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\(a,\frac{2}{3}\cdot x-\frac{4}{7}=\frac{1}{8}\)
\(\Leftrightarrow\frac{2}{3}\cdot x=\frac{1}{8}+\frac{4}{7}\)
\(\Leftrightarrow\frac{2}{3}\cdot x=\frac{7}{56}+\frac{32}{56}\)
\(\Leftrightarrow\frac{2}{3}\cdot x=\frac{39}{56}\)
\(\Leftrightarrow x=\frac{39}{56}:\frac{2}{3}=\frac{39}{56}\cdot\frac{3}{2}=\frac{39\cdot3}{56\cdot2}=\frac{117}{112}\)
\(b,\frac{2}{7}-\frac{8}{9}\cdot x=\frac{2}{3}\)
\(\Leftrightarrow\frac{8}{9}\cdot x=\frac{2}{7}-\frac{2}{3}\)
\(\Leftrightarrow\frac{8}{9}\cdot x=\frac{6}{21}-\frac{14}{21}\)
\(\Leftrightarrow\frac{8}{9}\cdot x=\frac{-8}{21}\)
\(\Leftrightarrow x=\frac{-8}{21}:\frac{8}{9}=\frac{-8}{21}\cdot\frac{9}{8}=\frac{-8\cdot9}{21\cdot8}=\frac{-1\cdot3}{7\cdot1}=\frac{-3}{7}\)
Làm nốt hai bài cuối đi nhé
Study well >_<
Mk k chép lại đề bài nha
a)\(\frac{2}{3}.x=\frac{1}{8}+\frac{4}{7}\)
\(\frac{2}{3}.x=\frac{7}{56}+\frac{32}{56}\)
\(\frac{2}{3}.x=\frac{39}{56}\)
\(x=\frac{39}{56}:\frac{2}{3}\)
\(x=\frac{39}{56}.\frac{3}{2}\)
\(x=\frac{117}{112}\)
Mk sợ sai lém!!!
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)
\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)
Vậy \(A>\frac{1}{10}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)
\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)
\(VayA>\frac{1}{100}=B\)
\(a,\frac{x-1}{12}=\frac{75}{x-1}\)
\(\Rightarrow\left(x-1\right)\left(x-1\right)=12\cdot75\)
\(\Rightarrow\left(x-1\right)^2=900\)
\(\Rightarrow x-1=\pm30\)
\(\Rightarrow\orbr{\begin{cases}x=31\\x=-29\end{cases}}\)
\(b,\left|12-|4x+1|\right|=5\)
\(\Rightarrow\orbr{\begin{cases}12-\left|4x+1\right|=5\\12-\left|4x+1\right|=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\left|4x+1\right|=7\\\left|4x+1\right|=17\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4x+1=7\text{ hoặc}4x+1=-7\\4x+1=17\text{ hoặc}4x+1=-17\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}or\text{ }x=-2\\x=4or\text{ }x=-\frac{9}{2}\end{cases}}\)
x - 1/12 = 75/x - 1
<=> (x - 1).(x - 1) = 12.75
<=> x2 - 2x + 1 = 900
<=> x2 - 2x + 1 - 900 = 0
<=> x2 - 2x + 1 - 899 = 0
<=> (x + 29)(x - 31) = 0
x + 29 = 0 hoặc x - 31 = 0
x = 0 - 29 x = 0 + 31
x = -29 x = 31
=> x = -29 hoặc x = 31
Bài 1:
a) b) c) sẽ có bạn giải cho em thôi vì nó dễ tính tay cũng đc
d) \(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{23.26}\)
\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\frac{6}{13}\)
\(=\frac{8}{13}\)
Bài 2:
a) b) c)
d)\(|\frac{5}{8}x+\frac{6}{7}|-\frac{4}{7}=\frac{10}{7}\)
\(\Leftrightarrow|\frac{5}{8}x+\frac{6}{7}|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{6}{7}=2\\\frac{5}{8}x+\frac{6}{7}=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{8}{7}\\\frac{5}{8}x=\frac{-20}{7}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{64}{35}\\x=\frac{-32}{7}\end{cases}}}\)
Vậy \(x\in\left\{\frac{64}{35};\frac{-32}{7}\right\}\)
Bài 1 :
a) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
Tìm x, biết:
a) \(5,2x+7\frac{2}{5}=6\frac{3}{4}\)
b) \(2,4:\left(\frac{-1}{2}-x\right)=1\frac{3}{5}\)
a) \(5,2x+7\frac{2}{5}=6\frac{3}{4}\) b) \(2,4:\left(\frac{-1}{2}-x\right)=1\frac{3}{5}\)
\(5,2x+\frac{37}{5}=\frac{27}{4}\) \(2,4:\left(\frac{-1}{2}-x\right)=\frac{8}{5}\)
\(5,2x\) = \(\frac{27}{4}-\frac{37}{5}\) \(\frac{-1}{2}-x\)= \(2,4:\frac{8}{5}\)
\(5,2x\) = \(\frac{-13}{20}\) \(\frac{-1}{2}-x\)= \(\frac{3}{2}\)
x = \(\frac{-13}{20}:5,2\) x = \(\frac{-1}{2}-\frac{3}{2}\)
x = -0,125 x = \(\frac{-4}{2}\)= -2
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)
\(\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}:\frac{11}{45}\)
\(x=\frac{23}{11}\)
\(\frac{3}{4}.x-1\frac{1}{2}+x=2,4\)
\(\frac{3}{4}.x-\frac{3}{2}+x=2,4\)
\(\frac{3}{4}.x-1.\frac{3}{2}+x.1=2,4\)
\(x.\left(\frac{3}{4}-\frac{3}{2}+1\right)=2,4\)
\(x.\frac{1}{4}=\frac{24}{10}\)
\(x=\frac{24}{10}:\frac{7}{4}\)
\(x.=\frac{24}{10}.\frac{4}{7}\)
\(x=\frac{48}{35}\)
\(\frac{3}{4}x-1\frac{1}{2}+x=2,4\)
<=>\(\frac{3}{4}x-\frac{3}{2}+x-2,4=0\)
,<=>\(\frac{7}{4}x-3,9=0\)
=>\(\frac{7}{4}x=3,9\)
=>\(x=\frac{78}{35}\)