so sánh
a, A=\(\frac{10^{17}-1}{10^{16}-1}vaB=\frac{10^{16}+2}{10^{15}+2}\)
b,\(C=\frac{2017^{15}+1}{2017^{16}+1}vaO=\frac{2017^{16}-1}{2017^{17}-1}\)
c,\(E=\frac{99^{15}-1}{99^{16}-1}vaF=\frac{99^{16}+2}{99^{17}+2}\)
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28 tháng 8 2019
a,\(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\) (1)
<=> \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
<=> \(\left(x+1\right)\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
=> x+1=0 (vì \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\ne0\))
<=> x=-1
Vậy pt (1) có tập nghiệm S\(=\left\{-1\right\}\)
b, \(\frac{x+6}{2015}+\frac{x+5}{2016}+\frac{x+4}{2017}=\frac{x+3}{2018}+\frac{x+2}{2019}+\frac{x+1}{2010}\)(2)
<=> \(\frac{x+6}{2015}+1+\frac{x+5}{2016}+1+\frac{x+4}{2017}+1=\frac{x+3}{2018}+1+\frac{x+2}{2019}+1+\frac{x+1}{2020}+1\)
<=> \(\frac{x+2021}{2015}+\frac{x+2021}{2016}+\frac{x+2021}{2017}-\frac{x+2021}{2018}-\frac{x+2021}{2019}-\frac{x+2021}{2020}=0\)
<=> \(\left(x+2021\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
=> x+2021=0(vì \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x=-2021
Vậy pt (2) có tập nghiệm S=\(\left\{-2021\right\}\)
c,\(\frac{x+6}{2016}+\frac{x+7}{2017}+\frac{x+8}{2018}=\frac{x+9}{2019}+\frac{x+10}{2020}+1\) (3)
<=> \(\frac{x+6}{2016}-1+\frac{x+7}{2017}-1+\frac{x+8}{2018}-1=\frac{x+9}{2019}-1+\frac{x+10}{2020}-1+1-1\)
<=> \(\frac{x-2010}{2016}+\frac{x-2010}{2017}+\frac{x-2010}{2018}=\frac{x-2010}{2019}+\frac{x-2010}{2020}\)
<=> \(\frac{x-2010}{2016}+\frac{x-2010}{2017}+\frac{x-2010}{2018}-\frac{x-2010}{2019}-\frac{x-2010}{2020}=0\)
<=> \(\left(x-2010\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
=> x-2010=0 (vì \(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x=2010
Vậy pt (3) có tập nghiệm S=\(\left\{2010\right\}\)
d, \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\) (4)
<=>\(\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=15-1-2-3-4-5\)
<=> \(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
<=> (x-100)(\(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\))=0
=> x -100=0(vì \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\))
<=> x=100
Vậy pt (4) có tập nghiệm S=\(\left\{100\right\}\)
VM
28 tháng 8 2019
a) \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)
\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=0-1\)
\(\Rightarrow x=-1\)
Vậy \(x=-1.\)
Mình chỉ làm câu a) thôi nhé.
Chúc bạn học tốt!
TL
0
0
PQ
1
8 tháng 2 2017
\(A=\frac{10^{15}+1}{10^{16}+1}\)
\(\Rightarrow10A=\frac{10^{16}+10}{10^{16}+1}=\frac{\left(10^{16}+1\right)+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(A=\frac{10^{16}+1}{10^{17}+1}\)
\(\Rightarrow10B=\frac{10^{17}+10}{10^{17}+1}=\frac{\left(10^{17}+1\right)+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
Vì \(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\left(Do10^{16}+1< 10^{17}+1\right)\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
17 tháng 8 2017
Ta có:
\(A=\frac{10^{15}+1}{10^{16}+1}\)
\(10A=\frac{10^{16}+10}{10^{16}+1}\)
\(B=\frac{10^{16}+1}{10^{17}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}\)
Ta so sánh \(10A\) và \(10B\)
Có:
\(10A:\) Mẫu - tử = 9
\(10B:\) Mẫu - tử = 9
Lại có:
\(\frac{10^{16}+10}{10^{16}+1}\) \(-1\)\(=\frac{9}{10^{16}+1}\)
\(\frac{10^{17}+10}{10^{17}+1}-1=\frac{9}{10^{17}+1}\)
Vì \(\frac{9}{10^{16}+1}\)\(>\frac{9}{10^{17}+1}\)nên \(10A>10B\)
\(\Rightarrow\)\(A>B\)
Vậy \(A>B\)
17 tháng 8 2017
Theo bải ra ta có:
A=\(\frac{10^{15}+1}{10^{16}+1}\)=> 10A =.\(\frac{10.\left(10^{15}+1\right)}{10^{16}+1}\)= \(\frac{10.10^{15}+1.10}{10^{16}+1}\)
= \(\frac{10.10^{15}+10}{10^{16}+1}\)=\(\frac{10^{16}+1+9}{10^{16}+1}\)= \(1+\frac{9}{10^{16}+1}\)
B= \(\frac{10^{16}+1}{10^{17}+1}\)=> 10B = \(\frac{10.\left(10^{16}+1\right)}{10^{17}+1}\)=\(\frac{10.10^{16}+1.10}{10^{17}+1}\)
= \(\frac{10.10^{16}+10}{10^{17}+1}\)= \(\frac{10^{17}+1+9}{10^{17}+1}\)= \(1+\frac{9}{10^{17}+1}\)
Vì 1=1 mà \(\frac{9}{10^{16}+1}\)> \(\frac{9}{10^{17}+1}\)nên => 10A > 10B => A>B
Vậy A>B.
NT
1
2 tháng 2 2015
Ta có:
10A=1016+10/1016+1=1+(9/1016+1)
10B=1017+10/1017+1=1+(9/1017+1)
Vì 9/1016+1 > 9/1017+1 nên 10A>10B,do đó A>B
25 tháng 3 2019
sai đề rồi bạn.\(\frac{a}{b}>\frac{a+c}{b+c}\) với \(a>b\) mới đúng nha.
Ta có:\(A=\frac{10^{17}+1}{10^{16}+1}>\frac{10^{17}+1+9}{10^{16}+1+9}=\frac{10^{17}+10}{10^{16}+10}=\frac{10\left(10^{16}+1\right)}{10\left(10^{15}+1\right)}=\frac{10^{16}+1}{10^{15}+1}\)
\(\Rightarrow A>B\)
23 tháng 2 2017
a) Ta có: \(10A=\frac{10^{16}+10}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
\(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\Rightarrow1+\frac{9}{10^{16}+1}>1+\frac{9}{10^{17}+1}\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
Vậy A > B
b) Ta có: \(\frac{1}{10}C=\frac{10^{1992}+1}{10^{1992}+10}=1+\frac{10^{1992}+1}{9}\)
\(\frac{1}{10}D=\frac{10^{1993}+1}{10^{1993}+10}=1+\frac{10^{1993}+1}{9}\)
\(\frac{10^{1992}+1}{9}< \frac{10^{1993}+1}{9}\Rightarrow1+\frac{10^{1992}+1}{9}< 1+\frac{10^{1993}+1}{9}\)
\(\Rightarrow\frac{1}{10}C< \frac{1}{10}D\)
\(\Rightarrow C< D\)
Vậy C < D
BT
13 tháng 4 2018
\(10A=\frac{10^{16}+10}{10^{16}+1}=\frac{10^{16}+1+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}=\frac{10^{17}+1+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
Nhận thấy: \(\frac{9}{10^{17}+1}< \frac{9}{10^{16}+1}\)=> 10B < 10A
=> A > B
DA
25 tháng 3 2019
A = ( 10^15+1 ) / ( 10^16+1 ) => 10A = ( 10^16+10 ) / ( 10^16+1 ) = 1 + ( 9/10^15+1 )
B = ( 10^16+1 ) / ( 10^17+1 ) => 10B = ( 10^17+10 ) / ( 10^17+1 ) = 1 + ( 9/10^16+1 )
Vì 10^15+1 < 10^16+1 nên 9/10^15+1 > 9/10^16+1 => 1 + ( 9/10^15+1 ) > 1 + ( 9/10^16+1 )
Vậy A > B
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