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10a=10^2017+10/10^2017+1
10b=10^2018+10/10^2018+1
cậu tự so sánh nhé vậy là dễ rồi
Ta có: \(A=\dfrac{10^{2016}+1}{10^{2017}+1}\Rightarrow10A=\dfrac{10\left(10^{2016}+1\right)}{10^{2017}+1}=\dfrac{10^{2017}+10}{10^{2017}+1}\)
\(=\dfrac{10^{2017}+1+9}{10^{2017}+1}=\dfrac{10^{2017}+1}{10^{2017}+1}+\dfrac{9}{10^{2017}+1}=1+\dfrac{9}{10^{2017}+1}\)
Tương tự ta cũng có: \(10B=1+\dfrac{9}{10^{2018}+1}\)
Lại có: \(10^{2017}< 10^{2018}\Rightarrow10^{2017}+1< 10^{2018}+1\)
\(\Rightarrow\dfrac{1}{10^{2017}+1}>\dfrac{1}{10^{2018}+1}\Rightarrow\dfrac{9}{10^{2017}+1}>\dfrac{9}{10^{2018}+1}\)
\(\Rightarrow1+\dfrac{9}{10^{2017}+1}>1+\dfrac{9}{10^{2018}+1}\Rightarrow10A>10B\Rightarrow A>B\)
a, \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy x = -1
b, \(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2014}+1\right)+\left(\dfrac{x+3}{2015}+1\right)=\left(\dfrac{x+2}{2016}+1\right)+\left(\dfrac{x+1}{2017}+1\right)\)\(\Leftrightarrow\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\Leftrightarrow\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}-\dfrac{x+2018}{2016}-\dfrac{x+2018}{2017}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\)
\(\Leftrightarrow xx+2018=0\Leftrightarrow x=-2018\)
Vậy x = -2018
Nguyễn Huy Tú, cho mk hỏi sao câu a bt đó lại bằng 0 vậy ? Mk ko hiểu lắm
Bài 1 :
a, Ta có :
\(\dfrac{a}{b}< \dfrac{c}{d}\Leftrightarrow ad< bc\)
\(\Leftrightarrow ad+ab< bc+ab\)
\(\Leftrightarrow a\left(b+d\right)< b\left(a+c\right)\)
\(\Leftrightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}\) \(\left(1\right)\)
Mà \(ad< bc\)
\(\Leftrightarrow ad+cd< bc+cd\)
\(\Leftrightarrow d\left(a+c\right)< c\left(b+d\right)\)
\(\Leftrightarrow\dfrac{a+c}{b+d}< \dfrac{c}{d}\) \(\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\rightarrowđpcm\)
b) \(\dfrac{-1}{3}=\dfrac{-16}{48}< \dfrac{-15}{48};\dfrac{-14}{48};\dfrac{-13}{48}< \dfrac{-12}{48}=\dfrac{-1}{4}\)
Ta thấy :
\(\left\{{}\begin{matrix}A=\dfrac{10^{2017}+1}{10^{2016}+1}>1\\B=\dfrac{10^{2018}+1}{10^{2017}+1}>1\end{matrix}\right.\)
Áp dụng tính chất \(\dfrac{a}{b}>1\Leftrightarrow\dfrac{a+m}{b+m}\) ta có :
\(B=\dfrac{10^{2018}+1}{10^{2017}+1}>\dfrac{10^{2018}+1+9}{10^{2017}+1+9}=\dfrac{10^{2018}+10}{10^{2017}+10}=\dfrac{10\left(10^{2017}+1\right)}{10\left(10^{2016}+1\right)}=\dfrac{10^{2017}+1}{10^{2016}+1}=A\)
\(\Leftrightarrow B>A\)
Nhân cả hai tử của \(A\)và \(B\)với 2 , ta được :
\(10A=10.\left(\frac{10^{2016}+1}{10^{2017}+1}\right)=\frac{10^{2017}+1+9}{10^{2017}+1}=1+\frac{9}{2^{2017}+1}\)
\(10B=10\left(\frac{10^{2017}+1}{10^{2018}+1}\right)=\frac{10^{2018}+10}{10^{2018}+1}=\frac{10^{2018}+1+9}{10^{2018}}=1+\frac{9}{10^{2018}+1}\)
Vì \(1=1;9=9\)
\(\Rightarrow\)Ta so sánh mẫu , ta có:
\(10^{2017}< 10^{2018}\)
\(\Rightarrow10^{2017}+1< 10^{2018}+1\)
\(\Rightarrow1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\)
\(\Rightarrow10A>10B\)
Hay \(A>B\)
Áp dung công thức \(a>b\Leftrightarrow\frac{a}{b}>\frac{a+m}{b+m}\)
\(B=\frac{10^{2017}+1}{10^{2016}+1}>\frac{10^{2017}+1+9}{10^{2016}+1+9}=\frac{10^{2017}+10}{10^{2016}+10}=\frac{10\left(10^{2016}+1\right)}{10\left(10^{2015}+1\right)}=\frac{10^{2016}+1}{10^{2015}+1}=A\)
\(\Leftrightarrow B>A\)
Giải:
Có:
\(A=\dfrac{2017^{2016-1}}{2017^{2017-1}}\) và \(B=\dfrac{2017^{2015+1}}{2017^{2016+1}}\)
\(\Rightarrow A=\dfrac{2017^{2016-1}}{2017^{2017-1}}=\dfrac{2017^{2015}}{2017^{2016}}=\dfrac{1}{2017}\)
\(\Rightarrow B=\dfrac{2017^{2015+1}}{2017^{2016+1}}=\dfrac{2017^{2016}}{2017^{2017}}=\dfrac{1}{2017}\)
Vậy \(A=B\)
Chúc bạn học tốt!
Ta có:
\(A=\dfrac{2017^{2016-1}}{2017^{2017-1}}=\dfrac{2017^{2015}}{2017^{2016}}=\dfrac{1}{2017}\)(1)
\(B=\dfrac{2017^{2015+1}}{2017^{2016+1}}=\dfrac{2017^{2016}}{2017^{2017}}=\dfrac{1}{2017}\)(2)
Từ (1) và (2) suy ra:
\(A=B\)
Chúc bạn học tốt!!!
P/s: Xem lại đề xem là +1 vs -1 ở dưới hay bên trên số mũ nha!!
ồ, lâu h ms gặp
a,
Dễ thấy \(\dfrac{2005^{2016}+1}{2005^{2017}+1}< 1\)
Áp dụng khi \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+n}{b+n}\left(n\in N^{\circledast}\right)\)
Ta có:
\(\dfrac{2005^{2016}+1}{2005^{2017}+1}< \dfrac{2005^{2016}+1+\left(2005^2-1\right)}{2005^{2017}+1+\left(2005^2-1\right)}=\dfrac{2005^{2016}+2005^2}{2005^{2017}+2005^2}=\dfrac{2005^2\left(2005^{2014}+1\right)}{2005^2\left(2005^{2015}+1\right)}=\dfrac{2005^{2014}+1}{2005^{2015}+1}\)
Vậy \(\dfrac{2005^{2016}+1}{2005^{2017}+1}< \dfrac{2005^{2014}+1}{2005^{2015}+1}\)
b,
\(\dfrac{19}{10}=\dfrac{10+9}{10}=\dfrac{10}{10}+\dfrac{9}{10}=1+\dfrac{9}{10}\\ \dfrac{49}{40}=\dfrac{40+9}{40}=\dfrac{40}{40}+\dfrac{9}{40}=1+\dfrac{9}{40}\)
Vì \(10< 40\Rightarrow\dfrac{9}{10}>\dfrac{9}{40}\Rightarrow1+\dfrac{9}{10}>1+\dfrac{9}{40}\Leftrightarrow\dfrac{19}{10}>\dfrac{49}{40}\)Vậy \(\dfrac{19}{10}>\dfrac{49}{40}\)
c,
\(\dfrac{13}{20}=\dfrac{20-7}{20}=\dfrac{20}{20}-\dfrac{7}{20}=1-\dfrac{7}{20}\\ \dfrac{33}{40}=\dfrac{40-7}{40}=\dfrac{40}{40}-\dfrac{7}{40}=1-\dfrac{7}{40}\)
Vì \(20< 40\Rightarrow\dfrac{7}{20}>\dfrac{7}{40}\Rightarrow1-\dfrac{7}{20}< 1-\dfrac{7}{40}\Leftrightarrow\dfrac{13}{20}< \dfrac{33}{40}\)
Vậy \(\dfrac{13}{20}< \dfrac{33}{40}\)
Áp dụng tính chất:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(\)Đặt: \(B=\dfrac{2005^{2016}+1}{2005^{2017}+1}< 1\)
\(\Rightarrow B< \dfrac{2005^{2016}+1+4020024}{2005^{2017}+1+4020024}\)
\(B< \dfrac{2005^{2016}+4020025}{2005^{2017}+4020025}\)
\(B< \dfrac{2005^2\left(2005^{2014}+1\right)}{2005^2\left(2005^{2015}+1\right)}\)
\(B< \dfrac{2005^{2014}+1}{2005^{2015}+1}=A\)
\(B< A\)
Bài 2:
a)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)
=> a = b = c
b)
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}\)
=> x = y = z (theo a)
Thay x = y = z vào biểu thức, ta có:
\(M=\dfrac{x^{333}.x^{666}}{x^{999}}=1\)
c)
\(ac=b^2\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
\(ab=c^2\Rightarrow\dfrac{b}{c}=\dfrac{c}{a}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\Rightarrow a=b=c\)
Thay a = b = c vào biểu thức, ta có:
\(M=\dfrac{a^{333}}{a^{111}.a^{222}}=1\)
Ta có:A=\(\dfrac{-21}{10^{2016}}\)+\(\dfrac{-12}{10^{2017}}\)
= \(\dfrac{-12}{10^{2016}}\)+\(\dfrac{-9}{10^{2016}}\)+\(\dfrac{-12}{10^{2017}}\).
B=\(\dfrac{-12}{10^{2016}}\)+\(\dfrac{-21}{10^{2017}}\)
=\(\dfrac{-12}{10^{2016}}\)+\(\dfrac{-9}{10^{2017}}\)+ \(\dfrac{-12}{10^{2017}}\)
Khi đó để so sánh A và B ta chỉ cần so sánh:\(\dfrac{-9}{10^{2016}}\)và \(\dfrac{-9}{10^{2017}}\)vì A và B cùng có:
\(\dfrac{-12}{10^{2016}}\)+\(\dfrac{-12}{10^{2017}}\).
Do:\(\dfrac{9}{10^{2016}}\)>\(\dfrac{9}{10^{2017}}\).
Suy ra:\(\dfrac{-9}{10^{2016}}\)<\(\dfrac{-9}{10^{2017}}\).
Từ đó ta suy ra được: A< B
bn nhìn kĩ trên là hiểu thôi, cụ thể:
\(\dfrac{9}{10^{2016}}\)>\(\dfrac{9}{10^{2017}}\).Nên số đối của chúng sẽ là dấu ngược lại.
\(\dfrac{-9}{10^{2016}}\)<\(\dfrac{-9}{10^{2017}}\)
Bn suy nghĩ kĩ thì được thôi, nếu chắc ăn hơn thì bn cứ VD đi