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\(\Leftrightarrow\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right).462-\frac{17}{x+1,05}=19.\)
\(\Leftrightarrow\left(\frac{1}{11}-\frac{1}{21}\right)462-\frac{17}{x+1,05}=19\)
\(\Leftrightarrow20-\frac{17}{x+1,05}=19\)
\(\Leftrightarrow1-\frac{17}{x+1,05}=0\)
\(\Leftrightarrow x+1,05=17\)
\(\Leftrightarrow x=15,95\)
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Tìm x :
a) \(x-5=49:7\)
\(\Leftrightarrow x-5=7\)
\(\Leftrightarrow x=7+5\)
\(\Leftrightarrow x=12\)
Vậy : \(x=12\)
b) \(2x+6=24\)
\(\Leftrightarrow2x=24-6=18\)
\(\Leftrightarrow x=18:2\)
\(\Leftrightarrow x=9\)
Vậy : \(x=9\)
c) \(\frac{1}{3}:x+\frac{1}{2}=5\)
\(\Leftrightarrow\frac{1}{3}:x=5-\frac{1}{2}=\frac{9}{5}\)
\(\Leftrightarrow x=\frac{1}{3}:\frac{9}{5}\)
\(\Leftrightarrow x=\frac{5}{27}\)
Vậy : \(x=\frac{5}{27}\)
d) \(\frac{1}{6}.x-\frac{1}{3}=2\)
\(\Leftrightarrow\frac{1}{6}.x=2-\frac{1}{3}=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{5}{3}:\frac{1}{6}\)
\(\Leftrightarrow x=10\)
Vậy : \(x=10\)
e) \(\frac{x}{27}=\frac{3}{x}\)
\(\Leftrightarrow x.x=27.3\)
\(\Leftrightarrow x^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=9\end{matrix}\right.\) mà \(x\in N\)
\(\Rightarrow x=9\)
Vậy : \(x=9\)
g) \(1200:24-\left(17-x\right)=36\)
\(\Leftrightarrow50-17+x=36\)
\(\Leftrightarrow33+x=36\)
\(\Leftrightarrow x=36-33\)
\(\Leftrightarrow x=3\)
Vậy : \(x=3\)
h) \(674-\left(12+x\right)=427\)
\(\Leftrightarrow12+x=674-427=247\)
\(\Leftrightarrow x=247-12\)
\(\Leftrightarrow x=230\)
Vậy : \(x=230\)
k) \(36.\left(x-9\right)=900\)
\(\Leftrightarrow x-9=900:36\)
\(\Leftrightarrow x-9=25\)
\(\Leftrightarrow x=25+9\)
\(\Leftrightarrow x=34\)
m) \(1,2:x+3,8:x=2,5\)
\(\Leftrightarrow\left(1,2-3,8\right):x=2,5\)
\(\Leftrightarrow-2,6:x=2,5\)
\(\Leftrightarrow x=\frac{-2,6}{2,5}=-\frac{26}{25}\)
Vậy : \(x=-\frac{26}{25}\)
n) \(\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right).x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right).x=\frac{1}{3}\)
\(\Leftrightarrow\left[\frac{1}{2}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\right].x=\frac{1}{3}\)
\(\Leftrightarrow\left[\frac{1}{2}.\left(1-\frac{1}{10}\right)\right].x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}.\frac{9}{10}.x=\frac{1}{3}\)
\(\Leftrightarrow\frac{9}{20}.x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}:\frac{9}{20}\)
\(\Leftrightarrow x=\frac{20}{27}\)
Vậy : \(x=\frac{20}{27}\)
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a) -5/2 - 1/2x = -19/5
1/2x=-5/2 - (-19/5)
1/2x=13/10
x=13/10:1/2
x=13/5
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Giải:
a) \(\dfrac{7}{x}< \dfrac{x}{4}< \dfrac{10}{x}\)
\(\Rightarrow7< \dfrac{x^2}{4}< 10\)
\(\Rightarrow\dfrac{28}{4}< \dfrac{x^2}{4}< \dfrac{40}{4}\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x=6\)
b) \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
\(...\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{8}{9}\left(1\right)\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5}\)
\(...\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}>\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{2}{5}\left(2\right)\)
Từ (1) và (2), ta có:
\(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\left(đpcm\right)\)
Bạn có thể viết thay dòng "Từ (1) và (2)" thành "Từ các điều kiện trên" bạn nhé !(bạn ko cần phải sửa, đây chỉ là gợi ý)
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#)Mời bạn tham khảo :
Câu hỏi của SKT_NTT - Toán lớp 5 - Học toán với OnlineMath
Link : https://olm.vn/hoi-dap/detail/57615098100.html
Mình sẽ gửi link này về chat riêng cho bạn nha !
Xét vế phải
\(\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+\frac{4}{6}+\frac{5}{5}+\frac{6}{4}+\frac{7}{3}+\frac{8}{2}+\frac{9}{1}\)
\(=\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+\frac{4}{6}+\frac{5}{5}+\frac{6}{4}+\frac{7}{3}+\frac{8}{2}+9\)
\(=\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+\frac{4}{6}+\frac{5}{5}+\frac{6}{4}+\frac{7}{3}+\frac{8}{2}+\left(1+1+...+1\right)\)
\(=\left(1+\frac{1}{9}\right)+\left(1+\frac{2}{8}\right)+\left(1+\frac{3}{7}\right)+\left(1+\frac{4}{6}\right)+\left(1+\frac{5}{5}\right)+\left(1+\frac{6}{4}\right)+\left(1+\frac{7}{3}\right)+\left(1+\frac{8}{2}\right)+1\)\(=\frac{10}{9}+\frac{10}{8}+\frac{10}{7}+\frac{10}{6}+\frac{10}{5}+\frac{10}{4}+\frac{10}{3}+\frac{10}{2}+\frac{10}{10}\)
\(=10\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)
Thay vào bài ,ta được:
\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\times x=10\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)\(\Rightarrow x=10\)
Vậy x=10
chúc bạn học tốt
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a. \(\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=8.2\)
\(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow\left(x+1\right)^2=2^4\)
\(\Leftrightarrow\left(x+1\right)=2^2\)
\(\Leftrightarrow\left(x+1\right)=4\)
\(\Leftrightarrow x=4-1=3\)
b. \(x:\left(9\frac{1}{2}-\frac{3}{2}\right)=\frac{0,4+\frac{2}{9}-\frac{2}{11}}{1,6+\frac{8}{9}-\frac{8}{11}}\)
\(\Leftrightarrow x:\left(\frac{10}{2}-\frac{3}{2}\right)=\frac{0,4+0,2-0,18}{1,6+0,8-0,72}\)
\(\Leftrightarrow x:\frac{7}{2}=\frac{\frac{21}{50}}{\frac{42}{25}}\)
\(\Leftrightarrow x=\frac{\frac{21}{50}}{\frac{42}{25}}.\frac{7}{2}\Leftrightarrow x=\frac{1}{4}.\frac{7}{2}=\frac{7}{8}\)
a ) \(\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Rightarrow\left(x+1\right).\left(x+1\right)=2.8\)
\(\Rightarrow\left(x+1\right)^2=16\)
\(\Rightarrow\orbr{\begin{cases}x+1=4\\x+1=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=4-1\\x=-4-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
Dấu " \(\orbr{\begin{cases}\\\end{cases}}\)là hoặc nha !!!