Tìm x
a) \(\left[\left(6\frac{3}{7}-\frac{0,75.x-2}{0,35}\right).2,8+1,75\right]:0,05=2,35\)
b) \(\frac{3}{5}-\frac{2}{7}< \frac{2}{3}.x+\frac{3}{4}< \frac{1}{2}+\frac{7}{9}\)
K
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ZT
1
17 tháng 9 2016
\(\Leftrightarrow\left(\left(6\frac{3}{7}-\frac{0,75x-2}{0,35}\right).2,8+1,75\right)=0,1175\)
\(\Leftrightarrow\left(\frac{45}{7}-\frac{0,75x-2}{0,35}\right).2,8=-1,6325\)
\(\Leftrightarrow\frac{45}{7}-\frac{0,75x-2}{0,35}=\frac{-653}{1120}\)
\(\Leftrightarrow\frac{0,75x-2}{0,35}=\frac{7853}{1120}\Leftrightarrow x=5,93875\)
DH
27 tháng 7 2020
\(\left[\left(6\frac{3}{7}-\frac{0,75x-2}{0,35}\right)\cdot2,8+1,75\right]:0,05=235\)
\(\Leftrightarrow\left(6\frac{3}{7}-\frac{0,75x-2}{0,35}\right)\cdot2,8+1,75=11,75\)
\(\Leftrightarrow\left(6\frac{3}{7}-\frac{0,75x-2}{0,35}\right)\cdot2,8=10\)
\(\Leftrightarrow\left(6\frac{3}{7}-\frac{0,75x-2}{0,35}\right)\cdot2,8=10\)
\(\Leftrightarrow6\frac{3}{7}-\frac{0,75x-2}{0,35}=\frac{25}{7}\)
\(\Leftrightarrow\frac{0,75x-2}{0,35}=\frac{20}{7}\)
\(\Leftrightarrow0,75x-2=1\)
\(\Leftrightarrow0,75x=3\)
\(\Leftrightarrow x=4\)
Vậy \(x=4\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)
Vậy \(x = \frac{{ - 2}}{3}\).
b)
\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)
Vậy\(x = \frac{1}{12}\).
c)
\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)
Vậy \(x = \frac{7}{3}\).
d)
\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)
Vậy \(x = \frac{{ - 9}}{{10}}\).
21 tháng 5 2022
a: Sửa đề: \(\left[\left(6\dfrac{3}{7}-\dfrac{0,75x-2}{0,35}\right)\cdot2.8+1.75\right]\cdot0.05=2.35\)
\(\Leftrightarrow\left(\dfrac{45}{7}-\dfrac{15x-40}{7}\right)\cdot2.8+1,75=47\)
\(\Leftrightarrow\dfrac{45-15x+40}{7}\cdot2,8=45,25\)
\(\Leftrightarrow\dfrac{85-15x}{7}=\dfrac{905}{56}\)
=>8(85-15x)=905
=>680-120x=905
=>120x=-225
hay x=-15/8
b: \(\dfrac{3}{5}-\dfrac{2}{7}< \dfrac{2}{3}x+\dfrac{3}{4}< \dfrac{1}{2}+\dfrac{7}{9}\)
\(\Leftrightarrow\dfrac{11}{35}-\dfrac{3}{4}< \dfrac{2}{3}x< \dfrac{23}{18}-\dfrac{3}{4}\)
\(\Leftrightarrow-\dfrac{61}{140}< \dfrac{2}{3}x< \dfrac{19}{36}\)
\(\Leftrightarrow-\dfrac{183}{280}< x< \dfrac{19}{24}\)
