Tính \(\dfrac{1}{\sqrt{25}+\sqrt{24}}+\dfrac{1}{\sqrt{24}+\sqrt{23}}+\dfrac{1}{\sqrt{23}+\sqrt{22}}+...+\dfrac{1}{\sqrt{2}+1}\)
K
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NL
28 tháng 5 2018
Câu b nhé:
Ta có:
\(\dfrac{1}{\sqrt{25}+\sqrt{24}}+\dfrac{1}{\sqrt{24}+\sqrt{23}}+\dfrac{1}{\sqrt{23}+\sqrt{22}}+...+\dfrac{1}{\sqrt{2}+\sqrt{1}}\\ =\dfrac{\sqrt{25}-\sqrt{24}}{\left(\sqrt{25}+\sqrt{24}\right)\left(\sqrt{25}-\sqrt{24}\right)}+\dfrac{\sqrt{24}-\sqrt{23}}{\left(\sqrt{24}+\sqrt{23}\right)\left(\sqrt{24}-\sqrt{23}\right)}+...+\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}+\sqrt{1}\right)\left(\sqrt{2}-\sqrt{1}\right)}\\ =\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}\\ =5-1=4\left(đpcm\right)\)
GH
2 tháng 7 2023
\(M=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}\\ =\dfrac{1}{\sqrt{2}\left(\sqrt{2}+1\right)}+\dfrac{1}{\sqrt{2.3}\left(\sqrt{3}+\sqrt{2}\right)}+....+\dfrac{1}{\sqrt{24.25}\left(\sqrt{25}+\sqrt{24}\right)}\\ =\dfrac{\sqrt{2}-1}{\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}.\sqrt{3}}+...+\dfrac{\sqrt{25}-\sqrt{24}}{\sqrt{25}.\sqrt{24}}\\ =1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\\ =1-\dfrac{1}{\sqrt{25}}=1-\dfrac{1}{5}=\dfrac{4}{5}\)
2 tháng 7 2023
\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\)
=1-1/5=4/5
9 tháng 12 2023
a: \(\dfrac{1}{\sqrt{5}-\sqrt{3}+2}\)
\(=\dfrac{\sqrt{5}-\sqrt{3}-2}{\left(\sqrt{5}-\sqrt{3}\right)^2-4}\)
\(=\dfrac{\sqrt{5}-\sqrt{3}-2}{8-2\sqrt{15}-4}=\dfrac{\sqrt{5}-\sqrt{3}-2}{4-2\sqrt{15}}\)
\(=\dfrac{\left(\sqrt{5}-\sqrt{3}-2\right)\left(4+2\sqrt{15}\right)}{16-60}\)
\(=\dfrac{4\sqrt{5}+2\cdot\sqrt{75}-4\sqrt{3}-2\sqrt{45}-8-4\sqrt{15}}{-44}\)
\(=\dfrac{-2\sqrt{5}+6\sqrt{3}-8-4\sqrt{15}}{-44}\)
\(=\dfrac{\sqrt{5}-3\sqrt{3}+4+2\sqrt{15}}{22}\)
b: Sửa đề: \(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{24}+\sqrt{25}}\)
\(=\dfrac{-1+\sqrt{2}}{2-1}+\dfrac{-\sqrt{2}+\sqrt{3}}{3-2}+...+\dfrac{-\sqrt{24}+\sqrt{25}}{25-24}\)
\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}+...+\left(-\sqrt{24}\right)+\sqrt{25}\)
=5-1
=4
AH
Akai Haruma
Giáo viên
3 tháng 8 2021
a. ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$
$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$
$\Leftrightarrow -\sqrt{x-1}=-17$
$\Leftrightarrow \sqrt{x-1}=17$
$\Leftrightarrow x-1=289$
$\Leftrightarrow x=290$
b. ĐKXĐ: $x\geq \frac{1}{2}$
PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$
$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$
$\Leftrihgtarrow \sqrt{2x-1}=2$
$\Leftrightarrow x=2,5$ (tm)
AH
Akai Haruma
Giáo viên
3 tháng 8 2021
c. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$
$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)
Vậy pt vô nghiệm
MN
30 tháng 9 2018
Xét :\(\dfrac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{2n+1}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n+1}}< \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n}}=\dfrac{\sqrt{n+1}}{2\sqrt{n\left(n+1\right)}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
Do đó :
S\(< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\right)\)\(=\dfrac{1}{2}\left(1-\dfrac{1}{5}\right)=\dfrac{2}{5}\)(dpcm)
10 tháng 6 2017
sai đề rồi bạn ơi, sửa đề
\(B=\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{24}-\sqrt{25}}\)
ta có: \(\dfrac{1}{\sqrt{n}-\sqrt{n+1}}=\dfrac{\sqrt{n}+\sqrt{n+1}}{n-n-1}=-\sqrt{n}-\sqrt{n+1}\)
áp dụng vào B, ta có:
\(B=-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{24}+\sqrt{25}\)
\(B=\sqrt{25}-\sqrt{1}=4\)
28 tháng 8 2023
a: \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{3}+1+\sqrt{3}-1\right)=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
b: \(\dfrac{1}{\sqrt{7-\sqrt{24}}+1}-\dfrac{1}{\sqrt{7+\sqrt{24}}+1}\)
\(=\dfrac{1}{\sqrt{6}-1+1}-\dfrac{1}{\sqrt{6}+1+1}\)
\(=\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{6}+2}=\dfrac{\sqrt{6}+2-\sqrt{6}}{\sqrt{6}\left(\sqrt{6}+2\right)}\)
\(=\dfrac{2}{\sqrt{6}\left(\sqrt{6}+2\right)}=\dfrac{2}{6+2\sqrt{6}}=\dfrac{1}{3+\sqrt{6}}=\dfrac{3-\sqrt{6}}{3}\)
TD
29 tháng 7 2018
Câu a, b, bạn có thể làm được suy nghĩ đi nha
c)
Ta có pt tổng quát :
\(\dfrac{1}{a\sqrt{a+1}+\left(a+1\right)\sqrt{a}}=\dfrac{1}{\sqrt{a\left(a+1\right)}\left(\sqrt{a}+\sqrt{\left(a+1\right)}\right)}=\dfrac{\sqrt{a+1}-\sqrt{a}}{\sqrt{a}\sqrt{a+1}}=\dfrac{1}{\sqrt{a}}-\dfrac{1}{\sqrt{a+1}}\)\(\Rightarrow C=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+.....+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}=1-\dfrac{1}{5}=\dfrac{4}{5}\)..........Kaito Kid.......
\(A=\dfrac{1}{\sqrt{25}+\sqrt{24}}+\dfrac{1}{\sqrt{24}+\sqrt{23}}+....+\dfrac{1}{\sqrt{2}+1}\)
\(A=\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+......+\sqrt{2}-1=\sqrt{25}-1=4\)
làm max tắt chả hiểu gì yêu cầu làm lại đầy đủ hơn nhá