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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
\(A=\left(3,1-2,5\right)-\left(-2,5+3,1\right)\)
\(A=3,1-2,5+2,5-3,1\)
\(A=\left(3,1-3,1\right)-\left(2,5-2,5\right)\)
\(A=0-0\)
\(A=0\)
\(B=\left(5,3-2,8\right)-\left(4+5,3\right)\)
\(B=5,3-2,8-4-5,3\)
\(B=\left(5,3-5,3\right)-\left(2,8+4\right)\\ B=0-6,8\\ B=-6,8\)
>> 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 :
\(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=\left(6-5-3\right)+\left(\dfrac{-2}{3}+\dfrac{-5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{-5}{2}\right)\)
\(=\left(-2\right)+0+\dfrac{-1}{2}\)
\(=\dfrac{-5}{2}\)
A= (3,1 - 2,5) - (-2,5 + 3,1)
=3.1-2.5+2.5-3.1
=0
B= (5,3 - 2,8) - (4 + 5,3)
=5.3 - 2.8-4-5.3
= -6.8
C= -(251.3 + 281) + 3.251 - (1-281)
=-251.3-281+3.251-1+281
=-1
D= -(3/5+3/4)−(−3/4+2/5)
= -3/5-3/4+3/4-2/5
= -5/5
=-1
\(A=\left(3,1-2,5\right)-\left(-2,5+3,1\right)=3,1-2,5+2,5-3,1=0\)
\(B=\left(5,3-2,8\right)-\left(4+5,3\right)=5,3-2,8-4-5,3=-6,8\)
\(C=-\left(215\cdot3+281\right)+3\cdot215-\left(1-281\right)=-215\cdot3-281+3\cdot215-1+281=1\)
\(D=-\left(\frac{3}{5}+\frac{3}{4}\right)-\left(-\frac{3}{4}+\frac{2}{5}\right)=-\frac{3}{5}-\frac{3}{4}+\frac{3}{4}-\frac{2}{5}=-1\)
a)hình như đề sai thì phải
sửa lại
\(\left(\dfrac{1}{7}-\dfrac{2}{5}\right).\dfrac{2016}{2017}+\left(\dfrac{13}{7}+\dfrac{2}{5}\right).\dfrac{2016}{2017}\)
=\(\dfrac{2016}{2017}.\left(\dfrac{1}{7}-\dfrac{2}{5}+\dfrac{13}{7}+\dfrac{2}{5}\right)\)
=\(\dfrac{2016}{2017}.2=\dfrac{4032}{2017}\)
\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\left(1\right)\\ \Leftrightarrow1+\dfrac{a}{b}+\dfrac{b}{a}+1-4\ge0\\ \Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}-2\ge0\left(2\right)\)
Áp dụng t/c \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\) nên (2) luôn đúng.Do đó:(1) đúng
Vậy...(đpcm)
\(A=\left(3,1-2,5\right)-\left(-2,5+3,1\right)\)
\(=3,1-2,5+2,5-3,1=0\)
\(B=\left(5,3-2,8\right)-\left(4+5,3\right)\)
\(=5,3-2,8-4-5,3=-6,8\)
\(C=-\left(251.3+281\right)+3.215-\left(1-281\right)\)
\(=-753-281+645-1+281\)
\(=-753+654-1=-100\)
\(D=-\left(\dfrac{3}{5}+\dfrac{3}{4}\right)-\left(\dfrac{-3}{4}+\dfrac{2}{5}\right)\)
\(=-\dfrac{3}{5}-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{2}{5}=\dfrac{-3}{5}-\dfrac{2}{5}=-1\)
\(A=\left(3,1-2,5\right)-\left(-2,5+3,1\right)\)
\(=3,1-2,5+2,5-3,1\)
\(=\left(3,1-3,1\right)-\left(2,5-2,5\right)\)
\(=0\)
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.