so sánh:
6 ... 7
7 ... 7
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a) Ta có: \(\dfrac{15}{7}>1\) (tử lớn hơn mẫu)
\(\dfrac{9}{14}< 1\) (tử nhỏ hơn mẫu)
Vậy: \(\dfrac{15}{7}>\dfrac{9}{14}\)
b) Ta có:
\(\dfrac{899}{900}=1-\dfrac{1}{900}\)
\(\dfrac{1235}{1236}=1-\dfrac{1}{1236}\)
Mà: \(\dfrac{1}{900}>\dfrac{1}{1236}\)
Vậy: \(\dfrac{1235}{1236}>\dfrac{899}{900}\)
c) Ta có:
\(\dfrac{77}{75}=1+\dfrac{2}{75}\)
\(\dfrac{37}{35}=1+\dfrac{2}{35}\)
Mà: \(\dfrac{2}{75}< \dfrac{2}{35}\)
Vậy: \(\dfrac{37}{35}>\dfrac{77}{75}\)
\(\left\{{}\begin{matrix}\dfrac{15}{7}=\dfrac{30}{14}\\\dfrac{9}{14}< \dfrac{30}{14}\end{matrix}\right.\Rightarrow\dfrac{15}{7}>\dfrac{9}{14}\)
\(\left\{{}\begin{matrix}\dfrac{899}{900}=\dfrac{899.1236}{900.1236}=\dfrac{\text{1111164}}{900.1236}\\\dfrac{1235}{1236}=\dfrac{1235.900}{900.1236}=\dfrac{\text{1111500}}{900.1236}>\dfrac{\text{1111164}}{900.1236}\end{matrix}\right.\Rightarrow\dfrac{1235}{1236}>\dfrac{899}{900}\)
\(\left\{{}\begin{matrix}\dfrac{77}{75}=\dfrac{539}{525}\\\dfrac{37}{35}=\dfrac{555}{525}>\dfrac{539}{525}\end{matrix}\right.\Rightarrow\dfrac{77}{73}< \dfrac{37}{35}\)
Ta có \(1-\frac{67}{77}=\frac{10}{77}\)
\(1-\frac{77}{83}=\frac{6}{83}\)
Dễ nhận thấy : \(\frac{10}{77}>\frac{6}{83}\)
=> \(\frac{66}{77}< \frac{77}{83}\)
Ta có
\(\sqrt{123-22\sqrt{2}}=11-\sqrt{2}\)
\(\sqrt[3]{77\sqrt{2}-115}=\sqrt{2}-5\)
\(\Rightarrow\sqrt{123-22\sqrt{2}}+\sqrt[3]{77\sqrt{2}-115}=11-\sqrt{2}+\sqrt{2}-5=6\)
6 < 7
7 = 7
6 < 7
7 = 7
hok tốt