Giải bằng máy tính CASIO kết quả chính xác bài tập sau :
\(\dfrac{1}{\sqrt{2}+\sqrt{3}}\)+\(\dfrac{1}{\sqrt{3}+\sqrt{4}}+....+\dfrac{1}{\sqrt{2014}+\sqrt{2015}}\)
Cho M=\(\dfrac{\sqrt{2}-\sqrt{1}}{1+2}+\dfrac{\sqrt{3}-\sqrt{2}}{2+3}+\dfrac{\sqrt{4}-\sqrt{3}}{3+4}+...+\dfrac{\sqrt{2015}-\sqrt{2014}}{2014+2015}\)
Hãy so sánh M với \(\dfrac{1}{2}\)
Tính
\(M=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+.......+\dfrac{1}{2015\sqrt{2014}+2014\sqrt{2015}}\)
1. Tính
a. \(\left(3\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{3}-3\sqrt{2}\right)\)
b. \(\dfrac{1}{\sqrt{2013}-\sqrt{2014}}-\dfrac{1}{\sqrt{2014}-\sqrt{2015}}\)
c. \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}\)
d. \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}\)
c: Ta có: \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}\)
\(=4+\sqrt{10}-4+\sqrt{10}\)
\(=2\sqrt{10}\)
d: Ta có: \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}\)
\(=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1\)
\(=2\sqrt{2}\)
a) \(=\left(2\sqrt{3}\right)^2-\left(3\sqrt{2}\right)^2=12-18=-6\)
b) \(=\dfrac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\dfrac{\sqrt{2014}+\sqrt{2015}}{2014-2015}=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}-\sqrt{2015}=-\sqrt{2013}-\sqrt{2015}\)
c) \(=4+\sqrt{10}-4+\sqrt{10}=2\sqrt{10}\)
d) \(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1=2\sqrt{2}\)
Tính giá trị biểu thức
\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2015\sqrt{2014}+2014\sqrt{2015}}\)
Giải:
\(\dfrac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}\) \(=\dfrac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)^2k-k^2\left(k+1\right)}\)
\(=\dfrac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)k\left(k+1-k\right)}=\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\)
Áp dụng vào biểu thức ta có:
\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}\) \(+...+\dfrac{1}{2015\sqrt{2014}+2014\sqrt{2015}}\)
\(=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2014}}-\dfrac{1}{\sqrt{2015}}\)
\(=1-\dfrac{1}{\sqrt{2015}}\)
tính giá trị của các biểu thức sau(casio)
a/\(B=\dfrac{1}{\sqrt{x}+\sqrt{x+1}}+\dfrac{1}{\sqrt{x+1}+\sqrt{x+2}}+\dfrac{1}{\sqrt{x+2}+\sqrt{x+3}}+....+\dfrac{1}{\sqrt{x+2015}+\sqrt{x+2016}}v\text{ới}x=2017\)
\(B=B_1+B_2+...+B_{2016}\)
\(B_1=\dfrac{\sqrt{x+1}-\sqrt{x}}{\left(\sqrt{x}+\sqrt{x+1}\right)\left(\sqrt{x+1}-\sqrt{x}\right)}=\dfrac{\sqrt{x+1}-\sqrt{x}}{x+1-x}\)
\(B_1=\sqrt{x+1}-\sqrt{x}\)
\(B_2=\sqrt{x+2}-\sqrt{x+1}\)
\(B_3=\sqrt{x+3}-\sqrt{x+2}\)
...
\(B_{2015}=\sqrt{x+2015}-\sqrt{x+2014}\)
\(B_{2016}=\sqrt{x+2016}-\sqrt{x+2015}\)
\(B=\sqrt{x+2016}-\sqrt{x}\)
\(B\left(2017\right)=\sqrt{2017+2016}-\sqrt{2017}\)
Giải bài đầy đủ giùm mình, đừng viết tắt.
1) \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}\)
2)\(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
3)\(\dfrac{2}{\sqrt{5}-2}+\dfrac{-2}{\sqrt{5}+2}\)
4)\(\dfrac{3}{1-\sqrt{2}}+\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)
5)\(\dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}-\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
1, \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}=\dfrac{3+2\sqrt{2}}{9-8}-\dfrac{3-2\sqrt{2}}{9-8}\)
\(=3+2\sqrt{2}-3+2\sqrt{2}=4\sqrt{2}\)
2, \(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+\sqrt{12}}\)
\(=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{6}\left(\sqrt{2}+\sqrt{3}\right)}=\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{6}.\left(-1\right)}-\dfrac{3\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}.\left(-1\right)}\)
\(=\dfrac{2\sqrt{3}+3\sqrt{2}-3\sqrt{2}+3\sqrt{3}}{-\sqrt{6}}=\dfrac{5\sqrt{3}}{-\sqrt{6}}=-5\sqrt{18}=-15\sqrt{2}\)
3, \(\dfrac{2}{\sqrt{5}-2}+\dfrac{-2}{\sqrt{5}+2}=\dfrac{2\left(\sqrt{5}+2\right)}{1}-\dfrac{2\left(\sqrt{5}-2\right)}{1}\)
\(=2\sqrt{5}+4-2\sqrt{5}+4=8\)
tương tự
\(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}=3+2\sqrt{2}-3+2\sqrt{2}=4\sqrt{2}\)
Chứng tỏ :
\(\dfrac{1}{\sqrt{x+2014}+\sqrt{y+2014}}-\dfrac{1}{\sqrt{2015-x}+\sqrt{2015-y}}+\dfrac{1}{\sqrt{2014-x}+\sqrt{2014-y}}\ne0\)
Thu gọn và cho bt tập xác định của biểu thức
A= \(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{3\sqrt{x}-2}{x-\sqrt{x}+1}-\dfrac{2x\sqrt{x}+2\sqrt{x-5}}{x\sqrt{x}+1}\)
B= \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
C= \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}-2}\)
D= \(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{3\sqrt{x}-1}{x-5\sqrt{x}+2}\)
E= \(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)
a, ĐKXĐ: \(x\ge0,\)
b, ĐKXĐ: \(x\ge0,x\ne1\)
c, ĐKXĐ: \(x\ge0,x\ne4\)
d,ĐKXĐ:\(x\ge0,x\ne9,x\ne4\)
e,ĐKXĐ:\(x\ge0,x\ne1,x\ne4\)
Tìm điều kiện để các biểu thức sau xác định
a)\(\sqrt{x+1}-\dfrac{1}{2}\)
b)\(2.\sqrt{1-2x}-\dfrac{\sqrt{3}-1}{4}\)
c)\(\sqrt{x+1}-\sqrt{x-2}\)
d)\(\sqrt{2-3x}-\sqrt{1-2x}\)
e)\(2.\sqrt{\sqrt{3}-2x}+\dfrac{1}{x-1}\)
f)\(\dfrac{1}{2}.\sqrt{x-\dfrac{\sqrt{3}}{2}}-\dfrac{1}{\sqrt{x}-1}\)
g)\(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+2}\)
h)\(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x^2+2}}\)
a, \(x+1\ge0\Leftrightarrow x\ge-1\)
b, \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)
c, \(\left\{{}\begin{matrix}x+1\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge2\end{matrix}\right.\Leftrightarrow x\ge2\)
d, \(\left\{{}\begin{matrix}2-3x\ge0\\1-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{2}{3}\\x\le\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\le\dfrac{1}{2}\)
e, \(\left\{{}\begin{matrix}\sqrt{3}-2x\ge0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\Leftrightarrow x\le\dfrac{\sqrt{3}}{2}\)
f, \(\left\{{}\begin{matrix}x-\dfrac{\sqrt{3}}{2}\ge0\\x\ge0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{\sqrt{3}}{2}\\x\ge0\\x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\)
g, \(\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}+2\ne0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)