\(P1=\left(\dfrac{-43}{51}\right).\left(\dfrac{-19}{80}\right)tinhp1\)
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9: \(=\dfrac{47}{51}\cdot\dfrac{17}{94}-\dfrac{47}{51}\cdot\dfrac{53}{91}-\dfrac{53}{91}\cdot\dfrac{91}{53}+\dfrac{53}{91}\cdot\dfrac{47}{51}\)
\(=\dfrac{1}{6}-1=-\dfrac{5}{6}\)
10: \(=\dfrac{13}{19}\cdot\dfrac{19}{26}-\dfrac{13}{19}\cdot\dfrac{71}{43}+\dfrac{71}{43}\cdot\dfrac{13}{19}-\dfrac{71}{43}\cdot\dfrac{86}{71}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}\)
\(A=\left(-\dfrac{43}{51}\right)\left(-\dfrac{19}{80}\right)\)
=>A>0(1)
\(B=\left(-\dfrac{7}{13}\right)\left(-\dfrac{4}{65}\right)\left(-\dfrac{8}{21}\right)\)
=>B<0(2)
C\(=-\dfrac{5}{10}.\left(-\dfrac{4}{10}\right).....\left(\dfrac{4}{10}\right)\left(\dfrac{5}{10}\right)=0\)
=>C=0(3)
Từ 1;2;3 =>A>C>B
\(A=\dfrac{-43}{51}.\dfrac{-19}{80}\Leftrightarrow A>0\left(1\right)\)
\(B=\left(\dfrac{-7}{13}\right).\left(-\dfrac{4}{65}\right).\left(\dfrac{-8}{31}\right)\Leftrightarrow B< 0\left(2\right)\)
\(C=\dfrac{-5}{10}.\dfrac{-4}{10}...........\dfrac{3}{10}.\dfrac{4}{10}.\dfrac{5}{10}\Leftrightarrow C=0\left(3\right)\)
Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow A>C>B\)
a) (1/7.x-2/7).(-1/5.x-2/5)=0
=> 1/7.x-2/7=0hoặc-1/5.x-2/5=0
*1/7.x-2/7=0
1/7.x=0+2/7
1/7.x=2/7
x=2/7:1/7
x=2
b)1/6.x+1/10.x-4/5.x+1=0
(1/6+1/10-4/5).x+1=0
(1/6+1/10-4/5).x=0-1
(1/6+1/10-4/5).x=-1
(-8/15).x=-1
x=-1:(-8/15) =15/8
P1 có 2 thừa số âm nhân với nhau nên P1 >0
P2 có 3 thừa số âm nhân với nhau nên P2 <0
P3 vì trong dấu ba chấm có thừa số 0/10 nên P3 =0
Vậy P2 < P3 < P1
KO khó lắm đâu. Mong bạn hiểu để bài sau tương tự thì làm được.
CHúc bạn học tốt.
\(\dfrac{\left(17\dfrac{8}{19}-16\dfrac{9}{18}\right).\left(17,5+16\dfrac{17}{51}-32\dfrac{15}{22}\right)}{\dfrac{7}{3.13}+\dfrac{7}{13.23}+\dfrac{7}{23.33}}\)
=\(\dfrac{\dfrac{35}{38}.\dfrac{38}{33}}{\dfrac{7}{10}\left(\dfrac{1}{3}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{33}\right)}\)
=\(\dfrac{\dfrac{35}{33}}{\dfrac{7}{10}.\left(\dfrac{1}{3}-\dfrac{1}{33}\right)}\)
=\(\dfrac{\dfrac{35}{33}}{\dfrac{7}{10}.\dfrac{10}{33}}\)
=\(\dfrac{\dfrac{35}{33}}{\dfrac{7}{33}}\)
=\(\dfrac{35}{33}:\dfrac{7}{33}\)
=\(\dfrac{35}{33}.\dfrac{33}{7}\)
=5
\(A=\dfrac{\left(1+17\right).\left(1+\dfrac{17}{2}\right)..........\left(1+\dfrac{17}{19}\right)}{\left(1+19\right).\left(1+\dfrac{19}{2}\right)..........\left(1+\dfrac{19}{17}\right)}\)
\(=\dfrac{18.\dfrac{19}{2}.............\dfrac{36}{19}}{20.\dfrac{21}{2}..........\dfrac{36}{17}}\)
\(=\dfrac{18.19.20.......36}{1.2.3...19}:\dfrac{20.21.....36}{1.2.3...17}\)
\(=\dfrac{1.2.3......36}{1.2.....36}\)
\(=1\)
a: \(=\dfrac{3}{8}\left(27+\dfrac{1}{5}-51-\dfrac{1}{5}\right)+19\)
\(=-24\cdot\dfrac{3}{8}+19=-9+19=10\)
b: \(=\left(35+\dfrac{1}{6}-46-\dfrac{1}{6}\right):\left(\dfrac{-4}{5}\right)\)
\(=\dfrac{-11\cdot5}{-4}=\dfrac{55}{4}\)
c: \(=\left(\dfrac{-15+8}{20}\right):\left[\dfrac{3}{7}+\dfrac{7}{3}\cdot\dfrac{12-5}{20}\right]\)
\(=\dfrac{-7}{20}:\left(\dfrac{3}{7}+\dfrac{49}{60}\right)\)
\(=-\dfrac{147}{523}\)
1/Ta co :
\(\dfrac{15}{7}.\left(\dfrac{1}{5}-\dfrac{46}{45}\right)+\dfrac{46}{45}.\left(\dfrac{15}{7}-\dfrac{45}{46}\right)\)
=\(\dfrac{15}{7}.\dfrac{1}{5}-\dfrac{15}{7}.\dfrac{46}{45}+\dfrac{46}{45}.\dfrac{15}{7}-\dfrac{46}{45}.\dfrac{45}{46}\)
=\(\dfrac{15}{7}.\left(\dfrac{1}{5}-\dfrac{46}{45}+\dfrac{46}{45}\right)-1\)
=\(\dfrac{15}{7}.\dfrac{1}{5}-1\)
=\(\dfrac{3}{7}-1=\dfrac{-4}{7}\)
2/Ta co
\(\dfrac{43}{47}.\left(\dfrac{18}{37}+\dfrac{47}{43}\right)-\dfrac{18}{3}.\left(\dfrac{43}{47}+\dfrac{37}{36}\right)\)
=\(\dfrac{43}{47}.\dfrac{18}{37}+\dfrac{43}{47}.\dfrac{47}{43}-\dfrac{18}{37}.\dfrac{43}{47}+\dfrac{18}{37}.\dfrac{37}{36}\)
=\(\dfrac{18}{37}.\left(\dfrac{43}{37}-\dfrac{43}{37}+\dfrac{37}{36}\right)+1\)
=\(\dfrac{18}{37}.\dfrac{37}{36}+1\)
=\(\dfrac{1}{2}+1=\dfrac{3}{2}\)
tick cho mk nha
Lời giải:
1.
$(\frac{5}{6})^{10}.(\frac{3}{10})^{10}=(\frac{5}{6}.\frac{3}{10})^{10}=(\frac{1}{4})^{10}$
$=\frac{1}{4^{10}}$
2.
$(\frac{4}{7})^{19}: (\frac{-12}{35})^{19}=(\frac{4}{7}: \frac{-12}{35})^{19}=(\frac{-5}{3})^{19}$
3.
$(\frac{-3}{7})^7:\frac{-3}{5}=\frac{(-3)^7}{7^7}.\frac{5}{-3}=\frac{5.(-3)^6}{7^7}=\frac{5.3^6}{7^7}$
\(\dfrac{-43}{51}\). \(\dfrac{-19}{80}\)=\(\dfrac{817}{4080}\)