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a) \(\left|2x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\orbr{\begin{cases}2x+\frac{3}{4}=\frac{1}{2}\\2x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{1}{2}-\frac{3}{4}\\2x=\frac{-1}{2}-\frac{3}{4}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{-1}{4}\\2x=\frac{-5}{4}\end{cases}}\) => \(\orbr{\begin{cases}x=\frac{-1}{8}\\x=\frac{-5}{8}\end{cases}}\)
Vậy \(x=\left\{\frac{-1}{8},\frac{-5}{8}\right\}\)
b) \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)= \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{\frac{9}{4}}\)
=> \(3x.\frac{9}{4}=2,7.\frac{1}{4}\)=> \(\frac{27x}{4}=\frac{27}{40}\)
\(27x.40=27.4\)
\(1080.x=108\)
\(x=\frac{1}{10}\)
Vậy \(x=\frac{1}{10}\)
c) \(\left|x-1\right|+4=6\)
\(\left|x-1\right|=6-4\)
\(\left|x-1\right|=2\)
\(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Vậy \(x=\left[3,-1\right]\)
d) \(\frac{x}{3}=\frac{y}{5}=>\frac{y}{5}=\frac{x}{3}=>\frac{y-x}{5-3}=\frac{24}{2}=12\)
e) \(\left(x^2-3\right)^2=16\)
\(\left(x^2-3\right)^2=4^2\)\(=>x^2-3=4\)
\(x^2=7=>x=\sqrt{7}\)
Vậy \(x=\sqrt{7}\)
f) \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{29}{60}-\frac{3}{4}\)
\(\frac{2}{5}x=-\frac{4}{15}\)
\(x=-\frac{4}{15}:\frac{2}{5}=-\frac{4}{15}.\frac{5}{2}=-\frac{2}{3}\)
Vậy \(x=-\frac{2}{3}\)
g) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)
\(\left(-\frac{1}{27}\right).x=\frac{1}{81}\)
\(x=\left(-\frac{1}{27}\right):\frac{1}{81}=\left(-\frac{1}{27}\right).81=-3\)
Vậy \(x=-3\)
k)\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{3}{4}-\frac{29}{60}\)
\(\frac{2}{5}x=\frac{4}{15}\)
\(x=\frac{2}{5}-\frac{4}{15}=>x=\frac{2}{15}\)
Vậy \(x=\frac{2}{15}\)
I) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
\(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}\)
\(\frac{3}{5}x=\frac{5}{14}\)
\(x=\frac{5}{14}:\frac{3}{5}=\frac{5}{14}.\frac{5}{3}=\frac{25}{42}\)
Vậy \(x=\frac{25}{42}\)
Dài đấy :))
a) \(\left|x-1\right|-\left(-2\right)^3=9\cdot\left(-1\right)^{100}\)
\(\Leftrightarrow\left|x-1\right|-\left(-8\right)=9\cdot1\)
\(\Leftrightarrow\left|x-1\right|+8=9\)
\(\Leftrightarrow\left|x-1\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
b) \(\frac{x-2}{-4}=\frac{-9}{x-2}\)( ĐKXĐ : \(x\ne2\))
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)=-4\cdot\left(-9\right)\)
\(\Leftrightarrow\left(x-2\right)^2=36\)
\(\Leftrightarrow\left(x-2\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}\left(tmđk\right)\)
c) \(\frac{x-5}{3}=\frac{-12}{5-x}\)( ĐKXĐ : \(x\ne5\))
\(\Leftrightarrow\frac{x-5}{3}=\frac{-12}{-\left(x-5\right)}\)
\(\Leftrightarrow\frac{x-5}{3}=\frac{12}{x-5}\)
\(\Leftrightarrow\left(x-5\right)\left(x-5\right)=3\cdot12\)
\(\Leftrightarrow\left(x-5\right)^2=36\)
\(\Leftrightarrow\left(x-5\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=6\\x-5=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=11\\x=-1\end{cases}}\left(tmđk\right)\)
d) \(8x-\left|4x+\frac{3}{4}\right|=x+2\)
\(\Leftrightarrow8x-x-2=\left|4x+\frac{3}{4}\right|\)
\(\Leftrightarrow7x-2=\left|4x+\frac{3}{4}\right|\)(*)
\(\left|4x+\frac{3}{4}\right|\ge0\Leftrightarrow4x+\frac{3}{4}\ge0\Leftrightarrow x\ge-\frac{3}{16}\)
Vậy ta xét hai trường hợp sau :
1. \(x\ge-\frac{3}{16}\)
(*) <=>\(7x-2=4x+\frac{3}{4}\)
\(\Leftrightarrow7x-4x=\frac{3}{4}+2\)
\(\Leftrightarrow3x=\frac{11}{4}\)
\(\Leftrightarrow x=\frac{11}{12}\)(tmđk)
2. \(x< -\frac{3}{16}\)
(*) <=> \(7x-2=-\left(4x+\frac{3}{4}\right)\)
\(\Leftrightarrow7x-2=-4x-\frac{3}{4}\)
\(\Leftrightarrow7x+4x=-\frac{3}{4}+2\)
\(\Leftrightarrow11x=\frac{5}{4}\)
\(\Leftrightarrow x=\frac{5}{44}\left(ktmđk\right)\)
Vậy x = 11/12
e) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2020}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{4040}\)
\(\Leftrightarrow x+1=4040\)
\(\Leftrightarrow x=4039\)
Bài 1:
Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Leftrightarrow2x=\frac{1440}{144}=10\)
\(\Rightarrow x=5\)
Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
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
a) \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{9}{45}+\frac{4}{5}\right)\)
\(=\frac{47}{47}+\left(\frac{1}{5}+\frac{4}{5}\right)\)
\(=1+1=2\)
b) \(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}\)
\(=12.\frac{4}{9}+\frac{4}{3}\)
\(=\frac{16}{3}+\frac{4}{3}\)
\(=\frac{20}{3}\)
c) \(12,5.\left(-\frac{5}{7}\right)+15.\left(-\frac{5}{7}\right)\)
\(=\left(-\frac{5}{7}\right).\left(12,5+15\right)\)
\(=\left(-\frac{5}{7}\right).27,5\)
\(=\left(-\frac{5}{7}\right).\frac{55}{2}\)
\(=-\frac{275}{14}\)
d) \(\frac{4}{5}.\left(\frac{7}{2}+\frac{1}{4}\right)^2\)
\(=\frac{4}{5}.\left(\frac{14}{4}+\frac{1}{4}\right)^2\)
\(=\frac{4}{5}.\left(\frac{15}{4}\right)^2\)
\(=\frac{4}{5}.\frac{225}{16}\)
\(=\frac{45}{4}\)
a)\(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
=\(\frac{21}{47}+\frac{1}{5}+\frac{26}{47}+\frac{4}{5}\)
=\(\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{1}{5}+\frac{4}{5}\right)\)
=\(\frac{47}{47}+\frac{5}{5}=1+1=2\)
b)\(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}\)
=\(12.\frac{4}{9}+\frac{4}{3}\)
=\(\frac{12}{1}.\frac{4}{9}+\frac{4}{3}=\frac{48}{9}+\frac{4}{3}\)
=\(\frac{16}{3}+\frac{4}{3}=\frac{20}{3}\)
c)\(12,5.\left(-\frac{5}{7}\right)+1,5.\left(-\frac{5}{7}\right)\)
=\(\left(-\frac{5}{7}\right).\left(12,5+1,5\right)\)
=\(\left(-\frac{5}{7}\right).14=\left(-\frac{5}{7}\right).\frac{14}{1}=-10\)
d)\(\frac{4}{5}.\left(\frac{7}{2}+\frac{1}{4}\right)^2\)
=\(\frac{4}{5}.\left(\frac{14}{4}+\frac{1}{4}\right)^2\)
=\(\frac{4}{5}.\left(\frac{15}{4}\right)^2\)
=\(\frac{4}{5}.\frac{225}{16}\)
=\(\frac{900}{80}=\frac{45}{4}\)
Nhớ tick cho mình nha!
a) \(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\frac{1}{3}:2x=\frac{-21}{4}\)
\(2x=\frac{-4}{63}\)
\(x=\frac{2}{63}\)
b) \(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}\)
Vậy.........
a,\(\frac{x}{3}=\frac{2}{5}\)
\(\Rightarrow x=\frac{2.3}{5}=\frac{6}{5}\)
b,\(\frac{-9}{x}=\frac{\frac{\left(-x\right)}{4}}{49}\)
\(\Rightarrow x=-9.49.\frac{-4}{x}\)
\(\Rightarrow x=\frac{1764}{x}\)
\(\Rightarrow x^2=1764=42^2\)
\(\Rightarrow x=\pm2\)
c.\(3^{x+2}-3^x=24\)
\(\Rightarrow3^x.3^2-3^x=24\)
\(\Rightarrow3^x.\left(9-1\right)=24\)
\(\Rightarrow3^x.8=24\)
\(\Rightarrow3^x=3\)
\(\Rightarrow x=1\)