S=\(\dfrac{\sqrt{2}-1}{1+2}+\dfrac{\sqrt{3}-\sqrt{2}}{2+3}+...+\dfrac{\sqrt{25}-\sqrt{24}}{24+25}< \dfrac{2}{5}\)
Tính:
\(M=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}\)
\(M=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}\\ =\dfrac{1}{\sqrt{2}\left(\sqrt{2}+1\right)}+\dfrac{1}{\sqrt{2.3}\left(\sqrt{3}+\sqrt{2}\right)}+....+\dfrac{1}{\sqrt{24.25}\left(\sqrt{25}+\sqrt{24}\right)}\\ =\dfrac{\sqrt{2}-1}{\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}.\sqrt{3}}+...+\dfrac{\sqrt{25}-\sqrt{24}}{\sqrt{25}.\sqrt{24}}\\ =1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\\ =1-\dfrac{1}{\sqrt{25}}=1-\dfrac{1}{5}=\dfrac{4}{5}\)
\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\)
=1-1/5=4/5
Với `n` làm cho biểu thức dưới đây có nghĩa, ta có:
`1/((n+1)sqrtn+nsqrt(n+1))=1/(sqrtn sqrt(n+1)(sqrt(n+1)+sqrt(n)))=(sqrt(n+1)-sqrt(n))/(sqrtn sqrt(n+1))=1/(sqrtn)-1/(sqrtn+1)`
Khi đó:
`M=\sum_{n=1}^(24)=1/((n+1)sqrtn+nsqrt(n+1))=1/(sqrtn)-1/(sqrtn+1)=1/(sqrt1)-1/(sqrt25)=1-1/5=4/5`
\(\dfrac{1}{\sqrt{5}-\sqrt{3}+2}\)
A=\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}\sqrt{3}}+...+\dfrac{1}{\sqrt[]{24}\sqrt{25}}\)
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
rút gọn biểu thức
K=\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}\)
M=\(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
N=\(\dfrac{1+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{1-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1} }=\frac{1}{\sqrt{n(n+1)}(\sqrt{n+1)+\sqrt{n}) } } =\frac{\sqrt{n+1}-\sqrt{n} }{\sqrt{n(n+1)} } =\frac{1}{\sqrt{n} }-\frac{1}{\sqrt{n+1} } \)
=>K=1-\( \frac{1}{5} \)=\(\frac{4}{5} \)
Tính:
a, A= \(\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}\)+ \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
b, B= \(\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)+ \(\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
c, C= \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}\)
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.......
Giải PT:
a) \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
b) \(\sqrt{18x-9}-0,5\sqrt{2x-1}+\dfrac{1}{2}\sqrt{25\left(2x-1\right)}+\sqrt{49\left(2x-1\right)}=24\)
c) \(\sqrt{36x-72}-15\sqrt{\dfrac{x-2}{25}}=4\left(5+\sqrt{x-2}\right)\)
d) \(\sqrt{\dfrac{1}{3x+2}}-\dfrac{1}{2}\sqrt{\dfrac{9}{3x+2}}+\sqrt{\dfrac{16}{3x+2}}-5\sqrt{\dfrac{1}{12x+8}}=1\)
e) \(\dfrac{1}{2}\sqrt{\dfrac{49x}{x+2}}-3\sqrt{\dfrac{x}{4x+8}}-\sqrt{\dfrac{x}{x+2}}-\sqrt{5}=0\)
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)
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
d. ĐKXĐ: $x>\frac{-2}{3}$
PT $\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{1}{2}\sqrt{9}.\sqrt{\frac{1}{3x+2}}+\sqrt{16}.\sqrt{\frac{1}{3x+2}}-5\sqrt{\frac{1}{4}}\sqrt{\frac{1}{3x+2}}=1$
$\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{3}{2}\sqrt{\frac{1}{3x+2}}+4\sqrt{\frac{1}{3x+2}}-\frac{5}{2}\sqrt{\frac{1}{3x+2}}=1$
$\Leftrightarrow \sqrt{\frac{1}{3x+2}}=1$
$\Leftrightarrow \frac{1}{3x+2}=1$
$\Leftrightarrow 3x+2=1$
$\Leftrightarrow x=-\frac{1}{3}$
THỰC HIỆN PHÉP TÍNH
1,\(\sqrt{1\dfrac{9}{16}}\)
2,\(\dfrac{\sqrt{12,5}}{0,5}\)
3,\(\sqrt{\dfrac{25}{64}}\)
4,\(\dfrac{\sqrt{230}}{\sqrt{2,3}}\)
5,\((\sqrt{\dfrac{2}{3}}+\sqrt{\dfrac{50}{3}}-\sqrt{24}).\sqrt{6}\)
1) \(\sqrt{1\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{5}{4}\)
2) \(\dfrac{\sqrt{12.5}}{0.5}=\sqrt{\dfrac{12.5}{0.25}}=5\sqrt{2}\)
3) \(\sqrt{\dfrac{25}{64}}=\dfrac{5}{8}\)
4) \(\dfrac{\sqrt{230}}{\sqrt{2.3}}=\sqrt{\dfrac{230}{2.3}}=\sqrt{100}=10\)
5) \(\left(\sqrt{\dfrac{2}{3}}+\sqrt{\dfrac{50}{3}}-\sqrt{24}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{5\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{6\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)
\(=0\cdot\sqrt{6}=0\)
Rút gọn
A=\(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{24}-\sqrt{25}}\)
B=\(\dfrac{5}{4+\sqrt{11}}+\dfrac{11-3\sqrt{11}}{\sqrt{11}-3}-\dfrac{4}{\sqrt{5}-1}+\sqrt{\left(\sqrt{5}-2\right)^2}\)
C=\(\dfrac{\sqrt{x}+1}{x\sqrt[]{x}+x+\sqrt{x}}:\dfrac{1}{x^2-\sqrt{x}}\) (với x>0; x#1)
D=\(\dfrac{\sqrt{x^2-10x+25}}{x-5}\)
\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-....-\frac{1}{\sqrt{24}-\sqrt{25}}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}-...-\frac{\sqrt{24}+\sqrt{25}}{(\sqrt{24}-\sqrt{25})(\sqrt{24}+\sqrt{25})}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{-1}-\frac{\sqrt{2}+\sqrt{3}}{-1}+\frac{\sqrt{3}+\sqrt{4}}{-1}-...-\frac{\sqrt{24}+\sqrt{25}}{-1}\)
\(=\frac{(1+\sqrt{2})-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-...-(\sqrt{24}+\sqrt{25})}{-1}\)
\(=\frac{1-\sqrt{25}}{-1}=4\)
\(B=\frac{5}{4+\sqrt{11}}+\frac{11-3\sqrt{11}}{\sqrt{11}-3}-\frac{4}{\sqrt{5}-1}+\sqrt{(\sqrt{5}-2)^2}\)
\(=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{\sqrt{11}(\sqrt{11}-3)}{\sqrt{11}-3}-\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\sqrt{5}-2\)
\(=\frac{5(4-\sqrt{11})}{5}+\sqrt{11}-\frac{4(\sqrt{5}+1)}{4}+\sqrt{5}-2\)
\(=4-\sqrt{11}+\sqrt{11}-(\sqrt{5}+1)+\sqrt{5}-2\)
\(=1\)
\(C=\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}: \frac{1}{x^2-\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}(x+\sqrt{x}+1)}.(x^2-\sqrt{x})\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}(x+\sqrt{x}+1)}.\sqrt{x}(\sqrt{x^3}-1)\)
\(=\frac{(\sqrt{x}+1)(\sqrt{x^3}-1)}{x+\sqrt{x}+1}\)
\(=\frac{(\sqrt{x}+1)(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1}\)
\(=(\sqrt{x}+1)(\sqrt{x}-1)=x-1\)
\(S=\dfrac{\sqrt{2}-\sqrt{1}}{1+2}+\dfrac{\sqrt{3}-\sqrt{2}}{2+3}+\dfrac{\sqrt{4}-\sqrt{3}}{3+4}+...+\dfrac{\sqrt{25}-\sqrt{29}}{29+25}< \dfrac{2}{5}\)
Xét :\(\dfrac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{2n+1}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n+1}}< \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n}}=\dfrac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
Do đó :
\(S< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{5}\right)=\dfrac{2}{5}\)
Rút gọn :
\(B=\dfrac{1}{\sqrt{1}-\sqrt{2}}+\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}+...+\dfrac{1}{\sqrt{24}-\sqrt{25}}\)
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:
\(B=-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{24}+\sqrt{25}\)
\(B=\sqrt{25}-\sqrt{1}=4\)