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\(A=\dfrac{12}{25}-\dfrac{4}{11}+\dfrac{13}{25}+1\dfrac{4}{9}-\dfrac{7}{11}\)
\(A=\dfrac{12}{25}+\dfrac{13}{25}-\dfrac{4}{11}-\dfrac{7}{11}+1\dfrac{4}{9}\)
\(A=\dfrac{12+15}{25}-\dfrac{4-7}{11}+\dfrac{13}{9}\)
\(A=\dfrac{25}{25}+\dfrac{3}{11}+\dfrac{13}{9}\)
\(A=1+\dfrac{3}{11}+\dfrac{13}{9}\)
\(A=\dfrac{11+3}{11}+\dfrac{13}{9}=\dfrac{14}{11}+\dfrac{13}{9}\)
\(A=\dfrac{126+141}{99}=\dfrac{42+47}{22}=\dfrac{89}{22}\)
\(=4\dfrac{1}{22}\)
a, \(\dfrac{2}{3}-4\left(\dfrac{1}{2}+\dfrac{3}{4}\right)=\dfrac{2}{3}-4.\dfrac{5}{4}=\dfrac{2}{3}-5=-\dfrac{13}{3}\)
b, \(-\dfrac{2}{3}.\dfrac{3}{11}+\dfrac{-16}{9}.\dfrac{3}{11}=\dfrac{3}{11}.\left(-\dfrac{2}{3}+\dfrac{-16}{9}\right)\)
\(=\dfrac{3}{11}.\dfrac{-22}{9}=-\dfrac{2}{3}\)
Câu c lấy máy tính tính nha!
c,
c) \(\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{10}{14}+\dfrac{8}{18}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}+\left(-\dfrac{17}{18}+\dfrac{8}{18}\right)+\left(-\dfrac{10}{14}+\dfrac{17}{14}\right)\)
\(=\dfrac{11}{125}+\left(-\dfrac{9}{18}\right)+\dfrac{7}{14}\)
\(=\dfrac{11}{125}+\left(-\dfrac{1}{2}\right)+\dfrac{1}{2}\)
\(=\dfrac{11}{125}+0=\dfrac{11}{125}\)
Ta có : a, 25/7 + 13/21 - 11/7 + 17/21 + 1/3 .
= ( 25/7 - 11/7 ) + ( 13/21 + 17/21 + 1/3 ) .
= 2 + ( 20/21 + 7/21 ) .
= 2 + 9/7 .
= 23/7 .
b, ( 1/3 + 12/67 + 13/41 ) - ( 79/67 - 28/41 ) .
= 1/3 + 12/67 + 13/41 - 79/67 + 28/41 .
= 1/3 + ( 12/67 - 79/67 ) + ( 13/41 + 28/41 ) .
= 1/3 - 1 + 1 .
= 1/3 .
c, ( 11/4 . -5/9 - 4/9 . 11/4 ) . 8/33 .
= [ 11/4 . ( -5/9 - 4/9 ) ] . 8/33 .
= [ 11/4 . ( - 1 ) ] . 8/33 .
= -11/4 . 8/33 .
= -2/3 .
d, 38/45 - ( 8/45 - 17/51 - 3/11 ) .
= 38/45 - 8/45 + 17/51 + 3/11 .
= 2/3 + 17/51 + 3/11 .
= 374/561 + 187/561 + 153/561 .
= 14/11 .
a: \(A=\dfrac{-13}{21}=\dfrac{-26}{42}\)
\(B=\dfrac{-9}{14}=\dfrac{-27}{42}\)
mà -26>-27
nên A>B
b: \(A=\dfrac{99}{101}=1-\dfrac{2}{101}\)
\(B=\dfrac{2011}{2013}=1-\dfrac{2}{2013}\)
mà 2/101>2/2013
nên A<B
\(a)\dfrac{-1}{4}.13\dfrac{9}{11}-0,25.6\dfrac{2}{11}\)
\(=\dfrac{-1}{4}.\dfrac{152}{11}-\dfrac{1}{4}.\dfrac{68}{11}\)
\(=-38-\dfrac{17}{11}\)
\(=\dfrac{-418}{11}+\dfrac{-17}{11}\)
\(=\dfrac{-435}{11}\)
\(b)\dfrac{31}{9}.\left|x\right|-\dfrac{5}{2}=\dfrac{8}{3}\)
\(\Leftrightarrow\dfrac{31}{9}.\left|x\right|=\dfrac{16}{6}+\dfrac{15}{6}\)
\(\Leftrightarrow\dfrac{31}{9}.\left|x\right|=\dfrac{31}{6}\)
\(\Leftrightarrow\left|x\right|=\dfrac{31}{6}.\dfrac{9}{31}\)
\(\Leftrightarrow\left|x\right|=1,5\)
\(\Leftrightarrow x\in\left\{1,5;-1,5\right\}\)
Vậy \(x\in\left\{1,5;-1,5\right\}\)
Lời giải:
a)\(\dfrac{a}{b}=\dfrac{3}{4}\Leftrightarrow4a=3b\)
Và \(4a.5=3b.5\Leftrightarrow20a=15b\Leftrightarrow\dfrac{20a}{3}=5b\)
Khi đó:
\(A=\dfrac{2a-5b}{a-3b}=\dfrac{2a-\dfrac{20}{3}a}{a-4a}=\dfrac{-\dfrac{14}{3}a}{-3a}=\dfrac{-14}{\dfrac{3}{-3}}=14\)
b) Ta có:
\(a-b=7\Leftrightarrow b=a-7\)
\(B=\dfrac{3a-b}{2a+7}+\dfrac{3b-a}{2b-7}=\dfrac{3a-\left(a-7\right)}{2a+7}+\dfrac{3\left(a-7\right)-a}{2\left(a-7\right)-7}\)
\(B=\dfrac{3a-a+7}{2a+7}+\dfrac{3a-21-a}{2a-14-7}\)
\(B=\dfrac{2a+7}{2a+7}+\dfrac{2a-21}{2a-21}=1+1=2\)
\(\dfrac{15}{11}+\dfrac{8}{13}-\dfrac{4}{11}+\dfrac{18}{13}\)
\(=\left(\dfrac{15}{11}-\dfrac{4}{11}\right)+\left(\dfrac{8}{13}+\dfrac{18}{13}\right)\)
\(=1+2\)
\(=3\)
\(\dfrac{15}{11}+\dfrac{8}{13}-\dfrac{4}{11}+\dfrac{18}{13}=\left(\dfrac{15}{11}-\dfrac{4}{11}\right)+\left(\dfrac{8}{13}+\dfrac{18}{13}\right)=\dfrac{11}{11}+\dfrac{26}{13}=1+2=3\)
a)\(\left(\dfrac{-3}{7}+\dfrac{5}{11}\right):\dfrac{-3}{5}+\left(\dfrac{-9}{7}+\dfrac{6}{11}\right):\dfrac{-3}{5}\)
=\(\dfrac{2}{77}\):\(\dfrac{-3}{5}+\left(\dfrac{-57}{77}\right):\dfrac{-3}{5}\)
=[\(\dfrac{2}{77}+\left(\dfrac{-57}{77}\right)\)]:\(\dfrac{-3}{5}\)
=\(\left(\dfrac{_{ }-5}{7}\right):\dfrac{-3}{5}\)
=\(\dfrac{25}{21}\)
b)\(\left(x-\dfrac{1}{4}\right)^3=27\)
⇔\(\left(x-\dfrac{1}{4^{ }}\right)^3=3^3\)
⇔\(x-\dfrac{1}{4}=3\)
⇔\(x=3+\dfrac{1}{4}\)
⇔\(x=\dfrac{13}{4}\)
Vậy \(x=\dfrac{13}{4}\)
\(a,A=\dfrac{\dfrac{3}{4}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\\ A=\dfrac{\dfrac{405}{572}}{\dfrac{645}{1001}}+\dfrac{\dfrac{5}{12}}{\dfrac{25}{24}}\\ A=\dfrac{189}{172}+\dfrac{2}{5}\\ A=\dfrac{1289}{860}\)
\(a-b=11\)
\(P=\dfrac{5a-b}{4a+11}+\dfrac{5b-a}{4b-11}=\dfrac{5a-b}{4a+a-b}+\dfrac{5b-a}{4b-\left(a-b\right)}\)
\(=\dfrac{5a-b}{5a-b}+\dfrac{5b-a}{5b-a}\)
\(=2\)
Vậy...