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\(A=\sqrt{2007}-\sqrt{2006}=\frac{\left(\sqrt{2007}-\sqrt{2006}\right)\left(\sqrt{2007}+\sqrt{2006}\right)}{\left(\sqrt{2007}+\sqrt{2006}\right)}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)(1)
\(B=\sqrt{2008}-\sqrt{2007}=\frac{\left(\sqrt{2008}-\sqrt{2007}\right)\left(\sqrt{2008}+\sqrt{2007}\right)}{\left(\sqrt{2008}+\sqrt{2007}\right)}=\frac{1}{\sqrt{2008}+\sqrt{2007}}\)(2)
Từ 1 và 2 => \(\frac{1}{\sqrt{2007}+\sqrt{2006}}>\frac{1}{\sqrt{2008}+\sqrt{2007}}\)
hay \(\sqrt{2007}-\sqrt{2006}>\sqrt{2008}-\sqrt{2007}\)
P/s tham khảo nha
a,hay \(\left(1995\cdot1997\right)^n\)và \(\left(1996\cdot1996\right)^n\)
hay so sánh \(1995\cdot1997\)và \(1996\cdot1996\)
ta có 1995*1997=1995*(1996+1)=1995*1996+1995
1996*1996=1996*(1995+1)=1996*1995+1996
vì 1995<1996 => \(\left(1995\cdot1997\right)^n\)<\(\left(1996\cdot1996\right)^n\)
a) Ta có: \(\frac{1}{5}\sqrt{150}=\frac{1}{5}\cdot5\sqrt{6}=\sqrt{6}=\frac{1}{3}\cdot\sqrt{6\cdot9}=\frac{1}{3}\sqrt{54}>\frac{1}{3}\sqrt{51}\)
b) Ta có: \(\frac{1}{2}\sqrt{6}=\sqrt{\frac{6}{4}}< \sqrt{\frac{36}{2}}=6\sqrt{\frac{1}{2}}\)
a) Vì \(5,\left(6\right)< 6\)\(\Rightarrow\)\(\frac{51}{9}< \frac{150}{25}\)
\(\Rightarrow\)\(\sqrt{\frac{51}{9}}< \sqrt{\frac{150}{25}}\)
\(\Rightarrow\)\(\frac{1}{3}\sqrt{51}< \frac{1}{5}\sqrt{150}\)
b) Vì \(1,5< 18\)\(\Rightarrow\)\(\frac{6}{4}< \frac{36}{2}\)
\(\Rightarrow\)\(\sqrt{\frac{6}{4}}< \sqrt{\frac{36}{2}}\)
\(\Rightarrow\)\(\frac{1}{2}\sqrt{6}< 6\sqrt{\frac{1}{2}}\)
Giả sử A > B
<=> 19 >\(5\sqrt{3}+6\sqrt{2}\)
<=> (6 + 3 - \(2\sqrt{3}\sqrt{6}\)
) + (10 - 5\(\sqrt{3}\))>0
<=> (\(\sqrt{6}-\sqrt{3}\))2 + (10 - \(5\sqrt{3}\))>0
Mà 10 - 5\(\sqrt{3}\)> 10 - 5\(\sqrt{4}\) = 0
Vậy A > B
a) Có \(\sqrt{2}< \sqrt{2,25}=1,5\)
\(\sqrt{6}< \sqrt{6,25}=2,5\);
\(\sqrt{12}< \sqrt{12,25}=3,5\);
\(\sqrt{20}< \sqrt{20,25}=4,5\)
=> \(P=\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}< 1,5+2,5+3,5+4,5=12\)
Vậy P < 12
Answer:
ý a, tham khảo bài làm của @xyzquynhdi
\(\sqrt{2}+\sqrt{3}+\sqrt{5}\)
\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\)
\(=\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+\left(\sqrt{5}\right)^2+2\sqrt{2}\sqrt{3}+2\sqrt{2}\sqrt{5}+2\sqrt{3}\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{2}+\sqrt{3}+\sqrt{5}\)
Ta có :
\(A=\dfrac{\left(\sqrt{2000}-\sqrt{1999}\right)\left(\sqrt{2000}+\sqrt{1999}\right)}{\left(\sqrt{2000}+\sqrt{1999}\right)}=\dfrac{1}{\sqrt{2000}+\sqrt{1999}}\)
\(B=\dfrac{\left(\sqrt{2001}-\sqrt{2000}\right)\left(\sqrt{2001}+\sqrt{2000}\right)}{\left(\sqrt{2001}+\sqrt{2000}\right)}=\dfrac{1}{\sqrt{2001}+\sqrt{2000}}\)
Do \(\sqrt{2000}+\sqrt{1999}< \sqrt{2001}+\sqrt{2000}\)
\(\Rightarrow A>B.\)
Bài làm:
Theo máy tính Vinacal 570ES PLUS II, ta có:
A>B
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