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bài 3 :
Số học sinh trung bình là :
\(1200\times\dfrac{5}{8}=750\) ( hs)
Số học sinh khá là :
\(750\times\dfrac{2}{5}=300\) (hs)
Số học sinh giỏi là :
\(1200-750-300=150\left(hs\right)\)
b) So với cả trường chứ ?
3b ) Tỉ số của hs giỏi so với toàn trường :150: 1200 = 0,125
Tỉ số phần trăm của hs giỏi so vs toàn trường là : 12,5%
a) \(\left(\dfrac{-3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+\dfrac{-9}{4}\right):\dfrac{3}{7}\)
\(=\left(\dfrac{-3}{4}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{-9}{4}\right):\dfrac{3}{7}\)
\(=-2:\dfrac{3}{7}=\dfrac{-14}{3}\)
\(\dfrac{7}{8}:\left(\dfrac{2}{9}-\dfrac{1}{18}\right)+\dfrac{7}{8}:\left(\dfrac{1}{36}-\dfrac{5}{12}\right)\)
\(=\dfrac{7}{8}:\dfrac{1}{6}+\dfrac{7}{8}:\dfrac{-7}{18}\)
\(=\dfrac{7}{8}:\left(\dfrac{1}{6}+\dfrac{-7}{18}\right)=\dfrac{7}{8}:\dfrac{-2}{9}=\dfrac{63}{-16}\)
3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
Trời ơi cái đề bài !!!
Thoy thì làm từng câu vậy
a) \(I=10101.\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{111111}\right)\)
\(I=10101.\left(\dfrac{10}{222222}+\dfrac{5}{222222}-\dfrac{8}{222222}\right)\)
\(I=10101.\left(\dfrac{15}{222222}-\dfrac{8}{222222}\right)\)
\(I=10101.\dfrac{7}{222222}\)
\(I=\dfrac{7}{22}\)
\(4\dfrac{1}{3}.\dfrac{4}{9}+13\dfrac{2}{3}.\dfrac{4}{9}\)\(=\dfrac{4}{9}\left(4\dfrac{1}{3}+13\dfrac{2}{3}\right)=\dfrac{4}{9}.18=8\)
\(5\dfrac{1}{4}.\dfrac{3}{8}+10\dfrac{3}{4}.\dfrac{3}{8}=\dfrac{3}{8}\left(5\dfrac{1}{4}+10\dfrac{3}{4}\right)=\dfrac{3}{8}.16=6\)
\(a.\)
\(\dfrac{1}{3}\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{1}{6}-2+\dfrac{3}{5}=-5x+x\)
\(\Rightarrow-4x=-\dfrac{37}{30}\)
\(\Rightarrow4x=\dfrac{37}{30}\)
\(\Rightarrow x=\dfrac{37}{120}\)
\(b.\)
\(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{3}{2}=x-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{3}{2}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{2}\)
\(c.\)
\(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\)
\(\Rightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\)
\(\Rightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)
\(\Rightarrow x=\dfrac{19}{21}\)
\(d.\)
\(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{39}{10}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{39}{10}+1\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{49}{10}\)
\(\Rightarrow x=\dfrac{49}{5}\)
\(e.\)
\(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\)
\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\)
\(\Rightarrow\left[{}\begin{matrix}4x+3=\dfrac{3}{5}\\4x+3=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=-\dfrac{12}{5}\\4x=-\dfrac{18}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{9}{10}\end{matrix}\right.\)
a) \(\dfrac{1}{3}.\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\Leftrightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)
\(\Leftrightarrow5x-x=-\dfrac{3}{5}-\dfrac{1}{6}+2\Leftrightarrow4x=\dfrac{37}{30}\Leftrightarrow x=\dfrac{\dfrac{37}{30}}{4}=\dfrac{37}{120}\)
vậy \(x=\dfrac{37}{120}\)
b) \(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\Leftrightarrow\dfrac{3}{2}x-x=\dfrac{-3}{4}+1\dfrac{1}{2}\Leftrightarrow\dfrac{1}{2}x=\dfrac{-3}{4}+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.2=\dfrac{6}{4}=\dfrac{3}{2}\) vậy \(x=\dfrac{3}{2}\)
c) \(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\Leftrightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\Leftrightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)
\(\Leftrightarrow x=\dfrac{\dfrac{19}{12}}{\dfrac{7}{4}}=\dfrac{19}{21}\) vậy \(x=\dfrac{19}{21}\)
d) \(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\Leftrightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)
\(\Leftrightarrow\dfrac{3}{2}x-x=1+\dfrac{5}{2}+\dfrac{7}{5}\Leftrightarrow\dfrac{1}{2}x=\dfrac{49}{10}\Leftrightarrow x=\dfrac{49}{10}.2=\dfrac{49}{5}\)
vậy \(x=\dfrac{49}{5}\)
e) \(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\Leftrightarrow\left|4x+3\right|=\dfrac{\dfrac{6}{15}}{\dfrac{2}{3}}=\dfrac{3}{5}\)
th1 : \(4x+3\ge0\Leftrightarrow4x\ge-3\Leftrightarrow x\ge\dfrac{-3}{4}\)
\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\Leftrightarrow4x+3=\dfrac{3}{5}\Leftrightarrow4x=\dfrac{3}{5}-3=\dfrac{-12}{5}\)
\(\Leftrightarrow x=\dfrac{\dfrac{-12}{5}}{4}=\dfrac{-3}{5}\left(tmđk\right)\)
th2: \(4x+3< 0\Leftrightarrow4x< -3\Leftrightarrow x< \dfrac{-3}{4}\)
\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\Leftrightarrow-\left(4x+3\right)=\dfrac{3}{5}\Leftrightarrow-4x-3=\dfrac{3}{5}\)
\(\Leftrightarrow4x=-3-\dfrac{3}{5}=\dfrac{-18}{5}\Leftrightarrow x=\dfrac{\dfrac{-18}{5}}{4}=\dfrac{-9}{10}\left(tmđk\right)\)
vậy \(x=\dfrac{-3}{5};x=\dfrac{-9}{10}\)
bạn cứ tính 2 vế là xong mà:
a) x\(\in\){1;2;3;4;5;6;7}
b) x=0