tìm \(x\in Q\),biết rằng
\(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
S
18 tháng 5 2017
Bài 3:
a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
3A = \(1-\frac{1}{2^6}\)
=> 3A < 1
=> A < \(\frac{1}{3}\)(đpcm)
b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)
4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
4B = \(3-\frac{1}{3^{99}}\)
=> 4B < 3
=> B < \(\frac{3}{4}\) (2)
Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)
30 tháng 4 2016
\(\left(\frac{1}{4}x-1\right)-\left(\frac{2}{3}x-1\right)+\left(\frac{4}{5}x-1\right)=\frac{2}{3}\)
\(\frac{1}{4}x-1-\frac{2}{3}x+1+\frac{4}{5}x-1\)\(=\frac{2}{3}\)
\(\left(\frac{1}{4}x-\frac{2}{3}x+\frac{4}{5}x\right)+1-1-1\)\(=\frac{2}{3}\)
\(\frac{23}{60}x-1\)\(=\frac{2}{3}\)
\(\frac{23}{60}x=\frac{2}{3}+1\)
\(\frac{23}{60}x=\frac{5}{3}\)
\(x=\frac{5}{3}:\frac{23}{60}=\frac{100}{23}\)
Vậy x=\(\frac{100}{23}\)
24 tháng 7 2020
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
...
\(\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}\)
Cộng vế theo vế
\(\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
\(\Rightarrow B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
Lại có \(\frac{7}{8}< 1\)
Theo tính chất bắc cầu => \(B< \frac{7}{8}< 1\)
\(\Rightarrow B< 1\left(đpcm\right)\)
TY
20 tháng 3 2019
\(\frac{x+1}{5}+\frac{x+2}{4}=\frac{x+3}{3}+\frac{x+4}{2}\)\(\Rightarrow\left(\frac{x+1}{5}+1\right)+\left(\frac{x+2}{4}+1\right)=\left(\frac{x+3}{3}+1\right)+\left(\frac{x+4}{2}+1\right)\)
\(\Rightarrow\frac{x+1+5}{5}+\frac{x+2+4}{4}=\frac{x+3+3}{3}+\frac{x+4+2}{2}\)
\(\Rightarrow\frac{x+6}{5}+\frac{x+6}{4}-\frac{x+6}{3}-\frac{x+6}{2}=0\)
\(\Rightarrow\left(x+6\right)\cdot\frac{1}{5}+\left(x+6\right)\cdot\frac{1}{4}-\left(x+6\right)\cdot\frac{1}{3}-\left(x+6\right)\cdot\frac{1}{2}=0\)
\(\Rightarrow\left(x+6\right)\cdot\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Rightarrow x+6=0\Rightarrow x=-6\)
Các bạn nhớ cho mk nha
21 tháng 7 2017
\(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}\)và x + y -z = 10
\(\frac{x}{2}=\frac{y}{3}=\frac{1}{4}.\frac{x}{2}=\frac{1}{4}.\frac{y}{3}\)\(=\frac{x}{8}=\frac{y}{12}\)
\(\frac{y}{4}=\frac{z}{5}=\frac{1}{3}.\frac{y}{4}=\frac{1}{3}.\frac{z}{5}=\frac{y}{12}=\frac{z}{15}\)
\(\Leftrightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)và x + y - z = 10
Theo tính chất dãy tỉ số bằng nhau:
\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)
* \(\frac{x}{8}=2\Rightarrow x=2.8=16\)
* \(\frac{y}{12}=2\Rightarrow y=2.12=24\)
* \(\frac{z}{5}=2\Rightarrow z=2.5=10\)
Vậy...
21 tháng 7 2017
Ý mk nhầm chút xíu nhé! Cko sorry!
* \(\frac{z}{15}=2\Rightarrow z=2.15=30\)
... :( Xl
P1
6 tháng 8 2017
Câu b thôi các bạn nhé, câu a mình ko cần nx với cả mình ghi sai dữ liệu câu a r
DP
6 tháng 8 2017
a, \(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot...\cdot\frac{30}{62}\cdot\frac{31}{64}=2x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}=2x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{2\cdot2\cdot3\cdot2\cdot4\cdot2\cdot5\cdot2\cdot....\cdot31\cdot2\cdot32\cdot2}=2x\)
\(\Leftrightarrow\frac{1}{2\cdot2\cdot2\cdot2\cdot....\cdot2\cdot2\cdot32}=2x\)
Có : (31 - 1) : 1 + 1 = 31 (thừa số 2)
\(\Rightarrow\frac{1}{2^{31}.32}=2x\)
\(\Rightarrow x=\frac{1}{2^{31}.32}\div2\)
b, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow x+1=x+4\)
\(\Leftrightarrow0=3\text{ (vô lý) }\)
29 tháng 4 2016
\(\frac{1}{4}x-1-\frac{2}{3}x+1+\frac{4}{5}x-1=\frac{2}{3}\)
\(\left(\frac{1}{4}x-\frac{2}{3}x+\frac{4}{5}x\right)+\left(1-1-1\right)=\frac{2}{3}\)
\(\frac{23}{60}x-1=\frac{2}{3}\)
\(\frac{23}{60}x=\frac{2}{3}+1\)
\(\frac{23}{60}x=\frac{2+3}{3}\)
\(\frac{23}{60}x=\frac{5}{3}\)
\(x=\frac{5}{3}\div\frac{23}{60}\)
\(x=\frac{5}{3}\times\frac{60}{23}\)
\(x=\frac{100}{23}\)
29 tháng 4 2016
\(\left(\frac{1}{4}x-1\right)-\left(\frac{2}{3}x-1\right)+\left(\frac{4}{5}x-1\right)=\frac{2}{3}\)
<=> \(\frac{1}{4}x-1-\frac{2}{3}x+1+\frac{4}{5}x-1=\frac{2}{3}\)
<=> \(\frac{1}{4}x-\frac{2}{3}x+\frac{4}{5}x-1+1-1=\frac{2}{3}\)
<=> \(\frac{23}{60x}=\frac{2}{3}\)=> x=\(\frac{40}{23}\)
Giải cụ thể theo cách lớp 7 đó...còn giải theo cách lớp 8 đơn giản hơn nhiều..nhưng làm theo lớp 8 sợ khó hiểu với lớp 7
>.<
27 tháng 8 2017
a) \(\frac{2}{3x}-\frac{3}{12}=\frac{4}{5}-\left(\frac{7}{x}-2\right)\)
\(\frac{2}{3x}+\left(\frac{7}{x}-2\right)=\frac{4}{5}+\frac{3}{12}\)
\(\frac{2}{3x}+\frac{7}{x}-2=\frac{21}{20}\)
\(\frac{2}{3x}+\frac{7}{x}=\frac{61}{20}\)
\(\frac{2}{3x}+\frac{21}{3x}=\frac{61}{20}\)
\(\frac{23}{3x}=\frac{61}{20}\)
\(3x=\frac{460}{61}\)
\(x=\frac{460}{183}\)
b) \(\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
\(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)
\(\frac{19}{10}:x=2\)
\(x=\frac{19}{20}\)
Ta có: \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\Rightarrow\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=\frac{-7}{20}\)
\(\Rightarrow\frac{1}{4}:\frac{-7}{20}=\frac{-5}{7}\)