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a: \(\Leftrightarrow x:\dfrac{3}{10}=\dfrac{33}{8}:\dfrac{1}{5}=\dfrac{165}{8}\)
\(\Leftrightarrow x=\dfrac{165}{8}\cdot\dfrac{3}{10}=\dfrac{99}{16}\)
b: \(\Leftrightarrow x\cdot\dfrac{1}{100}:4=\dfrac{11}{15}\)
\(\Leftrightarrow x\cdot\dfrac{1}{100}=\dfrac{44}{15}\)
hay x=880/3
1)\(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)
\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0\).Do \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\ne0\)
\(\Leftrightarrow x=-1\)
từ 15/5,1=-35/11,9
=> -15/-35 = 5,1/11,0
=> -15/11,0 = 5.1/-35
=> 11,9/5,1= -35/-15
=> 11,9/ -35 = 5,1/-15
a, 1,2 : 3,24 = 120/324 = 10/27
Ta có tỉ lệ thức: 1,2 : 3,24 = 10 : 27
b, 2 và 1/5 : 3/4 = 11/5 : 3/4 = 11/5 . 4/3 = 44/15
Ta có tỉ lệ thức: 2 và 1/5 : 3/4 = 44 : 15
c, 2/7 : 0,42 = 2/7 : 21/50 = 2/7 . 50/21 = 100/147
Ta có tỉ lệ thức: 2/7 : 0,42 = 100 : 147
e)
=> (x-2) . (x+7) = ( x-1 ) . ( x +4)
=> x2 +7x - 2x -14 = x2 - x + 4x - 4
x2 + 5x - 14 = x2 + 3x - 4
=> 5x - 14 = 3x - 4
=> 5x - 3x = 14-4
=> 2x = 10 => x = 10 : 2 => x = 5
c)
=>( x-1) . 7 = ( x + 5 ) . 6
=> 7x - 7 = 6x + 30
=> 7x - 6x= 30 + 7
=> x = 37
a,x=\(\frac{5}{2}\)
b,x=\(\frac{13}{176}\)
c,x=37
d, x=\(\frac{12}{5}\)
e, x=5
a)
c) 6,51.7 = 15,19.3 Suy ra 6,51: 15,19 = 3/7
d)
nên hai tỉ số trên không lập thành tỉ lệ thức
a) \(\frac{x}{7}=\frac{18}{14}\)
\(\Rightarrow\frac{x}{7}=\frac{9}{7}\)
\(\Rightarrow x=7\)
Vậy x=7
b)\(6:x=1\frac{3}{4}:5\)
\(\frac{6}{x}=\frac{7}{4}:5\)
\(\frac{6}{x}=\frac{7}{20}\)
\(\Rightarrow6.20=7x\)
\(\Rightarrow120=7.x\)
\(\Rightarrow x=\frac{120}{7}\)
Vậy \(x=\frac{120}{7}\)
\(C=\frac{5x^2+3y^2}{10x^2-3y^2}\)
Có \(\frac{x}{3}=\frac{y}{5}\Rightarrow\frac{x}{y}=\frac{3}{5}\)
Thay \(x=3;y=5\) ta có : \(\frac{5x^2+3y^2}{10x^2-3y^2}=\frac{5\cdot3^2+3\cdot5^2}{10\cdot3^2-3\cdot5^2}=8\)
Vậy \(C=8\)
Ta có:\(\frac{x+1}{11}+\frac{x+2}{10}=\frac{x+3}{9}+\frac{x+4}{8}\)
\(\Rightarrow1+\frac{x+1}{11}+1+\frac{x+2}{10}=1+\frac{x+3}{9}+1+\frac{x+4}{8}\)
\(\Rightarrow\frac{x+12}{11}+\frac{x+12}{10}=\frac{x+12}{9}+\frac{x+12}{8}\)
\(\Rightarrow\frac{x+12}{11}+\frac{x+12}{10}-\frac{x+12}{9}-\frac{x+12}{8}=0\)
\(\Rightarrow\left(x+12\right)\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)
Mà \(\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)>0\)
\(\Rightarrow x+12=0\Rightarrow x=-12\)
\(\frac{x+1}{11}+\frac{x+2}{10}=\frac{x+3}{9}+\frac{x+4}{8}\)
<=> \(\frac{x+1}{11}+\frac{x+2}{10}-\frac{x+3}{9}-\frac{x+4}{8}=0\)
<=> \(\left(\frac{x+1}{11}+1\right)+\left(\frac{x+2}{10}+1\right)-\left(\frac{x+3}{9}+1\right)-\left(\frac{x+4}{8}+1\right)=0\)<=> \(\frac{x+12}{11}+\frac{x+12}{10}-\frac{x+12}{9}-\frac{x+12}{8}=0\)
<=> \(\left(x+12\right)\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)
<=> x + 12 = 0.Vì \(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\ne0\)
<=> x = -12
\(\frac{x}{27}=-\frac{2}{3,6}\)
\(x=\frac{-2\times27}{3,6}\)
\(x=-15\)
***
\(\frac{-0,52}{x}=\frac{-9,36}{16,38}\)
\(x=\frac{-0,52\times16,38}{-9,36}\)
\(x=0,91\)
***
\(\frac{4\frac{1}{4}}{2\frac{7}{8}}=\frac{x}{1,61}\)
\(\frac{4,25}{2,875}=\frac{x}{1,61}\)
\(x=\frac{4,25\times1,61}{2,875}\)
\(x=2,38\)
\(a.\)
\(\frac{x}{27}=\frac{-2}{3,6}\)
\(\Rightarrow x=\frac{\left(-2\right).27}{3,6}\)
\(\Rightarrow x=\frac{-54}{3,6}\)
\(\Rightarrow x=-15\)
Vậy : \(x=-15\)
\(b.\)
\(-0,52:x=-9,36:16,38\)
\(\Rightarrow\frac{-0,52}{x}=-\frac{4}{7}\)
\(\Rightarrow x=\frac{-0,52.7}{-4}\)
\(\Rightarrow x=0,91\)
Vậy : \(x=0,91\)
\(c.\)
\(\frac{4\frac{1}{4}}{2\frac{7}{8}}=\frac{x}{1,61}\)
\(\Rightarrow\frac{\frac{17}{4}}{\frac{23}{8}}=\frac{x}{1,61}\)
\(\Rightarrow\frac{17}{4}.\frac{8}{23}=\frac{x}{1,61}\)
\(\Rightarrow\frac{34}{23}=\frac{x}{1,61}\)
\(\Rightarrow x=\frac{1,61.34}{23}\)
\(\Rightarrow x=\frac{119}{50}\)
Vậy : \(x=\frac{119}{50}\)