\(x-\dfrac{2}{3}=-\dfrac{5}{12}\)

\(\Rightarrow x=-\dfrac{5}{12}+\dfrac{2}{3}\)

\(\Rightarrow x=-\dfrac{5}{12}+\dfrac{8}{12}\)

\(\Rightarrow x=\dfrac{3}{12}=\dfrac{1}{4}\)

`----`

\(\dfrac{x}{15}=\dfrac{3}{4}+\dfrac{19}{-20}\)

\(\Rightarrow\dfrac{x}{15}=-\dfrac{1}{5}\)

\(\Rightarrow-\dfrac{3}{15}=-\dfrac{1}{5}\)

\(\Rightarrow x=-3\)

`-----`

\(\dfrac{x}{-2}=-\dfrac{8}{x}\)

\(\Rightarrow x^2=-2.\left(-8\right)\)

\(\Rightarrow x^2=16\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

a) $x-\genfrac{}{}{0ex}{}{2}{3}=\genfrac{}{}{0ex}{}{-5}{12}$

$x=\genfrac{}{}{0ex}{}{-5}{12}+\genfrac{}{}{0ex}{}{2}{3}$

$x=\genfrac{}{}{0ex}{}{-5}{12}+\genfrac{}{}{0ex}{}{8}{12}$

$x=\genfrac{}{}{0ex}{}{-5+8}{12}$

$x=\genfrac{}{}{0ex}{}{3}{12}$

$x=\genfrac{}{}{0ex}{}{1}{4}$

b) $\genfrac{}{}{0ex}{}{8}{5}:x=\genfrac{}{}{0ex}{}{-2}{3}$

$x=\genfrac{}{}{0ex}{}{8}{5}:\left(\genfrac{}{}{0ex}{}{-2}{3}\right)$

$x=\genfrac{}{}{0ex}{}{8}{5}.\left(\genfrac{}{}{0ex}{}{3}{-2}\right)$

$x=\genfrac{}{}{0ex}{}{-12}{5}$

c) $1-\genfrac{}{}{0ex}{}{3}{7}.x=-\genfrac{}{}{0ex}{}{2}{7}$

$\genfrac{}{}{0ex}{}{3}{7}.x=1-\left(-\genfrac{}{}{0ex}{}{2}{7}\right)$

$\genfrac{}{}{0ex}{}{3}{7}.x=\genfrac{}{}{0ex}{}{9}{7}$

$x=\genfrac{}{}{0ex}{}{9}{7}:\genfrac{}{}{0ex}{}{3}{7}$

$x=\genfrac{}{}{0ex}{}{9}{7}.\genfrac{}{}{0ex}{}{7}{3}$

$x=3$

a. \(\left(\dfrac{-2}{7}+\dfrac{4}{7}\right)+\dfrac{1}{7}=\dfrac{2}{7}+\dfrac{1}{7}=\dfrac{3}{7}\)

b. \(\left(\dfrac{-8}{19}+\dfrac{27}{19}\right)+\left(\dfrac{-4}{21}+\dfrac{-17}{21}\right)+\dfrac{-12}{16}=\dfrac{19}{19}+\dfrac{-21}{21}+\dfrac{-12}{16}\)

\(=1+\left(-1\right)+\dfrac{-12}{16}=\dfrac{-12}{16}\)

c. \(\dfrac{-3}{7}+\dfrac{-11}{13}+\dfrac{-2}{13}+\dfrac{3}{7}+1=\left(\dfrac{-3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{-11}{13}+\dfrac{-2}{13}\right)+1=0-1+1=0\)

-2/7 + 2/7 : 3/5

= -2/7 + 2/7. 3/5

= -2/7 + 35/6

= -12/42 + 245/42

= -233/42

−819+−421−1721+271919−8​+21−4​−2117​+1927​
 

= (-4/21 - 17/21) + (-8/19 + 27/19)

= - 13/21 + -19/19

= (-13/21)+ (-1)

= (-13/21) + (-1/1)

= (-13/21) + (-13/21)

= - 26/21 

65.313−65.161356​.133​−56​.1316​
 

=6/5. (3/13 - 16/13)

= 6/5. (-13/13)

=6/5. (-1)

=6/5. (-1/1)

=6/5. (-6/5)

=(-36/25)

a) c1 := -2/-5+-5/-6+4/5

=2/5+5/6+4/5

=12/30+25/30+24/30

=61/30

c2:=(2/5+4/5)+4/5

=(2/5+4/5)+5/6

=6/5+5/6

=36/30+25/30

=61/30

b)c1:=-3/-4+11/-15+-1/2

=3/4+-11/15+-1/2

=45/60+-44/30+-30/60

-29/30

c2:=3/4+(-11/15+-1/2)

=(3/4+-1/2)+-11/15

=(3/4+-2/4)+-11/5

=1/4+-11/15

=15/60+-44/60

=-29/60

A = \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + \(\dfrac{1}{7^2}\) +.................+ \(\dfrac{1}{2004^2}\)

A = \(\dfrac{1}{5.5}\) + \(\dfrac{1}{6.6}\) + \(\dfrac{1}{7.7}\)+..............+ \(\dfrac{1}{2004.2004}\)

Vì \(\dfrac{1}{5}>\dfrac{1}{6}>\dfrac{1}{7}>...........>\dfrac{1}{2004}\)

nên ta có : \(\dfrac{1}{5.5}>\dfrac{1}{5.6}>\dfrac{1}{6.6}>\dfrac{1}{6.7}>\dfrac{1}{7.7}>.....>\dfrac{1}{2004.2004}>\dfrac{1}{2004.2005}\)

\(\dfrac{1}{5.5}+\dfrac{1}{6.6}+\dfrac{1}{7.7}+...+\dfrac{1}{2004.2004}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+..+\dfrac{1}{2004.2005}\)

A > \(\dfrac{1}{5}\) \(-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+....+\dfrac{1}{2004}-\dfrac{1}{2005}\)

A > \(\dfrac{1}{5}\) - \(\dfrac{1}{2005}\) = \(\dfrac{1}{5}\) - \(\dfrac{12}{24060}\)

\(\dfrac{1}{65}\) = \(\dfrac{1}{5}\) - \(\dfrac{12}{65}\) 

Vì \(\dfrac{12}{65}\) > \(\dfrac{12}{24060}\) nên A>  \(\dfrac{1}{65}\) ( phân số nào có phần bù nhỏ hơn thì phân số đó lớn hơn)

Tương tự ta có :

A = \(\dfrac{1}{5.5}\) + \(\dfrac{1}{6.6}\)+ \(\dfrac{1}{7.7}\)+......+\(\dfrac{1}{2004.2004}\) >\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+.....\(\dfrac{1}{2003.2004}\)

A < \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) +......+ \(\dfrac{1}{2003}\) - \(\dfrac{1}{2004}\)

A < \(\dfrac{1}{4}-\dfrac{1}{2004}\) < \(\dfrac{1}{4}\)

\(\dfrac{1}{65}< \)A < \(\dfrac{1}{4}\) (đpcm)

Số học sinh khá của lớp 6A:

\(45.60:100=27\) học sinh

Số học sinh giỏi của lớp 6A:

\(27.\dfrac{1}{3}=9\) học sinh

a) Số học sinh trung bình của lớp 6A:

\(45-27-9=9\) học sinh

b) Tỉ số phần trăm số học sinh giỏi so với học sinh cả lớp 6A:

\(9:45=\dfrac{1}{5}\) 

Đáp số: a) 9 học sinh

             b) \(\dfrac{1}{5}\)

S O S

A = \(\dfrac{n+2}{n-1}=\dfrac{n-1+3}{n-1}=1+\dfrac{3}{n-1}\)

Để A là số nguyên thì \(3⋮n-1\)

\(\Leftrightarrow n-1\inƯ\left(3\right)\)

\(\Leftrightarrow n-1\in\left\{1;3;-1;-3\right\}\)

\(\Leftrightarrow n\in\left\{2;4;0;-2\right\}\)

có: A=\(\dfrac{n+2}{n-1}\)=\(\dfrac{n-1+3}{n-1}\)=\(1+\dfrac{3}{n-1}\)

Để A nhận giá trị nguyên thì 3/n-1 có giá trị nguyên

=> n-1ϵƯ(3)

Ta có bảng sau:

Vậy nϵ\(\left\{-2;0;2;4\right\}\)

\(A=\dfrac{-9}{10^{2010}}+\dfrac{-19}{10^{2011}}=\dfrac{-90}{10^{2011}}+\dfrac{-19}{10^{2011}}=\dfrac{-109}{10^{2011}}\)

\(B=\dfrac{-9}{10^{2011}}+\dfrac{-19}{10^{2010}}=\dfrac{-9}{10^{2011}}+\dfrac{-190}{10^{2011}}=\dfrac{-199}{10^{2011}}\)

Mà \(\dfrac{-109}{10^{2011}}>\dfrac{-199}{10^{2011}}\)

\(\Rightarrow A>B\).

\(\dfrac{9}{23}+\dfrac{53}{3}+\dfrac{2}{69}+\left(-15\right)\)

\(=\dfrac{9.3}{23.3}+\dfrac{53.23}{3.23}+\dfrac{2}{69}+\dfrac{-15.69}{1.69}\)

\(=\dfrac{27}{69}+\dfrac{1219}{69}+\dfrac{2}{69}+\dfrac{-345}{69}\)

\(=\dfrac{903}{69}\)

\(=\dfrac{301}{23}\).

\(-\dfrac{3}{4}+\dfrac{5}{7}+\dfrac{7}{2}\)

\(=-\dfrac{21}{28}+\dfrac{20}{28}+\dfrac{98}{28}\)

\(=\dfrac{97}{28}\)