Tìm x
(x - 23).(x + \(\frac{6}{7}\)) = 0
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\(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
\(=>\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}=>\orbr{\begin{cases}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{cases}=>\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{1}{3}\end{cases}}}}\)
Vậy x thuộc {-1/2 ; 1/3}
\(x.3\frac{1}{4}+\frac{-7}{6}.x-1\frac{2}{3}=\frac{5}{12}.2\)
\(x.\frac{13}{4}+\frac{-7}{6}.x-\frac{5}{3}=\frac{5}{6}\)
\(x.\left(\frac{13}{4}+\frac{-7}{6}\right)=\frac{5}{6}+\frac{5}{3}\)
\(x.\left(\frac{39}{12}+\frac{-14}{12}\right)=\frac{5}{6}+\frac{10}{6}\)
\(x.\frac{25}{12}=\frac{5}{2}\)
\(x=\frac{5}{2}:\frac{25}{12}\)
\(x=\frac{5}{2}.\frac{12}{25}\)
\(x=\frac{6}{5}\)
Ta có:
\(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}\)
\(\Leftrightarrow\left(\frac{x+7}{2010}+1\right)+\left(\frac{x+6}{2011}+1\right)=\left(\frac{x+5}{2012}+1\right)+\left(\frac{x+4}{2013}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2010}+\frac{x+2017}{2011}=\frac{x+2017}{2012}+\frac{x+2017}{2013}\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2010}+\frac{1}{2011}\right)=\left(x+2017\right)\left(\frac{1}{2013}+\frac{1}{2014}\right)\)
Suy ra \(x+2017=0\)
Vậy \(x=-2017\)
b) Dễ tự làm nhé
Bài 1 : Ta có:
\(\frac{7+\frac{7}{11}+\frac{7}{23}+\frac{7}{31}}{9+\frac{9}{11}+\frac{9}{23}+\frac{9}{31}}\)
= \(\frac{7.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}{9.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}\)
= \(\frac{7}{9}\)
Bài 2 :
\(\frac{x}{2}+\frac{3x}{4}+\frac{5x}{6}=\frac{10}{24}\)
=> \(\frac{12x+18x+20x}{24}=\frac{10}{24}\)
=> 50x = 10
=> x = 10 : 50
=> x = 1/5
Bài 3 : Để A nhận giá trị nguyên thì 3 \(⋮\)x + 3
<=> x + 3 \(\in\)Ư(3) = {1; -1; 3; -3}
Lập bảng :
x + 3 | 1 | -1 | 3 | -3 |
x | -2 | -4 | 0 | -6 |
Vậy
a/ \(\frac{6}{7}x=\frac{18}{23}\)
\(x=\frac{18}{23}:\frac{6}{7}=\frac{21}{23}\)
b/ \(2\frac{1}{2}x=\frac{5}{6}\)
\(=>\frac{5}{2}x=\frac{5}{6}\)
\(x=\frac{5}{6}:\frac{5}{2}=\frac{1}{3}\)
c/\(x:2\frac{3}{4}=9\frac{5}{8}\)
\(x:\frac{11}{4}=\frac{77}{8}\)
\(x=\frac{77}{8}\cdot\frac{11}{4}=\frac{847}{32}\)
d/\(7\frac{1}{7}\cdot\frac{1}{7}\cdot x=22\frac{1}{8}\)
\(\frac{50}{49}x=\frac{177}{8}\)
\(x=\frac{177}{8}:\frac{50}{49}=\frac{8673}{400}\)
\(a,\frac{6}{7}.x=\frac{18}{23}\) \(\Rightarrow x=\frac{18}{23}:\frac{6}{7}=\frac{18}{23}.\frac{7}{6}=\frac{21}{23}\)
\(b,2\frac{1}{2}.x=\frac{5}{6}\Rightarrow\frac{5}{2}.x=\frac{5}{6}\Rightarrow x=\frac{5}{6}:\frac{5}{2}=\frac{5}{6}.\frac{2}{5}=\frac{1}{3}\)
\(c,x:2\frac{3}{4}=9\frac{5}{8}\Rightarrow x:\frac{11}{4}=\frac{77}{8}\Rightarrow x=\frac{77}{8}.\frac{11}{4}=\frac{847}{32}\)
\(d,7\frac{1}{7}.\frac{1}{7}.x=22\frac{1}{8}\Rightarrow\frac{50}{49}.x=\frac{177}{8}\Rightarrow x=\frac{177}{8}:\frac{50}{49}=\frac{177}{8}.\frac{49}{50}=\frac{8673}{400}\)
a) Theo đề ta có: (1/8x-9)8=0 hoặc (1/23y-5)6=0
=>1/8x-9=0 hoặc 1/23y-5=0
=>1/8x=9 hoặc 1/23y=5
=>x=9:1/8=72 hoặc y=5:1/23=115
\(\left(x-23\right)\left(x+\frac{6}{7}\right)=0\\ x-23=0;x+\frac{6}{7}=0\\ x=0+23=23;x=0-\frac{6}{7}=-\frac{6}{7}\)
\(\left(x-23\right)\left(x+\frac{6}{7}\right)=0\)
\(x-23=0\Rightarrow x=23\)
\(x+\frac{6}{7}=0\rightarrow x=-\frac{6}{7}\)