Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a: \(=17+\dfrac{2}{31}-\dfrac{15}{17}-6-\dfrac{2}{31}=11-\dfrac{15}{17}=\dfrac{172}{17}\)
b: \(=31+\dfrac{6}{13}+5+\dfrac{9}{41}-36-\dfrac{9}{41}-36-\dfrac{6}{13}\)
=36
c: \(=27+\dfrac{51}{59}-7-\dfrac{51}{59}+\dfrac{1}{3}=20+\dfrac{1}{3}=\dfrac{61}{3}\)
a,
\(\dfrac{89}{-13}< 0< \dfrac{1}{123}\\ \Rightarrow\dfrac{89}{-13}< \dfrac{1}{123}\)
Vậy \(\dfrac{89}{-13}< \dfrac{1}{123}\)
b,
\(\dfrac{-13}{15}>\dfrac{-15}{15}=-1=\dfrac{-30}{30}>\dfrac{-31}{30}\)
Vậy \(\dfrac{-13}{15}>\dfrac{-31}{30}\)
c,
\(\dfrac{125}{123}=\dfrac{123}{123}+\dfrac{2}{123}=1+\dfrac{2}{123}\\ \dfrac{99}{97}=\dfrac{97}{97}+\dfrac{2}{97}=1+\dfrac{2}{97}\)
Vì \(\dfrac{2}{97}>\dfrac{2}{123}\Rightarrow1+\dfrac{2}{97}>1+\dfrac{2}{123}\Leftrightarrow\dfrac{99}{97}>\dfrac{125}{123}\)
Vậy \(\dfrac{99}{97}>\dfrac{125}{123}\)
d,
\(\dfrac{125}{126}< \dfrac{126}{126}=1=\dfrac{986}{986}< \dfrac{987}{986}\)
Vậy \(\dfrac{125}{126}< \dfrac{987}{986}\)
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, 1/3-3/4+3/5+1/4-2/9-1/36+1/15
=(1/3+3/5+1/15)-(3/4-1/4+2/9+1/36)
=1 - 3/4
=1/4
b, 3-1/4+2/3-5-1/3+6/5-6+7/4-3/2
=(3-5-6)-(1/4-7/4)+(2/3-1/3)+(6/5-3/2)
=-8 +3/2 +1/3 -3/10
=-97/15
a: \(\dfrac{17}{30}=\dfrac{1564}{30\cdot92}\)
\(\dfrac{51}{92}=\dfrac{1530}{30\cdot92}\)
mà 1564>1530
nên 17/30>51/92
b: \(\dfrac{-45}{47}>-1>-\dfrac{31}{30}\)
c: \(\dfrac{67}{22}=3+\dfrac{1}{22}\)
\(\dfrac{152}{51}=3+\dfrac{1}{51}\)
mà 1/22>1/51
nên 67/22>152/51
=>22/67<51/152
d: 17/39>17/41
nên -17/39<-17/41
=>-18/39<-17/39<-17/41
a: \(=0.5\cdot10-\dfrac{1}{7}+15=20-\dfrac{1}{7}=\dfrac{139}{7}\)
b: \(=6\cdot\dfrac{-2}{3}+12\cdot\dfrac{4}{9}+18\cdot\dfrac{-8}{27}\)
\(=-4+\dfrac{16}{3}-\dfrac{16}{3}=-4\)
c: \(=\left(\dfrac{5}{2}+\dfrac{3}{8}-\dfrac{5}{8}+\dfrac{2}{3}\right):\left(\dfrac{17}{2}+\dfrac{49}{4}-\dfrac{17}{8}+\dfrac{34}{15}\right)\)
\(=\dfrac{35}{12}:\dfrac{2507}{120}=\dfrac{350}{2507}\)
a: \(\Leftrightarrow\left(3x-2\right):\dfrac{7}{5}=\dfrac{17}{7}:\dfrac{13}{5}=\dfrac{85}{91}\)
\(\Leftrightarrow3x-2=\dfrac{85}{91}\cdot\dfrac{7}{5}=\dfrac{17}{13}\)
=>3x=43/13
hay x=43/39
b: \(\Leftrightarrow9x+207=121-8x\)
=>19x=-86
hay x=-86/19
c: \(\Leftrightarrow x^2-9=16\)
=>x2=25
=>x=5 hoặc x=-5
d: \(\Leftrightarrow\left|x\right|=\dfrac{1.64\cdot3.11}{8.51}\simeq0,6\)
=>x=0,6 hoặc x=-0,6
a/ \(x+\dfrac{3}{5}=\dfrac{4}{7}\)
\(x=\dfrac{4}{7}-\dfrac{3}{5}\)
\(x=-\dfrac{1}{35}\)
Vậy ....
b/ \(x-\dfrac{5}{6}=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}+\dfrac{5}{6}\)
\(x=1\)
Vậy ....
c/\(-\dfrac{5}{7}-x=\dfrac{-9}{10}\)
\(x=\dfrac{-5}{7}-\dfrac{-9}{10}\)
\(x=\dfrac{13}{70}\)
Vậy .....
d/ \(\dfrac{5}{7}-x=10\)
\(x=\dfrac{5}{7}-10\)
\(x=\dfrac{-65}{7}\)
Vậy ...
e/ \(x:\left(\dfrac{1}{9}-\dfrac{2}{5}\right)=\dfrac{-1}{2}\)
\(x:\dfrac{-13}{45}=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}.\dfrac{-13}{45}\)
\(x=\dfrac{13}{90}\)
Vậy ....
f/ \(\left(\dfrac{-3}{5}+1,25\right)x=\dfrac{1}{3}\)
\(0,65.x=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:0,65\)
\(x=\dfrac{20}{39}\)
Vậy ....
g/ \(\dfrac{1}{3}x+\left(\dfrac{2}{3}-\dfrac{4}{9}\right)=\dfrac{-3}{4}\)
\(\dfrac{1}{3}x+\dfrac{2}{9}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{3}x=\dfrac{-35}{36}\)
\(\Leftrightarrow x=\dfrac{-35}{12}\)
Vậy ...
\(A=17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11-\dfrac{15}{17}=\dfrac{172}{17}\)
\(B=\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{12}=36\dfrac{363}{533}-36\dfrac{6}{12}=\dfrac{193}{1066}\)
\(C=27\dfrac{51}{59}-\left(7\dfrac{51}{59}-\dfrac{1}{3}\right)=27\dfrac{51}{59}-7\dfrac{51}{59}+\dfrac{1}{3}=20+\dfrac{1}{3}=\dfrac{61}{3}\)
\(A=17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=\left(17\dfrac{2}{31}-6\dfrac{2}{31}\right)-\dfrac{15}{17}=11-\dfrac{15}{17}=\dfrac{172}{17}\)
\(B=\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{12}=36\dfrac{363}{533}-36\dfrac{1}{2}=\dfrac{193}{1066}\) (Casio :>)
\(C=27\dfrac{51}{59}-\left(7\dfrac{51}{59}-\dfrac{1}{3}\right)=27\dfrac{51}{59}-7\dfrac{51}{59}+\dfrac{1}{3}\)
\(=20+\dfrac{1}{3}=\dfrac{61}{3}\)