Bài 3: Tìm x, biết a) \(4\frac{1}{3}:\frac{x}{3}=6:0,3\) b) \(\left|x+\frac{97}{306}\right|+5=-1\)
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làm 1 bài mẫu nha
a) \(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{25}{6}\)
\(\Rightarrow\frac{1}{3}x=\frac{25}{9}\)
\(\Rightarrow x=\frac{25}{9}:\frac{1}{3}\)
\(\Rightarrow x=\frac{25}{3}\)
các bài sau dễ lắm
a) \(\left(\frac{1}{3}x\right):\frac{2}{3}=1\frac{2}{3}:\frac{2}{5}\)
\(\Rightarrow\frac{1}{3}x:\frac{2}{3}=\frac{25}{6}\)
\(\Rightarrow\frac{1}{3}x=\frac{5}{3}\)
\(\Rightarrow x=\frac{5}{3}:\frac{1}{3}\)
\(\Rightarrow x=5\)
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)
\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)
\(\frac{2}{5}-x=-3\)
\(x=\frac{2}{5}-\left(-3\right)\)
\(x=\frac{2}{5}+3\)
\(x=\frac{3}{5}-\frac{15}{5}\)
\(x=-\frac{12}{5}\)
Vay \(x=-\frac{12}{5}\)
b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)
\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)
\(-3+\frac{3}{x}=\frac{-25}{12}\)
\(\frac{3}{x}=\frac{-25}{12}+3\)
\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)
\(\frac{3}{x}=\frac{5}{6}\)
\(\frac{18}{6x}=\frac{5x}{6x}\)
Đèn dây , bạn tự làm tiếp nhé , de rồi chứ
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) \(\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}.\frac{1}{6}.\left(x-1,010\right)=\frac{1}{360}-\frac{1}{720}\)
\(\Rightarrow\frac{1}{2.3.4.5.6}.\left(x-1,01\right)=\frac{1}{720}\)
\(\Rightarrow\frac{1}{720}.\left(x-1,01\right)=\frac{1}{720}\)
\(\Rightarrow x-1,01=\frac{1}{720}:\frac{1}{720}\)
\(\Rightarrow x-1,01=1\)
\(\Rightarrow x=1+1,01\)
\(\Rightarrow x=2,01\)
\(\frac{1}{3}x:\frac{2}{3}=1\frac{3}{4}:\frac{2}{5}\)
\(\Rightarrow\frac{1}{3}x:\frac{2}{3}=\frac{35}{8}\)
\(\Rightarrow\frac{1}{3}x=\frac{35}{12}\)
\(\Rightarrow x=\frac{35}{4}\)
a) \(\left(\frac{1}{3}\cdot x\right):\frac{2}{3}=1\frac{3}{4}:\frac{2}{5}\)
\(=\left(\frac{1}{3}\cdot x\right):\frac{2}{3}=\frac{35}{8}\)
\(\Rightarrow\frac{1}{3}\cdot x=\frac{35}{8}\cdot\frac{2}{3}\)
\(\Rightarrow x=\frac{35}{12}:\frac{1}{3}=\frac{35}{4}\)
b) \(4,5:0,3=2,5:\left(0,1\cdot x\right)\)
\(=15=2,5:\left(0,1\cdot x\right)\)
\(\Rightarrow0,1\cdot x=2,25:15\)
\(\Rightarrow x=\frac{3}{20}:0,1=\frac{3}{2}=1,5\)
c) \(8:\left(\frac{1}{4}\cdot x\right)=2:0,02\)
\(8:\left(\frac{1}{4}\cdot x\right)=100\)
\(\Rightarrow\frac{1}{4}\cdot x=8:100\)
\(\Rightarrow x=\frac{2}{25}:\frac{1}{4}=\frac{8}{25}=0,32\)
d) \(3:2\frac{1}{4}=\frac{3}{4}:\left(6\cdot x\right)\)
\(=\frac{4}{3}=\frac{3}{4}:\left(6:x\right)\)
\(\Rightarrow6\cdot x=\frac{3}{4}:\frac{4}{3}\)
\(\Rightarrow x=\frac{9}{16}:6=\frac{3}{32}=0,09375\)
c, Trừ hai vế cho 6
Vế trái thì lấy từng số hạng trừ 1 là được
a) Ta có: \(4\frac{1}{3}:\frac{x}{3}=6:0,3\)
\(\Leftrightarrow\frac{13}{3}.\frac{3}{x}=6.\frac{10}{3}\)
\(\Leftrightarrow\frac{13}{x}=20\)
\(\Leftrightarrow x=\frac{13}{20}\)
Vậy \(x\in\left\{\frac{13}{20}\right\}\)
b) Ta có: \(\left|x+\frac{97}{306}\right|+5=-1\)
\(\Leftrightarrow\left|x+\frac{97}{306}\right|=-6\)
Vì \(\left|x+\frac{97}{306}\right|\ge0\forall x,\)\(-6< 0\)
mà \(\left|x+\frac{97}{306}\right|=-6\)
\(\Rightarrow\)Đa thức \(\left|x+\frac{97}{306}\right|=-6\)không có giá trị
Vậy \(x\in\varnothing\)