Bài tập:so sánh
a. \(2\sqrt{3}\) và \(3\sqrt{2}\)
b. \(2\sqrt{3}+1\)và 4
c.\(\sqrt{2015}-\sqrt{2014}\) và \(\sqrt{2014}-\sqrt{2013}\)
Tính gía trị biểu thức:
\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+....+\frac{1}{2014\sqrt{2013}+2013\sqrt{2014}}+\frac{1}{2015\sqrt{2014}+2014\sqrt{2015}}\)
Chứng minh \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\) rồi áp dụng với n = 1,2,....,2014
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) \(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1=2\sqrt{2}\)
so sánh : a) \(\sqrt{2014}-\sqrt{2013}và\sqrt{2015}-\sqrt{2014}\)
b) \(\sqrt{3}+\sqrt{5}và\sqrt{7}+\sqrt{2}\)
Rút gọn \(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}+\frac{1}{\sqrt{4}-\sqrt{5}}-...-\frac{1}{\sqrt{2013}+\sqrt{2014}}+\frac{1}{\sqrt{2014}-\sqrt{2015}}\)
\(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}+...-\frac{1}{\sqrt{2013}-\sqrt{2014}}+\frac{1}{\sqrt{2014}-\sqrt{2015}}\)
\(=\frac{\sqrt{2}+\sqrt{3}}{2-3}-\frac{\sqrt{3}+\sqrt{4}}{3-4}+...+\frac{\sqrt{2014}+\sqrt{2015}}{2014-2015}\)
\(=-\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{3}+\sqrt{4}-\left(\sqrt{4}+\sqrt{5}\right)+...+\sqrt{2014}+\sqrt{2015}\)
=\(-\sqrt{2}+\sqrt{2015}\)
So sánh 2 số:
\(a)\sqrt{2014}-\sqrt{2013};B=\sqrt{2015}-\sqrt{2014}\\ b)E=\frac{2014}{\sqrt{2015}}+\frac{2015}{\sqrt{2014}};F=\sqrt{2014}+\sqrt{2015}\)
RGBT:
E=\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{2015\sqrt{2014}+2014\sqrt{2015}}+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vô bài toán được
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
\(=1-\frac{1}{\sqrt{2016}}\)
bài 45:so sánh
a)3\(\sqrt{3}\) và \(\sqrt{12}\)
b)7 và 3\(\sqrt{5}\)
c)\(\dfrac{1}{3}\sqrt{51}\) và \(\dfrac{1}{5}\sqrt{150}\)
d)\(\dfrac{1}{2}\sqrt{6}\) và \(6\sqrt{\dfrac{1}{2}}\)
a) \(3\sqrt{3}=\sqrt{27}>\sqrt{12}\)
b) \(3\sqrt{5}=\sqrt{45}>\sqrt{27}\)
c) \(\dfrac{1}{3}\sqrt{51}=\sqrt{\dfrac{51}{9}}< \sqrt{\dfrac{54}{9}}=6=\sqrt{\dfrac{150}{25}}=\dfrac{1}{5}\sqrt{150}\)
d) \(\dfrac{1}{2}\sqrt{6}=\sqrt{\dfrac{6}{4}}=\sqrt{\dfrac{3}{2}}< \sqrt{\dfrac{36}{2}}=6\sqrt{\dfrac{1}{2}}\)
Giải phương trình, hệ phương trình:
a) \(\frac{\sqrt{x-2013}-1}{x-2013}+\frac{\sqrt{y-2014}-1}{y-2014}+\frac{\sqrt{z-2015}-1}{z-2015}=\frac{3}{4}\)
b) \(\left\{{}\begin{matrix}x^3+1=2y\\y^3+1=2x\end{matrix}\right.\)
c)\(\sqrt{x^2-3x+2}+\sqrt{x-3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)
d)\(5x-2\sqrt{x}\left(2+y\right)+y^2+1=0\)
c/ ĐKXĐ: \(x\ge3\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x-3}-\sqrt{x-2}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\left(\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{x-2}\right)-\left(\sqrt{\left(x-1\right)\left(x+3\right)}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}-\sqrt{x+3}=0\\\sqrt{x-1}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+3}\\\sqrt{x-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\left(vn\right)\\x=2< 3\left(ktm\right)\end{matrix}\right.\)
Vậy pt đã cho vô nghiệm
a/ ĐKXĐ: \(\left\{{}\begin{matrix}x>2013\\y>2014\\z>2015\end{matrix}\right.\)
\(\Leftrightarrow\frac{1}{4}-\frac{\sqrt{x-2013}-1}{x-2013}+\frac{1}{4}-\frac{\sqrt{y-2014}-1}{y-2014}+\frac{1}{4}-\frac{\sqrt{z-2015}-1}{z-2015}=0\)
\(\Leftrightarrow\frac{x-2013-4\sqrt{x-2013}+4}{4\left(x-2013\right)}+\frac{y-2014-4\sqrt{y-2014}+4}{4\left(y-2014\right)}+\frac{z-2015-4\sqrt{z-2015}+4}{4\left(z-2015\right)}=0\)
\(\Leftrightarrow\left(\frac{\sqrt{x-2013}-2}{2\sqrt{x-2013}}\right)^2+\left(\frac{\sqrt{y-2014}-2}{2\sqrt{y-2014}}\right)^2+\left(\frac{\sqrt{z-2015}-2}{2\sqrt{z-2015}}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2013}-2=0\\\sqrt{y-2014}-2=0\\\sqrt{z-2015}-2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2017\\y=2018\\z=2019\end{matrix}\right.\)
b/ Trừ vế cho vế 2 pt ta được:
\(x^3-y^3=2\left(y-x\right)\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+y^2-xy\right)+2\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+y^2-xy+2\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4}+2\right]=0\)
\(\Leftrightarrow x-y=0\Leftrightarrow x=y\)
Thay vào pt đầu:
\(x^3+1=2x\Leftrightarrow x^3-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\)
\(\Leftrightarrow...\)
Cho M=\(\frac{\sqrt{2}-\sqrt{1}}{1+1}+\frac{\sqrt{3}-\sqrt{2}}{2+3}+\frac{\sqrt{4}-\sqrt{3}}{3+4}+...+\frac{\sqrt{2015}-\sqrt{2014}}{2014+2015}\)
Hãy so sánh M với 1/2