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a.
\(\dfrac{x}{9}=\dfrac{4}{x}\)
\(\Rightarrow x^2=4.9\)
\(\Rightarrow x^2=36\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=6\end{matrix}\right.\)
b.
\(\dfrac{x+1}{3}=\dfrac{3}{x+1}\)
\(\Rightarrow\left(x+1\right)^2=3^2\)
\(\Rightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
\(\dfrac{x}{9}=\dfrac{4}{x}\)
\(x^2=4.9\)
\(x^2=36\)
\(x^2=6^2\)
\(\Rightarrow\left\{{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
\(---------\)
\(\dfrac{x+1}{3}=\dfrac{3}{x+1}\)
\(\left(x+1\right)^2=3.3=3^2\)
\(\Rightarrow\left(1\right):x+1=3\)
\(x=3-1\Rightarrow x=2.\)
\(\Rightarrow\left(2\right):x+1=-3\)
\(x=-3-1\Rightarrow x=-4\)
Từ \(\left(1\right)\) và \(\left(2\right)\), ta suy ra:
\(\Rightarrow x\in\left\{{}\begin{matrix}2\\-4\end{matrix}\right.\)
\(\frac{x-1}{3}+\frac{3x-5}{2}+\frac{2x}{9}+\frac{-5x+3}{9}=\frac{210}{420}\)
\(\Leftrightarrow\frac{x-1}{3}+\frac{3x-5}{2}+\frac{-3x+3}{9}=\frac{1}{2}\)
\(\Leftrightarrow\frac{6\left(x-1\right)}{18}+\frac{9\left(3x-5\right)}{18}+\frac{2\left(-3x+3\right)}{18}=\frac{9}{18}\)
\(\Rightarrow6x-6+27x-45-6x+6=9\)
\(\Leftrightarrow6x+27x-6x=9+6+45-6\)
\(\Leftrightarrow27x=54\)
\(\Rightarrow x=2\)
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 )
=> \(0,75< 1,02< 3\)
\(3.\)
\(\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(\Rightarrow\)\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1+\frac{x-3}{2009}-1-\frac{x-4}{2008}+1+2=0\)
\(\Rightarrow\)\(\frac{x-1}{2011}-\frac{2011}{2011}+\frac{x-2}{2010}-\frac{2010}{2010}+\frac{x-3}{2009}-\frac{2009}{2009}-\frac{x-4}{2008}+\frac{2008}{2008}=0\)
\(\Rightarrow\)\(\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\)\(x-2012\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)
\(\Rightarrow\)\(x=2012\)
Đáp án là C
3|x + 1| = 9 ⇒ |x + 1| = 3
⇒ x + 1 = 3 hay x = 2
Hoặc x + 1 = -3 hay x = -4
\(\dfrac{x+1}{3}=\dfrac{-3}{-9}\)
=>\(\dfrac{x+1}{3}=\dfrac{1}{3}\)
=>x+1=1
=>x=0