8\(\dfrac{8}{10}\) <x<8+\(\dfrac{2}{10}\)
Biết x là số thập phân.Tìm x
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\(\dfrac{5}{8}+\dfrac{6}{8}+\dfrac{9}{8}+\dfrac{11}{8}+\dfrac{10}{8}+\dfrac{55}{8}\)\(+\dfrac{4}{8}=?\)
\(...=\dfrac{5+6+9+11+10+55+4}{8}=\dfrac{100}{8}=\dfrac{25}{2}\)
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\(\dfrac{5+6+9+11+10+55+4}{8}=\dfrac{100}{8}\)
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\(\dfrac{100}{8}\) rút gọn được nhé bnn.
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Cho đẳng thức: (-8).1512.(-10) Tỉ lệ thức đc suy ra từ đẳng là:A.dfrac{-18}{12}dfrac{15}{-10}B.dfrac{15}{12}dfrac{-10}{-8}C.dfrac{-8}{-10}dfrac{15}{12}D.dfrac{-8}{15}dfrac{12}{-10}Mong mn giúp đỡ.Mk đang cần gấp.Thanks:)
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Cho đẳng thức: (-8).15=12.(-10) Tỉ lệ thức đc suy ra từ đẳng là:
A.\(\dfrac{-18}{12}\)=\(\dfrac{15}{-10}\)
B.\(\dfrac{15}{12}\)=\(\dfrac{-10}{-8}\)
C.\(\dfrac{-8}{-10}\)=\(\dfrac{15}{12}\)
D.\(\dfrac{-8}{15}\)=\(\dfrac{12}{-10}\)
Mong mn giúp đỡ.
Mk đang cần gấp.
Thanks:)
\(\text{B.}\dfrac{15}{12}=\dfrac{-10}{-8}\)
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\(\dfrac{4}{9}+\dfrac{5}{8}\)=
\(\dfrac{6}{10}+\dfrac{9}{2}=\)
\(\dfrac{10}{2}-\dfrac{5}{2}=\)
\(\dfrac{8}{8}x1=\)
\(\dfrac{4}{9}+\dfrac{5}{8}\)=\(\dfrac{32}{72}+\dfrac{45}{72}=\dfrac{77}{72}\)
\(\dfrac{6}{10}+\dfrac{9}{2}=\dfrac{6}{10}+\dfrac{45}{10}=\dfrac{51}{10}\)
\(\dfrac{10}{2}-\dfrac{5}{2}=\dfrac{5}{2}\)
\(\dfrac{8}{8}x1=\dfrac{8}{8}x\dfrac{1}{1}=\dfrac{8}{8}\)
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7 tháng 5 2021 lúc 13:27
So sánh:
a/ \(A=\dfrac{17^{18}+1}{17^{19}+1};B=\dfrac{17^{17}+1}{17^{18}+1}\)
b/ \(A=\dfrac{10^8-2}{10^8+2};B=\dfrac{10^8}{10^8+4}\)
c/ \(A=\dfrac{20^{10}+1}{20^{10}-1};B=\dfrac{20^{10}-1}{20^{10}-3}\)
GIÚP MÌNH VỚI
Giải:
a) A=1718+1/1719+1
17A=1719+17/1719+1
17A=1719+1+16/1719+1
17A=1+16/1719+1
Tương tự:
B=1717+1/1718+1
17B=1718+17/1718+1
17B=1718+1+16/1718+1
17B=1+16/1718+1
Vì 16/1719+1<16/1718+1 nên 17A<17B
⇒A<B
b) A=108-2/108+2
A=108+2-4/108+2
A=1+-4/108+2
Tương tự:
B=108/108+4
B=108+4-4/108+1
B=1+-4/108+1
Vì -4/108+2>-4/108+1 nên A>B
c)A=2010+1/2010-1
A=2010-1+2/2010-1
A=1+2/2010-1
Tương tự:
B=2010-1/2010-3
B=2010-3+2/2010-3
B=1+2/2010-3
Vì 2/2010-3>2/2010-1 nên B>A
⇒A<B
Chúc bạn học tốt!
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4, so sánh A và B:
a,A=\(\dfrac{3}{8^3}+\dfrac{7}{8^4}\);B=\(\dfrac{7}{8^3}+\dfrac{3}{8^4}\)
b,A=\(\dfrac{10^7+5}{10^7-8}\);B=\(\dfrac{10^8+6}{10^8-7}\)
c,A=\(\dfrac{10^{1992}+1}{10^{1991}+1}\);B=\(\dfrac{10^{1993}+1}{10^{1992}+1}\)
b: \(A=\dfrac{10^7-8+13}{10^7-8}=1+\dfrac{13}{10^7-8}\)
\(B=\dfrac{10^8-7+13}{10^8-7}=1+\dfrac{13}{10^8-7}\)
mà \(10^7-8< 10^8-7\)
nên A>B
c: \(\dfrac{1}{10}A=\dfrac{10^{1992}+1}{10^{1992}+10}=1-\dfrac{9}{10^{1992}+10}\)
\(\dfrac{1}{10}B=\dfrac{10^{1993}+1}{10^{1993}+10}=1-\dfrac{9}{10^{1993}+10}\)
mà \(\dfrac{9}{10^{1992}+10}>\dfrac{9}{10^{1993}+10}\)
nên A<B
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4, so sánh A và B:
a,A=\(\dfrac{3}{8^3}+\dfrac{7}{8^4}\);B=\(\dfrac{7}{8^3}+\dfrac{3}{8^4}\)
b,A=\(\dfrac{10^7+5}{10^7-8}\);B=\(\dfrac{10^8+6}{10^8-7}\)
c,A=\(\dfrac{10^{1992}+1}{10^{1991}+1}\);B=\(\dfrac{10^{1993}+1}{10^{1992}+1}\)
a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)
Vậy A < B
b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)
\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)
Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)
c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:
\(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)
Vậy A < B
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Cho M=\(\dfrac{10^8+2}{10^8-1}\) và N=\(\dfrac{10^8}{10^8-3}\).Em hãy so sánh M và N.
Giúp mình với
\(M=\dfrac{10^8+2}{10^8-1}=\dfrac{\left(10^8-1\right)+3}{10^8-1}=1+\dfrac{3}{10^8-1}\)
\(N=\dfrac{10^8}{10^8-3}=\dfrac{\left(10^8-3\right)+3}{10^8-3}=1+\dfrac{3}{10^8-3}\)
Vì \(1+\dfrac{3}{10^8-3}< 1+\dfrac{3}{10^8-1}\) nên \(M< N\)
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So sánh A và B :
a) Adfrac{20}{39}+dfrac{22}{27}+dfrac{18}{43}
Bdfrac{14}{39}+dfrac{22}{29}+dfrac{18}{41}
b) Adfrac{3}{8^3}+dfrac{7}{8^4}
Bdfrac{7}{8^3}+dfrac{3}{8^4}
c) Adfrac{10^7+5}{10^7-8}
Bdfrac{10^8+6}{10^8-7}
d) Adfrac{10^{1992}+1}{10^{1991}+1}
Bdfrac{10^{1993}+1}{10^{1992}+1}
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So sánh A và B :
a) \(A=\dfrac{20}{39}+\dfrac{22}{27}+\dfrac{18}{43}\)
\(B=\dfrac{14}{39}+\dfrac{22}{29}+\dfrac{18}{41}\)
b) \(A=\dfrac{3}{8^3}+\dfrac{7}{8^4}\)
\(B=\dfrac{7}{8^3}+\dfrac{3}{8^4}\)
c) \(A=\dfrac{10^7+5}{10^7-8}\)
\(B=\dfrac{10^8+6}{10^8-7}\)
d) \(A=\dfrac{10^{1992}+1}{10^{1991}+1}\)
\(B=\dfrac{10^{1993}+1}{10^{1992}+1}\)
d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B
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cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A
Suy ra B>A(chuc ban hoc goi nhe)
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\(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}\)
A = \(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}\)
A = \(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\)
A = - \(\dfrac{3}{5}\)
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