So sánh (không dùng bảng số hay máy tính cầm tay ) :
a. \(\frac{1}{7}\).\(\sqrt{51}\)với \(\frac{1}{9}\).\(\sqrt{150}\)
b. \(\sqrt{2017}\)-\(\sqrt{2016}\)với \(\sqrt{2016}\)-\(\sqrt{2015}\)
So sánh(không dùng bảng số hay máy tính cầm tay)
a)\(\dfrac{1}{7}\sqrt{51}\) với \(\dfrac{1}{9}\sqrt{150}\)
b)\(\sqrt{2017}-\sqrt{2016}\) với \(\sqrt{2016}-\sqrt{2015}\)
b: \(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2016}+\sqrt{2017}}\)
\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
mà \(\sqrt{2016}+\sqrt{2017}< \sqrt{2016}+\sqrt{2015}\)
nên \(\sqrt{2017}-\sqrt{2016}>\sqrt{2016}-\sqrt{2015}\)
so sánh (ko dùng bảng số hay máy tính cầm tay):
a) \(\frac{1}{7}\sqrt{51}với\frac{1}{9}\sqrt{150}\)
b) \(\sqrt{2017}-\sqrt{2016}với\sqrt{2016}-\sqrt{2015}\)
So sánh: \(\frac{2016}{\sqrt{2015}}+\frac{2015}{\sqrt{2016}}\text{ và }\sqrt{2015}+\sqrt{2016}\)(không dùng máy tính cầm tay)
Không dùng máy tính, hãy so sánh: \(\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}v\text{à}\sqrt{2016}+\sqrt{2017}\)
\(\frac{2016}{\sqrt{2016}}=\sqrt{2016}\)
\(\frac{2017}{\sqrt{2017}}=\sqrt{2017}\)
=> Bằng nhau
\(\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}-\sqrt{2016}-\sqrt{2017}=\left(\frac{2016}{\sqrt{2017}}-\sqrt{2017}\right)+\left(\frac{2017}{\sqrt{2016}}-\sqrt{2016}\right)\)
\(=\frac{2016-2017}{\sqrt{2017}}+\frac{2017-2016}{\sqrt{2016}}=\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}\)
vì \(2016< 2017\Rightarrow\sqrt{2016}< \sqrt{2017}\Rightarrow\frac{1}{\sqrt{2016}}>\frac{1}{\sqrt{2017}}\Rightarrow\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}>0\)
\(\Rightarrow\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}-\sqrt{2016}-\sqrt{2017}>0\Rightarrow\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}>\sqrt{2016}+\sqrt{2017}\)
Không dùng máy tính, hãy so sánh \(\sqrt{2017}-\sqrt{2016}\) và \(\sqrt{2016}-\sqrt{2015}\)
\(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2017}+\sqrt{2016}}\)
\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
2017>2015
=>căn 2017>căn 2015
=>\(\sqrt{2017}+\sqrt{2016}>\sqrt{2016}+\sqrt{2015}\)
=>\(\dfrac{1}{\sqrt{2017}+\sqrt{2016}}< \dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
=>\(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}-\sqrt{2015}\)
so sanh
\(\dfrac{1}{7}\sqrt{51}với\dfrac{1}{9}\sqrt{150}\)
\(\sqrt{2017}-\sqrt{2016}với\sqrt{2016}-\sqrt{2015}\)
\(\dfrac{1}{7}\sqrt{51}với\dfrac{1}{9}\sqrt{150}\)
<=> \(\dfrac{\sqrt{51}}{7}với\dfrac{\sqrt{150}}{9}\)
<=> \(9\sqrt{51}với7\sqrt{150}\)
<=> \(\sqrt{4131}với\sqrt{7350}\)
=> \(\sqrt{4131}< \sqrt{7350}\)
=> \(\dfrac{1}{7}\sqrt{51}< \dfrac{1}{9}\sqrt{150}\)
So sánh ( ko dung bảng số, máy tính )
a) \(\dfrac{1}{7}\sqrt{51}\) vs \(\dfrac{1}{9}\sqrt{150}\)
b) \(\sqrt{2017}-\sqrt{2016}\) vs \(\sqrt{2016}-\sqrt{2015}\)
Lời giải:
a)
Ta có: \(\frac{1}{7}\sqrt{51}< \frac{1}{7}\sqrt{64}=\frac{8}{7}\)
\(\frac{1}{9}\sqrt{150}> \frac{1}{9}\sqrt{144}=\frac{12}{9}=\frac{4}{3}=\frac{8}{6}> \frac{8}{7}\)
Do đó: \(\frac{1}{7}\sqrt{51}< \frac{1}{9}\sqrt{150}\)
b)
\(\sqrt{2017}-\sqrt{2016}=\frac{2017-2016}{\sqrt{2017}+\sqrt{2016}}=\frac{1}{\sqrt{2017}+\sqrt{2016}}< \frac{1}{\sqrt{2016}+\sqrt{2015}}\)
\(\sqrt{2016}-\sqrt{2015}=\frac{2016-2015}{\sqrt{2016}+\sqrt{2015}}=\frac{1}{\sqrt{2016}+\sqrt{2015}}\)
Do đó:
\(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}-\sqrt{2015}\)
1.So sánh A = \(\sqrt{2014}+\sqrt{2015}+\sqrt{2016}\) và B = \(\sqrt{2011}+\sqrt{2013}+\sqrt{2021}\) mà không dùng máy tính và bảng số.
2.Giải phương trình : \(\sqrt{\left(x-2015\right)^{14}}+\sqrt{\left(x-2016\right)^{10}}=1\)
So sánh Q=\(\frac{1-\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}}+\frac{1-\sqrt{3}+\sqrt{4}}{1+\sqrt{3}+\sqrt{4}}+...+\frac{1-\sqrt{2016}+\sqrt{2017}}{1+\sqrt{2016}+\sqrt{2017}}\)với R=\(\sqrt{2017}-1\)
Ta có:
\(\frac{1-\sqrt{n}+\sqrt{n+1}}{1+\sqrt{n}+\sqrt{n+1}}=\frac{\left(1-\sqrt{n}+\sqrt{n+1}\right)^2}{\left(1+\sqrt{n}+\sqrt{n+1}\right)\left(1-\sqrt{n}+\sqrt{n+1}\right)}=\frac{2n+2-2\sqrt{n}+2\sqrt{n+1}-2\sqrt{n\left(n+1\right)}}{2\left(1+\sqrt{n+1}\right)}\)
\(=\frac{\left[2n+2-2\sqrt{n}+2\sqrt{n+1}-2\sqrt{n\left(n+1\right)}\right]\left(1-\sqrt{n+1}\right)}{2\left(1+\sqrt{n+1}\right)\left(1-\sqrt{n+1}\right)}=\frac{-2n\sqrt{n+1}+2n\sqrt{n}}{-2n}=\sqrt{n+1}-\sqrt{n}\)
Suy ra:
\(Q=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2017}-\sqrt{2016}=\sqrt{2017}-\sqrt{2}< \sqrt{2017}-1=R\)
Vậy Q < R.