Bài 1: Tìm x, biết a) \(-\frac{3}{4}-\left(x+\frac{1}{2}\right)=1\frac{2}{3}\) b) \(3,15:0,4=2,1x:1,68\)
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a) \(-\frac{3}{4}-\left(x+\frac{1}{2}\right)=1\frac{2}{3}=\frac{5}{3}\Leftrightarrow-\left(x+\frac{1}{2}\right)=\frac{5}{3}+\frac{3}{4}=\frac{29}{12}\)
\(\Leftrightarrow x+\frac{1}{2}=-\frac{29}{12}\Leftrightarrow x=-\frac{29}{12}-\frac{1}{2}=-\frac{29}{12}-\frac{6}{12}=-\frac{35}{12}\)
b) \(3,15:0,4=2,1x:1,68\Leftrightarrow7,875=2,1x\div1,68\Leftrightarrow2,1x=7,875\times1,68=13,23\)
\(\Leftrightarrow x=13,23\div2,1\Leftrightarrow x=6,3\)
\(x-3\frac{1}{2}=\frac{3}{5}\)
\(x=\frac{3}{5}+3\frac{1}{2}\)
\(x=4\frac{1}{10}\)
Vậy x = \(4\frac{1}{10}\)
Nhớ k cho mình nhé! Thank you!!!
\(3,15\div0,4=2,1.x\div1,68\)
\(7,875=2,1.x\div1,68\)
\(2,1.x=7,875\times1,68\)
\(2,1.x=13,23\)
\(x=13,23\div2,1\)
\(x=6,3\)
Vậy x = 6,3
Nhớ k cho mình nhé! Thank you!!!
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
\(\left(2,1x+\frac{5}{3}\right)^6=\left(-3,2x+\frac{1}{2}\right)^6\Rightarrow2,1x+\frac{5}{3}=-3,2x+\frac{1}{2}\)
\(\Rightarrow2,1x+3,2x=\frac{1}{2}-\frac{5}{3}\)
\(5,3x=\frac{-7}{6}\)
\(x=\frac{-7}{6}:\frac{53}{10}=\frac{-35}{159}\)
Vậy \(x=\frac{-35}{159}\)
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) \(\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}.\frac{1}{6}.\left(x-1,010\right)=\frac{1}{360}-\frac{1}{720}\)
\(\Rightarrow\frac{1}{2.3.4.5.6}.\left(x-1,01\right)=\frac{1}{720}\)
\(\Rightarrow\frac{1}{720}.\left(x-1,01\right)=\frac{1}{720}\)
\(\Rightarrow x-1,01=\frac{1}{720}:\frac{1}{720}\)
\(\Rightarrow x-1,01=1\)
\(\Rightarrow x=1+1,01\)
\(\Rightarrow x=2,01\)
Bài 2 :
Ta có : \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\in R\)
\(\Rightarrow A=\frac{3}{4}+\left(x-\frac{1}{2}\right)^2\ge\frac{3}{4}\forall x\in R\)
Vậy Amin = \(\frac{3}{4}\) dấu "=" chỉ sảy ra khi x = \(\frac{1}{2}\)
a) Vì (2x - 5)2000 và (3y + 4)2002 đều có số mũ là chẵn => (2x - 5)2000 \(\ge\) 0; (3y + 4)2002 \(\ge\) 0
Mà tổng trên lại \(\le\) 0
=> (2x - 5)2000 = (3y + 4)2002 = 0
=> 2x - 5 = 3y + 4 = 0
=> x = 2,5; y = \(\frac{-4}{3}\)
b) x = 18 - 0,8 : \(\frac{1,5}{\frac{3}{2}.\frac{4}{10}.\frac{50}{2}}\)+ \(\frac{1}{4}\)+ \(\frac{1+0,5.4}{6-\frac{46}{23}}\)
= 18 - \(\frac{8}{10}:\frac{1,5}{15}+\frac{1}{4}+\frac{3}{4}\)
= \(18-8+1=11\)
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: \(x=\frac{63}{10}\)