2 tìm x biết
A)(5.x-15).37=74
B)[(x-9).2-1]:3+12=15
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1) 9-25=(7-x)-(25+7)
(7-x)-(25+7)=9-25
(7-x)-(25+7)=-16
(7-x)-32=-16
7-x=-16+32
7-x=16
x=7-16
x=-9
2) (27-514) -(486-73)+x=7
(27-514) -(486-73)+x=7
-487-413+x=7
-1011+x=7
x=7-(-1011)
x=7+1011
x=1018
3) 25+5+37-25+6-29-x=37
25+5+37-25+6-29-x=37
67-25+6-29-x=37
42+6-29-x=37
48-29-x=37
19-x=37
x=19-37
x=-18
4) 14+(-12)+x =10-/-15/+ /-3/
14+(-12)+x =10-15+3
14+(-12)+x =-5+3
14+(-12)+x =-2
2+x=-2
x=-2-2
x=-4
Bài 1:
a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)
hay \(x=\dfrac{50}{9}\)
Vậy: \(x=\dfrac{50}{9}\)
c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)
\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)
hay \(x=\dfrac{22}{15}\)
Vậy: \(x=\dfrac{22}{15}\)
d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)
\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)
hay \(x=\dfrac{15}{19}\)
Vậy:\(x=\dfrac{15}{19}\)
Bài 1:
a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)
\(=\dfrac{9}{20}-\dfrac{2}{9}\)
\(=\dfrac{41}{180}\)
b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)
\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)
\(=\dfrac{5}{6}\times\dfrac{12}{7}\)
\(=\dfrac{10}{7}\)
c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)
\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{7}{9}\times1\)
\(=\dfrac{7}{9}\)
Bài 2:
a) \(2\times\left(x-1\right)=4026\)
\(\left(x-1\right)=4026\div2\)
\(x-1=2013\)
\(x=2014\)
Vậy: \(x=2014\)
b) \(x\times3,7+6,3\times x=320\)
\(x\times\left(3,7+6,3\right)=320\)
\(x\times10=320\)
\(x=320\div10\)
\(x=32\)
Vậy: \(x=32\)
c) \(0,25\times3< 3< 1,02\)
\(\Leftrightarrow0,75< 3< 1,02\) ( S )
=> \(0,75< 1,02< 3\)
Bài 1:
a: \(=\dfrac{21}{7}\cdot\dfrac{15}{45}\cdot\dfrac{9}{18}=3\cdot\dfrac{1}{3}\cdot\dfrac{1}{2}=\dfrac{1}{2}\)
b: \(=\dfrac{11}{33}\cdot\dfrac{24}{8\cdot12}\cdot\dfrac{27}{9}=\dfrac{24}{96}=\dfrac{1}{4}\)
a, \(\frac{2}{3}:\frac{5}{7}:\frac{x}{9}=\frac{2}{7}:\frac{3}{5}:\frac{10}{9}\Leftrightarrow\frac{2\times7\times9}{3\times5\times x}=\frac{2\times5\times9}{7\times3\times10}\Leftrightarrow\frac{42}{5\times x}=\frac{3}{7}\Leftrightarrow294=15\times x\Leftrightarrow x=19,6\)
Các phần sau tương tự.
Dùng phương pháp nhân chéo và kiểm tra trình độ tính toán, khả năng rút gọn.
Thay kết quả vào x để kiểm tra lại bài.
a) ( 5x - 15) .37 =74
5x - 15= 74 : 37
5x - 15 = 2
5x = 2 + 15
5x = 17
x = 17: 5
x= \(\frac{17}{5}\)