Chừng tỏ √2+√6+√12+√20<12
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Với \(a,b>0;a\ne b\)ta có:
\(\left(\sqrt{a}-\sqrt{b}\right)^2>0\Leftrightarrow a-2\sqrt{ab}+b>0\Leftrightarrow2\left(a+b\right)>\left(\sqrt{a}+\sqrt{b}\right)^2\)
\(\Leftrightarrow\sqrt{a}+\sqrt{b}< \sqrt{2\left(a+b\right)}\)
Áp dụng ta được:
\(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}< \sqrt{2\left(2+6\right)}+\sqrt{2\left(12+20\right)}\)
\(=\sqrt{16}+\sqrt{64}=4+8=12\)
Ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: A=5/2+5/6+...+5/2450
=5(1/2+1/6+...+1/2450)
=5(1-1/2+1/2-1/3+...+1/49-1/50)
=5*49/50<5
![](https://rs.olm.vn/images/avt/0.png?1311)
Có: \(A=\frac{1}{2}+\frac{5}{6}+...+\frac{9899}{9900}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{9900}\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=99-\left(1-\frac{1}{100}\right)\)
\(=99-\frac{99}{100}< 99\)
\(\Rightarrow A< 99\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình làm câu b)
\(A=2+2^2+2^3+..+2^{20}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{19}+2^{20}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{19}\left(1+2\right)\)
\(=2.3+2^3.3+...+2^{19}.3=3\left(2+2^3+...+2^{19}\right)⋮3^{\left(đpcm\right)}\)
b,
A=(2+2^2)+(2^3+2^4)+...+2^20
=6+2^3.6+2^4.6+...2^19.6
=6.(1+2^3+2^4+...+2^19)÷3
Vì 6÷3 nên A÷3 (đmcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{20}-\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}\right)\)
\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT:25^6\cdot8^8=\left(5^2\right)^6\cdot\left(2^3\right)^8\)
\(=5^{12}\cdot2^{32}\)
\(=5^{12}\cdot\left(2^2\right)^{12}\)
\(=5^{12}\cdot4^{12}\)
\(=\left(5\cdot4\right)^{12}\)
\(=20^{12}=VP\)
\(\Rightarrowđpcm\)
Ta có \(25^6=\left(5^2\right)^6=5^{16}\)
\(8^8=\left(2^3\right)^8=2^{36}=\left(2^2\right)^{12}\)
<=> \(5^{12}.4^{12}=\left(5.4\right)^{12}=20^{12}\)
Good Luck To You