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Bài giải:
a) Ta có: \(-8⋮x\) và \(12⋮x\Rightarrow x\inƯC\left(-8;12\right)\)
\(Ư\left(-8\right)=\left\{-1;-2;-4;-8;1;2;4;8\right\}\)
\(Ư\left(12\right)=\left\{-1;-2;-3;-4;-6;-12;1;2;3;4;6;12\right\}\)
\(\RightarrowƯC\left(-8;12\right)=\left\{-1;-2;-4;1;2;4\right\}\)
Vậy: \(x\in\left\{-1;-2;-4;1;2;4\right\}\)
b)
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 ^^
B=(1-\(\frac{1}{2}\))x(1-\(\frac{1}{3}\))x(1-\(\frac{1}{4}\))x...x(1-\(\frac{1}{20}\))
B=\(\frac{1}{2}\)X\(\frac{2}{3}\)X\(\frac{3}{4}\)X...X\(\frac{19}{20}\)
B=\(\frac{1.2.3.4.4.5.7.8.9.10.11.12.13.14.15.16.17.18.19}{2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.17.18.19.20}\)
B=20
Vậy B=20
Không biết kết quả đúng ko nhưng cách làm thì đúng.
B= \(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x...x\left(1-\frac{1}{20}\right)\)
=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{19}{20}\)(Rút gọn cả tử xuống mẫu )
= \(\frac{1.2.3...19}{2.3.4...20}\)
=\(\frac{1}{20}\)
Vậy B= \(\frac{1}{20}\)
a) 45 ⋮ x
Vì 45 ⋮ x nên x E Ư( 45 )
= { 1;3;5;9;15;45 }
mà x E Ư(45)
=> x E { 1;3;5;9;15;45 }
b) 24 ⋮ x ; 36 ⋮ x ; 160 ⋮ x và x lớn nhất
Vì 24 ⋮ x ; 36 ⋮ x ; 160 ⋮ x nên x E ƯC ( 24;36;160)
mà x lớn nhất
=> x E ƯCLN ( 24;36;160 )
Ta có
24 = 23 . 3
36 = 22.32
160 = 25 . 5
=> ƯCLN ( 24;36;160 ) = 22 = 4
Đặt biểu thức là A, ta có:
\(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{20}.\left(1+2+3+...+20\right)\)
\(A=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+...+\frac{1}{20}.20.21:2=\frac{2}{2}+\frac{3}{2}+...+\frac{21}{2}\)
\(A=\frac{2+3+4+...+21}{2}=\frac{230}{2}=115\)
\(2+\left(x.\frac{1}{4}\right)=\frac{3}{5}\)
=> \(\left(x.\frac{1}{4}\right)=\frac{3}{5}-2\)
=> \(x.\frac{1}{4}=-\frac{7}{5}\)
=> \(x=\left(-\frac{7}{5}\right):\frac{1}{4}\)
=> \(x=-\frac{28}{5}\)
Vậy \(x=-\frac{28}{5}.\)
\(\frac{5}{6}:\left(\frac{2}{3}.x\right)=\frac{20}{3}\)
=> \(\left(\frac{2}{3}.x\right)=\frac{5}{6}:\frac{20}{3}\)
=> \(\frac{2}{3}.x=\frac{1}{8}\)
=> \(x=\frac{1}{8}:\frac{2}{3}\)
=> \(x=\frac{3}{16}\)
Vậy \(x=\frac{3}{16}.\)
Chúc bạn học tốt!
2+(x.\(\frac{1}{4}\))=\(\frac{3}{5}\)
=>x.\(\frac{1}{4}\)=\(\frac{3}{5}\)-2
=>x.\(\frac{1}{4}\)=\(\frac{-7}{5}\)
=>x=\(\frac{-7}{5}\):\(\frac{1}{4}\)
=>x=\(\frac{-28}{5}\)
\(\frac{5}{6}\):(\(\frac{2}{3}\).x)=\(\frac{20}{3}\)
\(\frac{2}{3}\).x=\(\frac{5}{6}\):\(\frac{20}{3}\)
\(\frac{2}{3}\).x=\(\frac{1}{8}\)
x=\(\frac{1}{8}\):\(\frac{2}{3}\)
x=\(\frac{3}{16}\)
Không viết lại đề bài
\(\Leftrightarrow\frac{x-20}{9}+\frac{x-21}{10}+\frac{x-26}{15}+3=0\)
\(\Leftrightarrow\left(\frac{x-20}{9}+1\right)+\left(\frac{x-21}{10}+1\right)+\left(\frac{x-26}{15}+1\right)=0\)
\(\Leftrightarrow\left(\frac{x-20+9}{9}\right)+\left(\frac{x-21+10}{10}\right)+\left(\frac{x-26+15}{15}\right)=0\)
\(\Leftrightarrow\frac{x-11}{9}+\frac{x-11}{10}+\frac{x-11}{15}=0\)
\(\Leftrightarrow\left(x-11\right)\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{15}\right)=0\)
Vì \(\frac{1}{9}+\frac{1}{10}+\frac{1}{15}\ne0\)
\(\Rightarrow x-11=0\Leftrightarrow x=11\)
Ta có\(\frac{x-20}{9}+\frac{x-21}{10}+\frac{x-26}{15}=-3\)
=> \(\left(\frac{x-20}{9}+1\right)+\left(\frac{x-21}{10}+1\right)+\left(\frac{x-26}{15}+1\right)=-3+1+1+1\)
=> \(\frac{x-11}{9}+\frac{x-11}{10}+\frac{x-11}{15}=0\)
=> \(\left(x-11\right)\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{15}\right)=0\)
Vì \(\frac{1}{9}+\frac{1}{10}+\frac{1}{15}\ne0\)
=> x - 11 = 0
=> x = 11
Vậy x = 11
