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\(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}=\dfrac{x+1}{5}+\dfrac{x+1}{6}\)
\(\Leftrightarrow\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}-\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)
Mà \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\ne0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy ..
\(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}=\dfrac{x+1}{5}+\dfrac{x+1}{6}\)
=> \(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}-\dfrac{x+1}{5}-\dfrac{x+1}{6}\)= 0
(x + 1).(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\)) = 0
Ta thấy \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\) > 0
=> x + 1 = 0
x = 0 - 1
x = -1
a)hình như đề sai thì phải
sửa lại
\(\left(\dfrac{1}{7}-\dfrac{2}{5}\right).\dfrac{2016}{2017}+\left(\dfrac{13}{7}+\dfrac{2}{5}\right).\dfrac{2016}{2017}\)
=\(\dfrac{2016}{2017}.\left(\dfrac{1}{7}-\dfrac{2}{5}+\dfrac{13}{7}+\dfrac{2}{5}\right)\)
=\(\dfrac{2016}{2017}.2=\dfrac{4032}{2017}\)
>> Mình không chép lại đề bài nhé ! <<
Cách 1 :
\(A=\left(\dfrac{36-4+3}{6}\right)-\left(\dfrac{30+10-9}{6}\right)-\left(\dfrac{18-14+15}{6}\right)=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}=-\dfrac{15}{6}=-\dfrac{5}{2}\)
Cách 2 :
\(A=6-\dfrac{2}{3}+\dfrac{1}{2}-5+\dfrac{5}{3}-\dfrac{3}{2}-3-\dfrac{7}{3}+\dfrac{5}{2}\)
\(A=\left(6-5-3\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}-\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
\(A=-2-0-\dfrac{1}{2}=-\dfrac{5}{2}\)
Cách 1 :
\(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(=\left(\dfrac{36}{6}-\dfrac{4}{6}+\dfrac{3}{6}\right)-\left(\dfrac{30}{6}+\dfrac{10}{6}-\dfrac{9}{6}\right)-\left(\dfrac{18}{6}-\dfrac{14}{6}+\dfrac{15}{6}\right)\)
\(=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}\)
\(=-\dfrac{5}{2}\)
Cách 2 :
\(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=\left(6-5-3\right)+\left(\dfrac{-2}{3}+\dfrac{-5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{-5}{2}\right)\)
\(=\left(-2\right)+0+\dfrac{-1}{2}\)
\(=\dfrac{-5}{2}\)
Đăng từng bài một thôi bạn!
1)\(\left(-\dfrac{5}{13}\right)^{2017}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(-\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}.\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).1^{2016}\)
\(=-\dfrac{5}{13}\)
a/ \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{3}=0\)
\(\Leftrightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)
Vậy ..............
b, \(\dfrac{-12}{-37}=\dfrac{12}{37}< \dfrac{12}{36}=\dfrac{13}{39}< \dfrac{13}{38}\)
\(\Leftrightarrow\dfrac{13}{38}>\dfrac{-12}{-37}\)
a)\(\text{|}x+\dfrac{3}{4}\text{|}-\dfrac{1}{3}=0\)
=>\(\text{|}x+\dfrac{3}{4}\text{|}=\dfrac{1}{3}\)
=>\(x+\dfrac{3}{4}=-\dfrac{1}{3}\)hoặc\(x+\dfrac{3}{4}=\dfrac{1}{3}\)
=>\(x=-\dfrac{13}{12}\)hoặc\(x=-\dfrac{5}{12}\)
Vậy...
b)\(\dfrac{13}{38}\) và \(\dfrac{-12}{-37}\)
Ta có:\(\dfrac{-12}{-37}=\dfrac{12}{37}< \dfrac{12}{36}=\dfrac{1}{3}=\dfrac{13}{39}< \dfrac{13}{38}\)
=>\(\dfrac{13}{38}>\dfrac{-12}{-37}\)
\(A=\left(3,1-2,5\right)-\left(-2,5+3,1\right)\)
\(=3,1-2,5+2,5-3,1=0\)
\(B=\left(5,3-2,8\right)-\left(4+5,3\right)\)
\(=5,3-2,8-4-5,3=-6,8\)
\(C=-\left(251.3+281\right)+3.215-\left(1-281\right)\)
\(=-753-281+645-1+281\)
\(=-753+654-1=-100\)
\(D=-\left(\dfrac{3}{5}+\dfrac{3}{4}\right)-\left(\dfrac{-3}{4}+\dfrac{2}{5}\right)\)
\(=-\dfrac{3}{5}-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{2}{5}=\dfrac{-3}{5}-\dfrac{2}{5}=-1\)
\(A=\left(3,1-2,5\right)-\left(-2,5+3,1\right)\)
\(=3,1-2,5+2,5-3,1\)
\(=\left(3,1-3,1\right)-\left(2,5-2,5\right)\)
\(=0\)
Vậy \(A=0.\)
\(B=\left(5,3-2,8\right)-\left(4+5,3\right)\)
\(=5,3-2,8-4-5,3\)
\(=\left(5,3-5,3\right)-\left(2,8+4\right)\)
\(=-\dfrac{34}{5}\)
Vậy \(B=\dfrac{-34}{5}.\)
\(C=-\left(251.3+281\right)+3.215-\left(1-281\right)\)
\(=-251.3-281+3.215-1+281\)
\(=-753-\left(281-281\right)+645\)
\(=-753+645=-108\)
Vậy \(C=-108.\)
\(D=-\left(\dfrac{3}{5}+\dfrac{3}{4}\right)-\left(-\dfrac{3}{4}+\dfrac{2}{5}\right)\)
\(=-\dfrac{3}{5}-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{2}{5}\)
\(=\left(\dfrac{-3}{5}-\dfrac{2}{5}\right)-\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
\(=-1\)
Vậy D = -1.
A= \(\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}\)+ \(\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-\dfrac{9}{10}}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)
A= \(\dfrac{5}{13}\)+ \(\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{3}{10}\right)}{1\left(\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{3}{10}\right)}\)
A= \(\dfrac{5}{13}+3\) = \(\dfrac{44}{13}\)
\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\left(1\right)\\ \Leftrightarrow1+\dfrac{a}{b}+\dfrac{b}{a}+1-4\ge0\\ \Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}-2\ge0\left(2\right)\)
Áp dụng t/c \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\) nên (2) luôn đúng.Do đó:(1) đúng
Vậy...(đpcm)
a) \(\dfrac{12}{25}+\dfrac{5}{13}+\dfrac{13}{25}-\dfrac{18}{13}+\dfrac{3}{5}\)
\(=\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)+\dfrac{3}{5}\)
\(=1+\left(-1\right)+\dfrac{3}{5}\)
\(=0+\dfrac{3}{5}\)
\(=\dfrac{3}{5}\)
b) \(\dfrac{2}{5}:\left(\dfrac{5}{2}-\dfrac{3}{4}\right)\)
\(=\dfrac{2}{3}:\dfrac{3}{4}\)
\(=\dfrac{8}{21}\)
tích mình nha
a) \(\dfrac{12}{25}+\dfrac{5}{13}+\dfrac{13}{25}-\dfrac{18}{13}+\dfrac{3}{5}\)
= \(\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)+\dfrac{3}{5}\)
= 1 + (-1) + \(\dfrac{3}{5}\)
= 0 + \(\dfrac{3}{5}\) = \(\dfrac{3}{5}\)
b) \(\dfrac{2}{3}:\left(\dfrac{5}{2}-\dfrac{3}{4}\right)\)
=\(\dfrac{2}{3}:\left(\dfrac{10}{4}-\dfrac{3}{4}\right)\)
=\(\dfrac{2}{3}:\dfrac{7}{4}\)
=\(\dfrac{2}{3}.\dfrac{4}{7}\)
=\(\dfrac{8}{21}\)
Chúc bạn học tốt!!!