a: \(\dfrac{12}{49}< \dfrac{13}{49}< \dfrac{13}{47}\)

b: \(\dfrac{12}{47}>\dfrac{19}{47}>\dfrac{19}{77}\)

a/ so sánh bằng 12/47

b/ chưa tìm ra

a) Ta chọn phân số \(\dfrac{12}{47}\) là phân số trung gian, ta có :

\(\dfrac{12}{49}< \dfrac{12}{47};\dfrac{13}{47}>\dfrac{12}{47}\Rightarrow\dfrac{12}{49}< \dfrac{13}{47}\)

a)=

b)>

c)=

`a)75<77`

`=>1/75>1/77`

`=>37/75>37/77>35/77`

`b)7779>7569`

`=>1/7779<1/7569`

`=>3734/7569>3734/7779>2099/7779`

\(\sqrt{Ai}giúpmkvớiak\)

Chọn C hoặc A

Lời giải:

c. So sánh trung gian.

Ta thấy: $\frac{23}{48}< \frac{23}{46}=\frac{1}{2}$
$\frac{47}{92}> \frac{46}{92}=\frac{1}{2}$

$\Rightarrow \frac{23}{48}< \frac{1}{2}< \frac{47}{92}$

\(a,\dfrac{11}{49}< \dfrac{11}{46};\dfrac{11}{46}< \dfrac{13}{46}\\ Nên:\dfrac{11}{49}< \dfrac{13}{46}\\ b,\dfrac{62}{85}< \dfrac{62}{80};\dfrac{62}{80}< \dfrac{73}{80}\\ Nên:\dfrac{62}{85}< \dfrac{73}{80}\\ c,\dfrac{n}{n+3}< \dfrac{n}{n+2};\dfrac{n}{n+2}< \dfrac{n+1}{n+2}\\ Nên:\dfrac{n}{n+3}< \dfrac{n+1}{n+2}\)

c) \(\dfrac{27}{26}\)và\(\dfrac{38}{37}\)

Ta có: \(\dfrac{27}{26}=1+\dfrac{1}{26}\); \(\dfrac{38}{37}=1+\dfrac{1}{37}\)

Vì \(\dfrac{1}{26}>\dfrac{1}{37}\) nên \(\dfrac{27}{26}>\dfrac{38}{37}\)

c: \(\dfrac{27}{26}-1=\dfrac{1}{26}\)

\(\dfrac{38}{37}-1=\dfrac{1}{37}\)

mà \(\dfrac{1}{26}>\dfrac{1}{37}\)

nên \(\dfrac{27}{26}>\dfrac{38}{37}\)

Ta có:
\(\dfrac{37}{51}< \dfrac{43}{51}\)
\(\dfrac{43}{51}< \dfrac{43}{49}\)
Do đó \(\dfrac{37}{51}< \dfrac{43}{49}\)

.

\(\left\{{}\begin{matrix}a=\dfrac{35}{49}=\dfrac{5}{7}\\b=\sqrt{\dfrac{5^2}{7^2}}=\dfrac{5}{7}\\c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\dfrac{5+35}{7+49}=\dfrac{5}{7}\\d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\dfrac{5-35}{7-49}=\dfrac{5}{7}\end{matrix}\right.\)

\(\Rightarrow a=b=c=d=\dfrac{5}{7}\)

\(a=\dfrac{35}{49};b=\dfrac{5}{7}\\ c,=\dfrac{5+35}{7+49}=\dfrac{12}{14}=\dfrac{6}{7}\\ d,=\dfrac{5-35}{7-49}\)

Áp dụng t/c dtsbn:

\(\dfrac{5}{7}=\dfrac{35}{49}=\dfrac{5+35}{7+49}=\dfrac{5-35}{7-49}\) hay \(a=b=c=d\)