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a,x/9=5/3 = x=15
b,17/x=85/105 = x = 21
c,6/8=15/x = x = 20
d,x/8=2/3 = x = 6
2,
a) \(315-\left(135-x\right)=215\)
\(\Rightarrow135-x=315-215\)
\(\Rightarrow135-x=100\)
\(\Rightarrow x=135-100\)
\(\Rightarrow x=35\)
b) \(x-320:32=25\cdot16\)
\(\Rightarrow x-10=5^2\cdot4^2\)
\(\Rightarrow x-10=20^2\)
\(\Rightarrow x-10=400\)
\(\Rightarrow x=410\)
c) \(3\cdot x-2018:2=23\)
\(=3\cdot x-1009=23\)
\(\Rightarrow3\cdot x=1032\)
\(\Rightarrow x=1032:3\)
\(\Rightarrow x=344\)
d) \(280-9\cdot x-x=80\)
\(\Rightarrow280-x\cdot\left(9+1\right)=80\)
\(\Rightarrow280-10\cdot x=80\)
\(\Rightarrow10\cdot x=280-80\)
\(\Rightarrow10\cdot x=200\)
\(\Rightarrow x=20\)
e) \(38\cdot x-12\cdot x-x\cdot16=40\)
\(\Rightarrow x\cdot\left(38-12-16\right)=40\)
\(\Rightarrow x\cdot10=40\)
\(\Rightarrow x=40:10\)
\(\Rightarrow x=4\)
a) x + 10 = 20
<=> x = 20 - 10 = 10
Vậy x = 10
b) 2x + 15 = 35
<=> 2x = 35 - 15 = 20
<=> x = 10
Vậy x = 10
c) 3(x + 2) = 15
<=> x + 2 = 15 : 3 = 5
<=> x = 5 - 2 = 3
Vậy x = 3
d) 10x + 15.11 = 20.10
<=> 10x + 165 = 200
<=> 10x = 200 - 165 = 35
<=> x = 35 : 10 = 3,5
Vậy x = 3,5
e) 4(x + 2) = 3.4
<=> x + 2 = 3
<=> x = 3 - 2 = 1
Vậy x = 1
f) 33x + 135 = 26.9
<=> 33x + 135 = 234
<=> 33x = 234 - 135 = 99
<=> x = 99 : 33 = 3
Vậy x = 3
g) 2x + 15 + 16 + 17 = 100
<=> 2x + 48 = 100
<=> 2x = 100 - 48 = 52
<=> x = 52 : 2 = 26
Vậy x = 26
h) 2(x + 9 + 10 + 11) = 4.12.5
<=> x + 30 = 120
<=> x = 120 - 30 = 90
Vậy x = 90
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:\(x=\dfrac{15}{19}\)
Câu 2:
a: \(=\left(-2\right)^3\cdot\left(-3\right)^3\cdot5^3=30^3\)
b: \(=3^3\cdot\left(-2\right)^3\cdot\left(-7\right)\cdot\left(-7\right)^2\)
\(=\left(3\cdot2\cdot7\right)^3=42^3\)
a: \(x-\dfrac{10}{3}=\dfrac{7}{15}\cdot\dfrac{3}{5}\)
=>\(x-\dfrac{10}{3}=\dfrac{21}{75}=\dfrac{7}{25}\)
=>\(x=\dfrac{7}{25}+\dfrac{10}{3}=\dfrac{21+250}{75}=\dfrac{271}{75}\)
b: \(x+\dfrac{3}{22}=\dfrac{27}{121}\cdot\dfrac{9}{11}\)
=>\(x+\dfrac{3}{22}=\dfrac{243}{1331}\)
=>\(x=\dfrac{243}{1331}-\dfrac{3}{22}=\dfrac{123}{2662}\)
c: \(\dfrac{8}{23}\cdot\dfrac{46}{24}-x=\dfrac{1}{3}\)
=>\(\dfrac{8}{24}\cdot\dfrac{46}{23}-x=\dfrac{1}{3}\)
=>\(\dfrac{2}{3}-x=\dfrac{1}{3}\)
=>\(x=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
d: \(1-x=\dfrac{49}{65}\cdot\dfrac{5}{7}\)
=>\(1-x=\dfrac{49}{7}\cdot\dfrac{5}{65}=\dfrac{7}{13}\)
=>\(x=1-\dfrac{7}{13}=\dfrac{6}{13}\)
a, \(\dfrac{4x}{7}=\dfrac{1}{5}+\dfrac{2}{3}=\dfrac{3+10}{15}=\dfrac{13}{15}\Rightarrow60x=91\Leftrightarrow x=\dfrac{91}{60}\)
b, \(\dfrac{5}{7}:x=\dfrac{1}{6}+\dfrac{4}{5}=\dfrac{5+24}{30}=\dfrac{29}{30}\Leftrightarrow x=\dfrac{5}{7}:\dfrac{29}{30}=\dfrac{150}{203}\)
a)\(\frac{x}{15}=\frac{2}{5}\)=> x=\(\frac{15.2}{5}=6\)
b) \(\frac{3}{x^7}hay\frac{3}{x.7}\)