so sánh các số sau : \(a=\dfrac{35}{49};b=\sqrt{\dfrac{5^2}{7^2}};c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}};d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}\)
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Ta có:
\(\dfrac{37}{-49}< 0;\dfrac{-12}{-35}=\dfrac{12}{35}>0\)
\(\Rightarrow\dfrac{37}{-49}< \dfrac{-12}{-35}\)
Vậy...
a) Ta có \(\dfrac{23}{27}>\dfrac{23}{29};\dfrac{23}{29}>\dfrac{22}{29}\)
Vậy \(\dfrac{23}{27}>\dfrac{22}{29}\)
b) Ta có \(\dfrac{15}{25}=1-\dfrac{2}{5}\)
\(\dfrac{25}{49}=1-\dfrac{24}{49}\)
Vì \(\dfrac{2}{5}=\dfrac{24}{60}< \dfrac{24}{49}\)
Vậy \(\dfrac{15}{25}>\dfrac{25}{49}\)
Ta có:
\(\dfrac{-49}{211}< 0;\dfrac{13}{1999}>0\)
⇒ \(\dfrac{-49}{211}< \dfrac{13}{1999}\)
\(a,\dfrac{-15}{17}=-1+\dfrac{2}{17}\\ -\dfrac{19}{21}=-1+\dfrac{2}{21}\\ Vì:\dfrac{2}{17}>\dfrac{2}{21}\Rightarrow-1+\dfrac{2}{17}>-1+\dfrac{2}{21}\Rightarrow-\dfrac{15}{17}>-\dfrac{19}{21}\\ b,-\dfrac{24}{35}=-1+\dfrac{11}{35};-\dfrac{19}{30}=-1+\dfrac{11}{30}\\ Vì:\dfrac{11}{35}< \dfrac{11}{30}\Rightarrow-1+\dfrac{11}{35}< -1+\dfrac{11}{30}\\ \Rightarrow-\dfrac{24}{35}< -\dfrac{19}{30}\)
a) \(\dfrac{27}{35}>\dfrac{19}{35}>\dfrac{19}{41}\)
\(\Rightarrow\dfrac{27}{35}>\dfrac{19}{41}\)
b) \(\dfrac{120}{121}< \dfrac{120+1}{121+1}=\dfrac{121}{122}\)
\(\Rightarrow\dfrac{120}{121}< \dfrac{121}{122}\)
a) HS tự thực hiện
b) $\frac{5}{6}$ < 1 ; $\frac{3}{2} > 1$
$\frac{9}{{19}}$ < 1 ; $\frac{7}{7}$ = 1
$\frac{{49}}{{46}}$ > 1 ; $\frac{{32}}{{71}}$ < 1
c) Ba phân số bé hơn 1 là: $\frac{2}{7};\,\,\,\frac{{11}}{{25}};\,\,\,\frac{{37}}{{59}}$
Ba phân số lớn hơn 1 là: $\frac{7}{2};\,\,\,\frac{{15}}{7};\,\,\,\,\frac{{33}}{{12}}$
Ba phân số bằng 1 là: $\frac{9}{9};\,\,\,\,\frac{{25}}{{25}};\,\,\,\,\frac{{47}}{{47}}$
21/35=3/5; 12/20=3/5
=>21/35=12/20
24/35<1<21/7
35/49=5/7; 40/64=5/8
=>35/49>40/64
2424/4848=1/2
9/11=1-2/11
13/15=1-2/15
mà 2/11>2/15
nên 9/11<13/15
19/15>11/15
\(\left\{{}\begin{matrix}a=\dfrac{35}{49}=\dfrac{5}{7}\\b=\sqrt{\dfrac{5^2}{7^2}}=\dfrac{5}{7}\\c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\dfrac{5+35}{7+49}=\dfrac{5}{7}\\d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\dfrac{5-35}{7-49}=\dfrac{5}{7}\end{matrix}\right.\)
\(\Rightarrow a=b=c=d=\dfrac{5}{7}\)
\(a=\dfrac{35}{49};b=\dfrac{5}{7}\\ c,=\dfrac{5+35}{7+49}=\dfrac{12}{14}=\dfrac{6}{7}\\ d,=\dfrac{5-35}{7-49}\)
Áp dụng t/c dtsbn:
\(\dfrac{5}{7}=\dfrac{35}{49}=\dfrac{5+35}{7+49}=\dfrac{5-35}{7-49}\) hay \(a=b=c=d\)