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1)
\(\frac{3}{4}.x+\frac{x}{5}=\frac{1}{6}\)
\(x.\left(\frac{3}{4}+\frac{1}{5}\right)=\frac{1}{6}\)
\(x.\frac{19}{20}=\frac{1}{6}\)
\(x=\frac{1}{6}:\frac{19}{20}\)
\(x=\frac{10}{57}\)
2)
\(x+3\frac{1}{2}+x=24\frac{1}{4}\)
\(2x+3\frac{1}{2}=24\frac{1}{4}\)
\(2x=24\frac{1}{4}-3\frac{1}{2}\)
\(2x=\frac{83}{4}\)
\(x=\frac{83}{4}:2\)
\(x=\frac{83}{8}\)
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\(a,\frac{3}{7}+\left|2x-\frac{1}{2}\right|=\frac{4}{5}\)
\(\Rightarrow\left|2x-\frac{1}{2}\right|=\frac{13}{35}\)
\(\Rightarrow2x-\frac{1}{2}=\pm\left(\frac{13}{35}\right)\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=\frac{13}{35}\\2x-\frac{1}{2}=\frac{-13}{35}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=\frac{61}{70}\\2x=\frac{9}{70}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{61}{140}\\x=\frac{9}{140}\end{cases}}\)
~Study well~
#KSJ
\(b,\frac{3}{4}-4\times\left|2x+1\right|=\frac{1}{2}\)
\(\Rightarrow4\times\left|2x+1\right|=\frac{1}{4}\)
\(\Rightarrow\left|2x+1\right|=\frac{1}{16}\)
\(\Rightarrow2x+1=\pm\left(\frac{1}{16}\right)\)
\(\Rightarrow\orbr{\begin{cases}2x+1=\frac{1}{16}\\2x+1=\frac{-1}{16}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=\frac{-15}{16}\\2x=\frac{-17}{16}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-15}{32}\\x=\frac{-17}{32}\end{cases}}\)
~Study well~
#KSJ
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\(\frac{8}{23}\cdot\frac{46}{24}-x=\frac{1}{3}\)
=> \(\frac{2}{3}-x=\frac{1}{3}\)
=> \(x=\frac{1}{3}\)
\(\frac{10}{12}\div x=\frac{28}{9}\cdot\frac{3}{56}\)
=> \(\frac{10}{12}\div x=\frac{1}{6}\)
=> \(x=\frac{60}{12}=5\)
\(\frac{x-12}{4}=\frac{1}{2}\)
=> \(\left(x-12\right)\cdot2=4\cdot1\)
=> \(2x-24=4\)
=> \(2x=28\)
=> \(x=14\)
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\(a,\frac{3}{4}-\frac{-1}{2}=x+\frac{1}{8}\Leftrightarrow x+\frac{1}{8}=\frac{5}{4}\Leftrightarrow x=\frac{5}{4}-\frac{1}{8}=\frac{9}{8}\)
\(b,\frac{x}{3}+\frac{3}{4}=\frac{2}{24}\Leftrightarrow\frac{x}{3}=\frac{2}{24}-\frac{3}{4}\Leftrightarrow\frac{x}{3}=\frac{-2}{3}\Leftrightarrow x=-2\)
\(c,\frac{-2}{3}-\left(-x\right)=\frac{1}{4}\Leftrightarrow-\left(-x\right)=\frac{1}{4}-\frac{-2}{3}\Leftrightarrow x=\frac{11}{12}\)
\(d,\frac{-5}{7}+x=\frac{-1}{2}+\frac{1}{3}\Leftrightarrow x=\frac{-1}{2}+\frac{1}{3}-\frac{-5}{7}=\frac{23}{42}\)
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ở câu 1 ở mỗi phẫn số chúng ta cộng thêm 1, tổng là ta cộng thêm 5. Lấy 5 + -5=0. Rồi ta được tất cả tử là x+200,đặt chung ra ngoài,từ đó tính x=-200
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a) 2 - ( \(5\frac{3}{8}\)x X - \(\frac{5}{24}\)) = \(\frac{5}{12}\)
\(5\frac{3}{8}\)x X - \(\frac{5}{24}\)= \(\frac{19}{12}\)
\(5\frac{3}{8}\)x X = \(\frac{43}{24}\)
X = \(\frac{1}{3}\)
b) \(1\frac{2}{9}\): ( \(3\frac{1}{3}\)x X + \(\frac{1}{6}\)) = \(\frac{22}{23}\)
\(3\frac{1}{3}\)x X + \(\frac{1}{6}\) = \(\frac{23}{18}\)
\(3\frac{1}{3}\)x X = \(\frac{10}{9}\)
X =\(\frac{1}{3}\)
C) \(\frac{4}{5}\)x X - \(\frac{1}{2}\)x X + \(\frac{3}{4}\)x X = \(\frac{7}{40}\)
( \(\frac{4}{5}-\frac{1}{2}+\frac{3}{4}\)) x X = \(\frac{7}{40}\)
\(\frac{21}{20}\) x X = \(\frac{7}{40}\)
X =\(\frac{1}{6}\)
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dễ mà bạn đây là bài cơ bản lớp 6 dấy
câu a nhé bạn bạn nếu ko làm kiểu khó thì đổi về phaan số bình thường nà sau đó tính trong ngoặc trước rồi tính xoong bỏ dấu ngoặc nhưng ko đổi dấu né thế lad đc tương tự như các câu dưới
a)\(8\frac{2}{3}:2\frac{1}{6}-2\frac{27}{51}=\frac{26}{3}.\frac{6}{13}-\frac{43}{17}=4-\frac{43}{17}=\frac{25}{17}\)
b)\(\frac{27}{20}.\frac{15}{4}+\frac{19}{8}=\frac{119}{16}\)
c)\(\left(\frac{1}{12}+\frac{5}{6}\right)+\left(\frac{13}{35}+\frac{23}{35}\right)=\frac{11}{12}+\frac{36}{35}=\frac{817}{420}\)
d)\(\frac{24}{37}.\left(\frac{13}{18}+\frac{2}{9}+\frac{1}{18}\right)=\frac{24}{37}.1=\frac{24}{37}\)
\(x+3\frac{1}{2}+x=24\frac{1}{4}\)
<=>\(2x+\frac{7}{2}=\frac{97}{4}\)
<=>2x=97/4-7/2
<=>2x=83/4
<=>x=83/4:2
<=>x=83/8