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dài :vv
a) \(\left|2x-5\right|=4\Leftrightarrow\hept{\begin{cases}2x-5=4\\2x-5=-4\end{cases}\Leftrightarrow\hept{\begin{cases}2x=9\\2x=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{9}{2}\\x=\frac{1}{2}\end{cases}}}\)
b) \(\frac{1}{3}-\left|\frac{5}{4}-2x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left|\frac{5}{4}-2x\right|=\frac{1}{12}\Leftrightarrow\hept{\begin{cases}\frac{5}{4}-2x=\frac{1}{12}\\\frac{5}{4}-2x=-\frac{1}{12}\end{cases}\Leftrightarrow\hept{\begin{cases}2x=\frac{7}{6}\\2x=\frac{4}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{7}{12}\\x=\frac{2}{3}\end{cases}}}\)
Bài 1 :
a) \(|2x-5|=4\)
\(\Rightarrow\orbr{\begin{cases}2x-5=4\\2x-5=-4\end{cases}\Rightarrow}\orbr{\begin{cases}2x=9\\2x=1\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x=\frac{1}{2}\end{cases}}}\)
b) \(\frac{1}{3}-\left|\frac{5}{4}-2x\right|=\frac{1}{4}\)
\(\Rightarrow\left|\frac{5}{4}-2x\right|=\frac{1}{12}\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{12}\\\frac{5}{4}-2x=-\frac{1}{12}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=\frac{7}{6}\\2x=\frac{4}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=\frac{2}{3}\end{cases}}}\)
c) \(\left|\frac{-2}{3}\right|+\left|x-\frac{1}{3}\right|=\left|-1\right|-\left|\frac{-1}{3}\right|\)
\(\Rightarrow\frac{2}{3}+\left|x-\frac{1}{3}\right|=1-\frac{1}{3}\)
\(\Rightarrow\frac{2}{3}+\left|x-\frac{1}{3}\right|=\frac{2}{3}\)
\(\Rightarrow\left|x-\frac{1}{3}\right|=0\Rightarrow x-\frac{1}{3}=0\Rightarrow x=\frac{1}{3}\)
d) \(\left|-\frac{1}{2}\right|-\left|x+\frac{1}{4}\right|=\left|-\frac{3}{4}\right|\)
\(\Rightarrow\frac{1}{2}-\left|x+\frac{1}{4}\right|=\frac{3}{4}\)
\(\Rightarrow\left|x+\frac{1}{4}\right|=-\frac{1}{4}\)
Vì \(\left|x\right|\ge0\Rightarrow\)ko có gtri nào của x thỏa mãn đề bài
Bài 2 :
a) \(\left|x-1\right|=3x+2\)
\(\Rightarrow\orbr{\begin{cases}x-1=3x+2\\x-1=-3x-2\end{cases}\Rightarrow\orbr{\begin{cases}x-3x=2+1\\x+3x=-2+1\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}-2x=3\\4x=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{-1}{4}\end{cases}}\)
b|) \(\left|9+x\right|=2x\Rightarrow\orbr{\begin{cases}9+x=2x\\9+x=-2x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-2x=-9\\x+2x=-9\end{cases}\Rightarrow\orbr{\begin{cases}-x=-9\\3x=-9\end{cases}\Rightarrow}\orbr{\begin{cases}x=9\\x=-3\end{cases}}}\)
c) \(\left|x+6\right|-9=2x\Rightarrow\left|x+6\right|=2x+9\)
\(\Rightarrow\orbr{\begin{cases}x+6=2x+9\\x+6=-2x-9\end{cases}\Rightarrow}\orbr{\begin{cases}x-2x=9-6\\x+2x=-9-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-x=3\\3x=-15\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}}\)
Cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt ^^
a) \(\left(6\frac{2}{7}x+\frac{3}{7}\right)\cdot\frac{11}{5}-\frac{3}{7}=-2\)
=> \(\left(\frac{44}{7}x+\frac{3}{7}\right)\cdot\frac{11}{5}=-\frac{11}{7}\)
=> \(\frac{44}{7}x+\frac{3}{7}=-\frac{5}{7}\)
=> \(\frac{44}{7}x=-\frac{8}{7}\)
=> \(\frac{44x}{7}=-\frac{8}{7}\)
=> 44x = -8 => 11x = -2 => \(x=-\frac{2}{11}\)
b) \(3\frac{1}{4}x+\left(-\frac{7}{6}\right)-1\frac{2}{3}=\frac{5}{12}\)
=> \(\frac{13}{4}x-\frac{7}{6}-1\frac{2}{3}=\frac{5}{12}\)
=> \(\frac{13}{4}x-\frac{7}{6}=\frac{25}{12}\)
=> \(\frac{13}{4}x=\frac{13}{4}\)
=> x = 1
c) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
=> \(\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{1}{3}\end{cases}}\)
d) \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
=> \(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)
=> \(\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)
e) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=\frac{-24}{27}\)
=> \(\left(3x-\frac{7}{9}\right)^3=-1\frac{5}{27}-\left(-\frac{24}{27}\right)=-\frac{32}{27}+\frac{24}{27}=-\frac{8}{27}\)
=> \(\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
=> \(3x-\frac{7}{9}=-\frac{2}{3}\)
=> \(x=\frac{-\frac{2}{3}+\frac{7}{9}}{3}=\frac{1}{27}\)
g) \(\frac{x}{1\cdot2}+\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+...+\frac{x}{99\cdot100}=\frac{99}{100}\)
=> \(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{99}-\frac{x}{100}=\frac{99}{100}\)
=> \(\frac{x}{1}-\frac{x}{100}=\frac{99}{100}\)
=> \(\frac{100x-x}{100}=\frac{99}{100}\)
=> \(\frac{99x}{100}=\frac{99}{100}\)
=> x = 1
h) \(\frac{x}{3}+\frac{x}{6}+\frac{x}{10}+\frac{x}{15}=3x-1\)
=> \(\frac{2x}{6}+\frac{2x}{12}+\frac{2x}{20}+\frac{2x}{30}=3x-1\)
=> \(\frac{2x}{2\cdot3}+\frac{2x}{3\cdot4}+\frac{2x}{4\cdot5}+\frac{2x}{5\cdot6}=3x-1\)
=> \(2\left(\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+\frac{x}{4\cdot5}+\frac{x}{5\cdot6}\right)=3x-1\)
=> \(2\left(\frac{x}{2}-\frac{x}{6}\right)=3x-1\)
=> \(2\left(\frac{3x}{6}-\frac{x}{6}\right)=3x-1\)
=> \(2\cdot\frac{2x}{6}=3x-1\)
=> \(\frac{x}{3}=\frac{3x-1}{2}\)
=> 2x = 3(3x - 1)
=> 2x - 9x + 3 = 0
=> -7x = -3
=> x = 3/7
Bài đây tính nhanh nhé ミ★ʟuғғʏ☆мũ☆ʀơм★彡 chứ không phải quy đồng lên đâu :)
a) \(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}\)
\(A=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}=35-5\frac{7}{32}=35-\frac{167}{32}=\frac{953}{32}\)
b) \(B=\frac{-3}{7}\cdot\frac{5}{9}+\frac{4}{9}:\frac{-7}{3}+2\frac{3}{7}\)
\(B=\frac{-3}{7}\cdot\frac{5}{9}+\frac{4}{9}\cdot\frac{-3}{7}+2\frac{3}{7}\)
\(B=\frac{-3}{7}\left(\frac{5}{9}+\frac{4}{9}\right)+2\frac{3}{7}\)
\(B=\frac{-3}{7}+\frac{17}{7}=\frac{14}{7}=2\)
c) \(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right)\cdot\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]\cdot\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{2}{8}\right):\frac{7}{12}\right]\cdot\frac{4}{5}\)
\(C=6\frac{3}{8}\cdot\frac{4}{5}=\frac{51}{8}\cdot\frac{4}{5}=\frac{51}{2}\cdot\frac{1}{5}=\frac{51}{10}\)
d) \(D=\frac{54\cdot107-53}{53\cdot107+54}=\frac{\left(53+1\right)\cdot107-53}{53\cdot107+54}=\frac{53\cdot107+107-53}{53\cdot107+54}=\frac{53\cdot107+54}{53\cdot107+54}=1\)
Ta có: \(x^2-\frac23=-\frac29\)
=>\(x^2=-\frac29+\frac23=-\frac29+\frac69=\frac49\)
=>\(\left[\begin{array}{l}x=\frac23\\ x=-\frac23\end{array}\right.\)
x mũ 2 = -2/9+2/3
x mũ 2 = -2/9 + 6/9
x mũ 2 = 4/9
x mũ 2 = 2 mũ 2/3 mũ 2
x= 2/3