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Bài 1 : \(a,\left|x-3,5\right|=7,5\)
\(\Rightarrow\orbr{\begin{cases}x-3,5=7,5\\x-3,5=-7,5\end{cases}}\Rightarrow\orbr{\begin{cases}x=11\\x=-4\end{cases}}\)
\(b,\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{cases}}\)
\(c,3,6-\left|x-0,4\right|=0\)
\(\Rightarrow\left|x-0,4\right|=3,6\)
\(\Rightarrow\orbr{\begin{cases}x-0,4=3,6\\x-0,4=-3,6\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-3,2\end{cases}}\)
\(d,\left|x-\frac{1}{2}\right|-\frac{1}{3}=1\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=\frac{4}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{4}{3}\\x-\frac{1}{2}=-\frac{4}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{6}\\x=-\frac{5}{6}\end{cases}}\)
a) \(\frac{x}{5}=\frac{2}{3}\)
\(\Rightarrow\)\(x=\frac{2.5}{3}=\frac{10}{3}\)
Vậy....
b) \(\frac{x+3}{15}=\frac{1}{5}\)
\(\Leftrightarrow\)\(5\left(x+3\right)=15\)
\(\Leftrightarrow\)\(x+3=3\)
\(\Leftrightarrow\)\(x=0\)
Vậy....
1) Tính :
a) \(\left(2008.2009.2010.2011\right).\left(1+\frac{1}{2}:\frac{2}{3}-\frac{4}{3}\right)\)
\(=\left(2008.2009.2010.2011\right).\left(1+\frac{1}{3}-\frac{4}{3}\right)\)
\(=\left(2008.2009.2010.2011\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)
\(=\left(2008.2009.2010.2011\right).0\)
\(=0\)
2) Tìm x
a) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\)
\(\Rightarrow x+1=2013\)
\(\Rightarrow x=2012\)
b) \(\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}.\frac{1}{6}.\left(x-1,010\right)=\frac{1}{360}-\frac{1}{720}\)
\(\Rightarrow\frac{1}{2.3.4.5.6}.\left(x-1,01\right)=\frac{1}{720}\)
\(\Rightarrow\frac{1}{720}.\left(x-1,01\right)=\frac{1}{720}\)
\(\Rightarrow x-1,01=\frac{1}{720}:\frac{1}{720}\)
\(\Rightarrow x-1,01=1\)
\(\Rightarrow x=1+1,01\)
\(\Rightarrow x=2,01\)
tung từng vế một thôi
bạn nhác quá éo chịu suy nghĩ
bài này dễ vl
Bài 1:
a, \(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2010}{2011}\)
\(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(1-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\frac{1}{5x+6}=1-\frac{2010}{2011}\)
\(\frac{1}{5x+6}=\frac{1}{2011}\)
=> 5x + 6 = 2011
5x = 2011 - 6
5x = 2005
x = 2005 : 5
x = 401
b, \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\frac{7}{x}=\frac{29}{45}-\frac{8}{45}\)
\(\frac{7}{x}=\frac{7}{15}\)
=> x = 15
c, ghi lại đề
d, ghi lại đề
Bài 2:
\(\frac{1}{n}-\frac{1}{n+a}=\frac{n+a}{n\left(n+a\right)}-\frac{n}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)
a, \(\frac{\left(2^3.5.7\right)\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)\(=\frac{2^3.5.7.5^2.7^3}{2^2.5^2.7^4}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=10\)
b, \(\frac{4}{77}+\frac{4}{165}+\frac{4}{285}\)
\(=\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}\)
\(=\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}\)
\(=\frac{1}{7}-\frac{1}{19}\)
\(=\frac{19}{133}-\frac{7}{133}=\frac{12}{133}\)
Bài 2:
\(a,\left(x+\frac{2}{3}\right).\frac{-3}{5}+\frac{4}{7}=1\frac{4}{7}.x\)
\(\Rightarrow\frac{-3}{5}x+\frac{-2}{5}+\frac{4}{7}=\frac{11}{7}.y\)
\(\Rightarrow\frac{-3}{5}x+\frac{6}{35}=\frac{11}{7}.y\)
Từ đây làm nốt
b, \(\left|5x-2\right|\le0\)
\(\Rightarrow\left|5x\right|\le2\)( x \(\ge0\))
Mà không có số x nào nhân với 5 bé hơn hoặc bằng 2
\(\Rightarrow\)x không có giá trị thỏa mãn
c đề bài sai, chỉ tìm x chứ làm gì có y
d, \(\left(x-3\right).\left(2y+1\right)=7\)
TH1:
x - 3 = 1
x = 1 + 3
x = 4
2y + 1 = 7
2y = 7 - 1 = 6
y = 6 : 2 = 3
TH2:
x - 3 = 7
x = 7 + 3 = 10
2y + 1 = 1
2y = 1 - 1 = 0
y = 0 : 2 = 0
TH3:
x - 3 = -1
x = -1 + 3
x = 2
2y+ 1 = -7
2y = -7 - 1 = -8
y = (-8) : 2 = -4
TH4:
x - 3 = -7
x = -7 + 3
x = -4
2y + 1 = -1
2y = (-1) - 1
2y = -2
y = (-2) : 2 = -1
Vậy ......
\(a,\)\(-\frac{3}{5}\cdot x=\frac{1}{4}+0,75\)
\(-\frac{3}{5}\cdot x=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)
\(x=1\div\left(-\frac{3}{5}\right)\)
\(x=-\frac{5}{3}\)
\(b,\)\(\left(\frac{1}{7}-\frac{1}{3}\right)\cdot x=\frac{28}{5}\times\left(\frac{1}{4}-\frac{1}{7}\right)\)
\(\left(\frac{3}{21}-\frac{7}{21}\right)\cdot x=\frac{28}{5}\cdot\left(\frac{7}{28}-\frac{4}{28}\right)\)
\(-\frac{4}{21}\cdot x=\frac{28}{5}\cdot\frac{3}{28}\)
\(-\frac{4}{21}\cdot x=\frac{3}{5}\)
\(x=\frac{3}{5}\div\left(-\frac{4}{21}\right)\)
\(x=-\frac{63}{20}\)
\(c,\)\(\frac{5}{7}\cdot x=\frac{9}{8}-0,125\)
\(\frac{5}{7}\cdot x=\frac{9}{8}-\frac{1}{8}\)
\(\frac{5}{7}\cdot x=1\)
\(x=1\div\frac{5}{7}\)
\(x=\frac{7}{5}\)
\(d,\)\(\left(\frac{2}{11}+\frac{1}{3}\right)\cdot x=\left(\frac{1}{7}-\frac{1}{8}\right)\cdot36\)
\(\left(\frac{6}{33}+\frac{11}{33}\right)\cdot x=\left(\frac{8}{56}-\frac{7}{56}\right)\cdot36\)
\(\frac{17}{33}\cdot x=\frac{1}{56}\cdot36\)
\(\frac{17}{33}\cdot x=\frac{9}{14}\)
\(x=\frac{9}{14}\div\frac{17}{33}\)
\(x=\frac{9}{14}\cdot\frac{33}{17}=\frac{297}{238}\)
hình như trong bài tìm x dấu phải là dấu => chứ nhỉ.
Bài 1: Tìm x
a) Ta có: \(-\frac{3}{4}-\left(x+\frac{1}{2}\right)=1\frac{2}{3}\)
\(\Leftrightarrow\frac{-3}{4}-x-\frac{1}{2}=\frac{5}{3}\)
\(\Leftrightarrow-x-\frac{5}{4}=\frac{5}{3}\)
\(\Leftrightarrow-x=\frac{5}{3}+\frac{5}{4}=\frac{35}{12}\)
hay \(x=-\frac{35}{12}\)
Vậy: \(x=-\frac{35}{12}\)
b) Ta có: \(3.15:0.4=2.1x:1.68\)
\(\Leftrightarrow\frac{63}{20}:\frac{2}{5}=\frac{21}{10}x:\frac{168}{100}\)
\(\Leftrightarrow x\cdot\frac{21}{10}:\frac{168}{100}=\frac{63}{8}\)
\(\Leftrightarrow x\cdot\frac{21}{10}=\frac{63}{8}\cdot\frac{168}{100}=\frac{1323}{100}\)
\(\Leftrightarrow x=\frac{1323}{100}:\frac{21}{10}=\frac{63}{10}\)
Vậy: \(x=\frac{63}{10}\)