\(\dfrac{3}{4a}=\dfrac{4}{a}=\dfrac{13}{20}\Rightarrow a=??\)

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10 tháng 7 2021

Ủa đề có lộn không em ? 

10 tháng 7 2021

sai ở đâu ạ??

29 tháng 7 2018

A)0,25:(10,3-9,8)-3/4

=1/4:(103/10-49/5)-3/4

=1/4:1/2-3/4

=1/2-3/4

=2/4-3/4

=-1/4

B)-5/9.13/28-13/28.4/9

=-5/9-4/9.13/28

=-1.13/28

=-13/28

c)6/7+5/8:5-3/16

=6/7+1/8-3/16

=55/56-3/16

=89/112

d)-5/7.2/11+-5/7.9/11+1/5/7

=-5/7.(2/11+9/11)+12/7

=-5/7.1+12/7

=-5/7+12/7

=1

e)-7/12-8/15+11/20

=-67/60+11/20

=-17/30

f)-17/25.20/33+-17/25.13/33+-3/25

=-17/25.(20/33+13/33)-3/25

=-17/25.1-3/25

=-17/25-3/25

=-4/5

CHÚC BẠN HỌC TỐT...............

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hihihihihihi

13 tháng 7 2017

\(A.\dfrac{-5}{13}.\dfrac{-4}{13}.\dfrac{-3}{13}.....\dfrac{4}{13}.\dfrac{5}{13}\)

\(A=\dfrac{-5.-4.-3.-2.-1.0.1.2.3.4}{13^{10}}\)

\(A=\dfrac{0}{13^{10}}=0\)

13 tháng 7 2017

\(A=\dfrac{-5}{13}\cdot\dfrac{-4}{13}\cdot\dfrac{-3}{13}\cdot...\cdot\dfrac{4}{13}\cdot\dfrac{5}{13}\)

\(A=\dfrac{\left(-5\cdot5\right)\cdot\left(-4\cdot4\right)\cdot...\cdot\left(-1\cdot1\right)\cdot0}{13\cdot13\cdot13\cdot...\cdot13}\)

\(A=\dfrac{0}{13\cdot13\cdot13\cdot...\cdot13}\)

\(A=0\)

3 tháng 8 2018

\(a)\dfrac{3}{4}+\dfrac{6}{12}-\dfrac{5}{24}\)

\(=\dfrac{18}{24}+\dfrac{12}{24}+\left(-\dfrac{5}{24}\right)\)

\(=\dfrac{18+12+\left(-5\right)}{24}\)

\(=\dfrac{25}{24}\)

\(b)\dfrac{-5}{7}.\dfrac{2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}\)

\(=\dfrac{5}{7}.\dfrac{-2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}\)

\(=\dfrac{5}{7}\left(\dfrac{-2}{13}+\dfrac{-11}{13}+\dfrac{13}{13}\right)\)

\(=\dfrac{5}{7}.0=0\)

\(c)\dfrac{27}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}\)

\(=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)

\(=1+1+\dfrac{1}{2}\)

\(=2\dfrac{1}{2}\)

\(d)\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}.\dfrac{20}{15}+\dfrac{3}{7}\)

\(=\dfrac{315}{714}+\dfrac{238}{714}+\dfrac{38}{51}+\dfrac{306}{714}\)

\(=\dfrac{315}{714}+\dfrac{238}{714}+\dfrac{532}{714}+\dfrac{306}{714}\)

\(=\dfrac{1391}{714}\)

3 tháng 8 2018

a)\(\dfrac{3}{4}+\dfrac{6}{12}-\dfrac{5}{24}=\dfrac{18}{24}+\dfrac{12}{24}-\dfrac{5}{24}=\dfrac{25}{24}\)

b)\(\dfrac{-5}{7}.\dfrac{2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}=\dfrac{5}{7}\left(\dfrac{-2}{13}-\dfrac{11}{13}+1\right)=\dfrac{5}{7}.0=0\)

c)\(\dfrac{27}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}=1+1+\dfrac{1}{2}=2,5\)

d)\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}.\dfrac{20}{15}+\dfrac{3}{7}=\dfrac{15}{34}+\left(\dfrac{1}{3}+\dfrac{38}{51}+\dfrac{3}{7}\right)=\dfrac{15}{34}+\dfrac{538}{357}=\dfrac{1391}{714}\)

Từ \(\dfrac{b}{c}=\dfrac{3}{4}\) =>\(\dfrac{b}{3}=\dfrac{c}{4}\) => \(\dfrac{b}{12}=\dfrac{c}{16}\) (1)

Từ \(\dfrac{a}{1}=\dfrac{b}{4}\) =>\(\dfrac{a}{3}=\dfrac{b}{12}\) (2)

Từ (1) và (2) => \(\dfrac{a}{3}=\dfrac{b}{12}=\dfrac{c}{16}=\dfrac{4a}{12}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\dfrac{a}{3}=\dfrac{b}{12}=\dfrac{c}{16}=\dfrac{4a}{12}=\dfrac{4a+b-c}{12+12-16}=\dfrac{8}{8}=1\)

=>\(\left\{{}\begin{matrix}a=3\cdot1=3\\b=12\cdot1=12\\c=16\cdot1=16\end{matrix}\right.\)

2 tháng 8 2018

bài này bạn cứ đặt a=bk, c=dk là được dễ tính lắm sao đó thì thay vào rồi rút gọn là được khi đó bạn sẽ chứng minh được dễ dàng hihi

3 tháng 8 2018

bạn giải luôn giúp mình nha Huyền Anh Lê

28 tháng 2 2017

giúp mình với mọi người ơi

28 tháng 2 2017

a) \(A=\left(\dfrac{1}{2^3}.3.\dfrac{13}{3}\right)\left(a^{3+2+1}\right)\left(x^{1+3}\right)\left(y^{1+2}\right)=\dfrac{13}{8}.a^6.x^4.y^3\)

\(B=\left[2^k.\left(-\dfrac{1}{2}\right)^2\right]\left(x^{2k+2}\right)\left(y^{3k+2.2}\right)\left(z^{4k+}\right)=2^{k-2}.x^{2\left(k+1\right)}.y^{3k+4}.z^{4k}\)

27 tháng 6 2017

\(A=\left(\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}\right)+\left(\dfrac{-6}{13}+\dfrac{1}{2}+1\dfrac{1}{3}\right)\)

\(A=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}-\dfrac{6}{13}+\dfrac{1}{2}+\dfrac{4}{3}\)

\(A=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{7}{13}+\dfrac{6}{13}\right)+\left(\dfrac{4}{3}-\dfrac{1}{3}\right)\)

\(A=1-1+1=1\)

\(B=\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right).\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}\)

\(B=\dfrac{-3}{2}:\dfrac{3}{-4}.\dfrac{-9}{2}-\dfrac{1}{4}\)

\(B=2.\dfrac{-9}{2}-\dfrac{1}{4}\)

\(=-9-\dfrac{1}{4}=\dfrac{-37}{4}\)

27 tháng 6 2017

\(a,A=\left(\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}\right)+\left(-\dfrac{6}{13}+\dfrac{1}{2}+1\dfrac{1}{3}\right)\)

\(A=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}+\dfrac{-6}{13}+\dfrac{1}{2}+\dfrac{4}{3}\)

\(A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{7}{13}-\dfrac{6}{13}\right)+\left(-\dfrac{1}{3}+\dfrac{4}{3}\right)\)

\(A=-1+1=0\)

\(b,B=\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right)\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}\)

\(B=\left(-\dfrac{3}{2}.\dfrac{-4}{3}\right).\dfrac{-9}{2}-\dfrac{1}{4}\)

\(B=8.\dfrac{-9}{2}-\dfrac{1}{4}\)

\(B=-36-\dfrac{1}{4}\)

B = \(-\dfrac{145}{4}\)

27 tháng 6 2017

a, \(\dfrac{3}{4}+x=\dfrac{8}{13}\)

\(x=\dfrac{8}{13}-\dfrac{3}{4}\)

\(x=-\dfrac{7}{52}\)

b,\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{1}{4}\)

\(x=\dfrac{1}{4}-\dfrac{2}{5}\)

\(x=-\dfrac{3}{20}\)

c, \(2x\left(x-\dfrac{1}{7}\right)=0\)

\(2x-\dfrac{1}{7}=0\)

\(x-\dfrac{1}{7}=0:2\)

\(x-\dfrac{1}{7}=0\)

\(x=0-\dfrac{1}{7}\)

\(x=\dfrac{1}{7}\)

d, \(\dfrac{3}{4}+\dfrac{1}{4}\div x=\dfrac{2}{5}\)

\(\left(\dfrac{3}{4}+\dfrac{1}{4}\right):x=\dfrac{2}{5}\)

\(1:x=\dfrac{2}{5}\)

\(x=1:\dfrac{2}{5}\)

\(x=\dfrac{5}{2}\)

27 tháng 6 2017

a) \(\dfrac{3}{4}+x=\dfrac{8}{13}\)\(\Leftrightarrow\) \(x=\dfrac{8}{13}-\dfrac{3}{4}=\dfrac{-7}{52}\) vậy \(x=\dfrac{-7}{52}\)

b) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\) \(\Leftrightarrow\) \(\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\) \(\Leftrightarrow\) \(x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=\dfrac{-3}{20}\) vậy \(x=\dfrac{-3}{20}\)

c) \(2x\left(x-\dfrac{1}{7}\right)=0\) \(\Leftrightarrow\) \(2x^2-\dfrac{2}{7}x=0\)

\(\Delta\) = \(\left(\dfrac{-2}{7}\right)^2-4.2.0=\dfrac{4}{49}>0\)

\(\Rightarrow\) phương trình có 2 nghiệm phân biệt

\(x_1=\dfrac{\dfrac{2}{7}+\sqrt{\dfrac{4}{49}}}{4}=\dfrac{1}{7}\)

\(x_2=\dfrac{\dfrac{2}{7}-\sqrt{\dfrac{4}{49}}}{4}=0\)

vậy \(x=0;x=\dfrac{1}{7}\)