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Ta có:
+) \(\frac{2013.2012-1}{2013.2012}=1-\frac{1}{2013.2012}\)
+) \(\frac{2012.2011-1}{2012.2011}=1-\frac{1}{2012.2011}\)
Vì \(\frac{1}{2013.2012}< \frac{1}{2012.2011}\Rightarrow1-\frac{1}{2013.2012}>1-\frac{1}{2012.2011}\)
Vậy \(\frac{2013.2012-1}{2013.2012}>\frac{2012.2011-1}{2012.2011}\)
a; \(\dfrac{x-1}{12}\) = \(\dfrac{5}{3}\)
\(x-1\) = \(\dfrac{5}{3}\) \(\times\) 12
\(x\) - 1 = 20
\(x\) = 20 + 1
\(x\) = 21
b; \(\dfrac{-x}{8}\) = \(\dfrac{-50}{x}\)
-\(x\).\(x\) = -50.8
-\(x^2\) = -400
\(x^2\) = 400
\(\left[{}\begin{matrix}x=-20\\x=20\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-20; 20}
c; \(\dfrac{x}{3}\) = \(\dfrac{14}{x+1}\)
\(x\).(\(x\)+1) = 14.3
\(x^2\) + \(x\) = 42
\(x^2\) + \(x\) - 42 = 0
\(x^2\) - 6\(x\) + 7\(x\) - 42 = 0
\(x\).(\(x\) - 6) + 7.(\(x\) - 6) = 0
(\(x\) - 6).(\(x\) + 7) = 0
\(\left[{}\begin{matrix}x-6=0\\x+7=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=6\\x=-7\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-7; 6}
d; \(x-\dfrac{2}{9}\) = \(\dfrac{1}{6}\)
\(x\) = \(\dfrac{1}{6}\) + \(\dfrac{2}{9}\)
\(x\) = \(\dfrac{7}{18}\)
Vậy \(x\) = \(\dfrac{7}{18}\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{40.43}+\frac{1}{43.46}\)
\(=3.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{43}-\frac{1}{46}\right)\)
\(=3.\left(1-\frac{1}{46}\right)\)
\(=3.\frac{45}{46}\)
\(=\frac{135}{46}\)
~Học tốt~
7 ngay 6 dem khach san 5 sao tang 4 phing 3 2 nguoi 1 giuong o quan o ao
7 ngay 6 lan 5 gio 4 phut nga 3 ,2 thang 1 chai o say o ve
125.(-8).(-25).9.4.1002:3
=125.(-8).(-25).4.1002.9:3
=(-1000).(-100).10000.3
=3000000000
Chúc bn học tốt
Lời giải:
a.
$\frac{8}{23}.\frac{46}{24}-\frac{2}{5}x=\frac{1}{3}$
$\frac{2}{3}-\frac{2}{5}x=\frac{1}{3}$
$\frac{2}{5}x=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$
$x=\frac{1}{3}: \frac{2}{5}=\frac{5}{6}$
b.
$\frac{10}{12}: \frac{2}{3}x=\frac{28}{9}.\frac{3}{56}$
$\frac{5}{4}x=\frac{1}{6}$
$x=\frac{1}{6}: \frac{5}{4}=\frac{2}{15}$
c.
$\frac{x-1}{24}=\frac{2}{x+1}$
$(x-1)(x+1)=2.24$
$x^2-1=48$
$x^2=49=7^2=(-7)^2$
$\Rightarrow x=7$ hoặc $x=-7$
d.
$(\frac{3}{4}x+\frac{1}{4}-\frac{1}{3}): (2+\frac{1}{6}-\frac{1}{4})=\frac{7}{46}$
$(\frac{3}{4}x-\frac{1}{12}):\frac{23}{12}=\frac{7}{46}$
$\frac{3}{4}x-\frac{1}{12}=\frac{7}{46}.\frac{23}{12}=\frac{7}{24}$
$\frac{3}{4}x=\frac{7}{24}+\frac{1}{12}=\frac{3}{8}$
$x=\frac{3}{8}: \frac{3}{4}=\frac{1}{2}$
e.
$2\frac{1}{2}x+0,5x=2\frac{1}{4}$
$2,5x+0,5x=2,25$
$x(2,5+0,5)=2,25$
$3x=2,25$
$x=2,25:3=0,75$
f.
$\frac{1}{3}x+\frac{2}{5}(x-1)=0$
$\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0$
$x(\frac{1}{3}+\frac{2}{5})=\frac{2}{5}$
$x.\frac{11}{15}=\frac{2}{5}$
$x=\frac{2}{5}: \frac{11}{15}=\frac{6}{11}$
g.
$x-3\frac{1}{2}x=-2\frac{6}{7}$
$x(1-3\frac{1}{2})=\frac{-20}{7}$
$x.\frac{-5}{2}=\frac{-20}{7}$
$x=\frac{-20}{7}: \frac{-5}{2}=\frac{8}{7}$
h.
$2(\frac{1}{2}x-\frac{1}{3})-\frac{3}{2}=\frac{1}{4}$
$2(\frac{1}{2}x-\frac{1}{3})=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}$
$\frac{1}{2}x-\frac{1}{3}=\frac{7}{4}:2=\frac{7}{8}$
$\frac{1}{2}x=\frac{7}{8}+\frac{1}{3}=\frac{29}{24}$
$x=\frac{29}{24}: \frac{1}{2}=\frac{29}{12}$
i.
$-2\frac{1}{3}x+1\frac{3}{4}x+3\frac{2}{3}=3\frac{1}{2}$
$x(-2\frac{1}{3}+1\frac{3}{4})=3\frac{1}{2}-3\frac{2}{3}$
$x.\frac{-7}{12}=\frac{-1}{6}$
$x=\frac{-1}{6}: \frac{-7}{12}=\frac{2}{7}$