}{2}=38+21\)
\(\frac{x\div3-16}{2}=59\)
\(x\div3-16=59.2\)
\(x\div3-16=118\)
\(x\div3=118+16\)
\(x\div3=134\)
\(x=134.3\)
\(x=402\)
b) \(\frac{1}{4}+\frac{1}{3}\div\frac{1}{x}=\frac{11}{12}\)
\(\frac{1}{3}\div\frac{1}{x}=\frac{11}{12}-\frac{1}{4}\)
\(\frac{1}{3}\div\frac{1}{x}=\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}\div\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{2}\)
Vậy x = ....
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\(\left(\times-\frac{1}{5}\right):\left(\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{110}\right)=\frac{1}{5}\)
\(\Rightarrow\left(\times-\frac{1}{5}\right):\left(\frac{1}{1\times2}+\cdot\cdot\cdot+\frac{1}{10\times11}\right)=\frac{1}{5}\)
\(\Rightarrow\left(\times-\frac{1}{5}\right):\left(1-\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{10}-\frac{1}{11}\right)=\frac{1}{5}\)
\(\Rightarrow\left(\times-\frac{1}{5}\right):\left(1-\frac{1}{11}\right)=\frac{1}{5}\)
\(\Rightarrow\left(\times-\frac{1}{5}\right):\frac{10}{11}=\frac{1}{5}\)
\(\Rightarrow\left(\times-\frac{1}{5}\right)=\frac{1}{5}\times\frac{10}{11}\)
\(\Rightarrow\times-\frac{1}{5}=\frac{2}{11}\)
\(\Rightarrow\times=\frac{2}{11}+\frac{1}{5}\)
\(\Rightarrow\times=\frac{21}{55}\)
\(\left(x-\frac{1}{5}\right):\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=\frac{1}{5}\)
\(\Rightarrow\left(x-\frac{1}{5}\right):\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{10\times11}\right)=\frac{1}{5}\)
\(\Rightarrow\left(x-\frac{1}{5}\right):\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=\frac{1}{5}\)
\(\Rightarrow\left(x-\frac{1}{5}\right):\left(1-\frac{1}{11}\right)=\frac{1}{5}\)
\(\Rightarrow\left(x-\frac{1}{5}\right):\frac{10}{11}=\frac{1}{5}\)
\(\Rightarrow x-\frac{1}{5}=\frac{1}{5}\times\frac{10}{11}\)
\(\Rightarrow x-\frac{1}{5}=\frac{2}{11}\)
\(\Rightarrow x=\frac{2}{11}+\frac{1}{5}\)
\(\Rightarrow x=\frac{21}{55}\)
Vậy \(x=\frac{21}{55}\)
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giai câu a
a) ta có (\(\frac{2}{11.13}\)+\(\frac{2}{13.15}\)+.....+\(\frac{2}{19.21}\))*462 - x =19
(\(\frac{1}{11}\)-\(\frac{1}{13}\)+\(\frac{1}{13}\)-\(\frac{1}{15}\)+....+\(\frac{1}{19}\)-\(\frac{1}{21}\)) * 462 -x=19
(\(\frac{1}{11}\)-\(\frac{1}{21}\))*462-x=19
<=> \(\frac{12\left(x+2\right)}{x\left(x+2\right)}\)= \(\frac{12x}{\left(x+2\right)x}\)+ \(\frac{\left(x+2\right)x}{\left(x+2\right)x}\)
<=> 12x + 24 = 12x + x2 + 2x
<=> 12x - 12x -x2 + 2x + 24 = 0
<=> - x2 + 2x + 24 = 0
<=> - x2 - 4x + 6x + 24 = 0
<=> - x(x - 4) + 6(x + 4) = 0
<=> (x + 4)(- x + 6) = 0
<=> x + 4 = 0 hoặc - x + 6 = 0
<=> x = -4 hoặc x = - 6
Cho tam giác ABC trên cạnh AB lấy điểm D sao cho AD gấp đôi DB trên AC lấy điểm E sao cho AE gấp đôi EC , BE cắt CD tại G .So sánh diện tích GDB với điện tích GEC