\(.S=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+...+\frac{1}{\sqrt{2009}+\sqrt{2011}}\)
Tính S
\(s=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+...+\frac{1}{\sqrt{2009}+\sqrt{2011}}\)
Tính tổng:
\(S=\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{2011}+\sqrt{2013}}\)
Viết ngược lại đề bài nha rồi trục căn thức
\(2S=\sqrt{2013}-\sqrt{2011}+\sqrt{2011}-\sqrt{2009}+....+\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}\)
\(S=\frac{\sqrt{2013}-\sqrt{3}}{2}\)
Tính :
a ) \(S=\frac{1}{\sqrt{1}\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+.....+\)\(\frac{1}{\sqrt{2017}+\sqrt{2019}}\)
b ) \(S=\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+....+\frac{1}{\sqrt{100}+\sqrt{102}}\)
c ) \(S=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.....+\frac{1}{\sqrt{100}+\sqrt{101}}\)
d ) \(S=\frac{1}{\sqrt{3}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}}+....+\frac{1}{\sqrt{2016}+\sqrt{2019}}\)
......................?
mik ko biết
mong bn thông cảm
nha ................
Giải pt :\(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\)
giải pt:\(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\)
Đặt: \(\hept{\begin{cases}\sqrt{x-2009}=a\\\sqrt{y-2010}=b\\\sqrt{z-2011}=c\end{cases}}\)
Ta có: \(\frac{1}{a}-\frac{1}{a^2}+\frac{1}{b}-\frac{1}{b^2}+\frac{1}{c}-\frac{1}{c^2}-\frac{3}{4}=0\)
\(\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{c^2}-\frac{1}{c}+\frac{3}{4}=0\)
\(\Leftrightarrow\left(\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}\right)+\left(\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}\right)+\left(\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow a=b=c=\frac{1}{2}\)
Thay vào tìm x;y;z
Đặt: \(\hept{\begin{cases}\sqrt{x-2009}=a\\\sqrt{y-2010}=b\\\sqrt{z-2011}=c\end{cases}}\)
Ta có: \frac{1}{a}-\frac{1}{a^2}+\frac{1}{b}-\frac{1}{b^2}+\frac{1}{c}-\frac{1}{c^2}-\frac{3}{4}=0a1−a21+b1−b21+c1−c21−43=0
\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{c^2}-\frac{1}{c}+\frac{3}{4}=0⇔a21−a1+b21−b1+c21−c1+43=0
\Leftrightarrow\left(\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}\right)+\left(\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}\right)+\left(\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}\right)=0⇔(a21−a1+41)+(b21−b1+41)+(c21−c1+41)=0
\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0⇔(a1−21)2+(b1−21)2+(c1−21)2=0
\Leftrightarrow a=b=c=\frac{1}{2}⇔a=b=c=21
Thay vào tìm x;y;z
Tính
M=\(\frac{\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2005.2007.2009}}{\frac{1}{1\sqrt{3}+3\sqrt{1}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+7\sqrt{5}}+...+\frac{1}{2007\sqrt{2009}+2009\sqrt{2007}}}\)
Xét tử số có dạng : \(\frac{1}{\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)}=\frac{1}{4}\left[\frac{1}{\left(2n+1\right)\left(2n+2\right)}-\frac{1}{\left(2n+2\right)\left(2n+3\right)}\right]\) với \(n\in N\)
Ta có : \(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2005.2007.2009}\)
\(=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\frac{1}{4}.\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\frac{1}{4}\left(\frac{1}{5.7}-\frac{1}{7.9}\right)+...+\frac{1}{4}\left(\frac{1}{2005.2007}-\frac{1}{2007.2009}\right)\)
\(=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{2005.2007}-\frac{1}{2007.2009}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{2007.2009}\right)\)
Xét mẫu số có dạng : \(\frac{1}{\left(2n+1\right)\sqrt{2n+3}+\left(2n+3\right)\sqrt{2n+1}}=\frac{1}{\sqrt{2n+1}.\sqrt{2n+3}\left(\sqrt{2n+1}+\sqrt{2n+3}\right)}\)
\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{\sqrt{2n+1}.\sqrt{2n+3}\left[\left(2n+3\right)-\left(2n+1\right)\right]}=\frac{1}{2}.\left(\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{2n+3}}\right)\)với \(n\in N\)
Áp dụng : \(\frac{1}{1\sqrt{3}+3\sqrt{1}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+7\sqrt{5}}+...+\frac{1}{2007\sqrt{2009}+2009\sqrt{2007}}\)
\(=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}+...+\frac{1}{\sqrt{2007}}-\frac{1}{\sqrt{2009}}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{2009}}\right)\)
Suy ra : \(M=\frac{\frac{1}{4}\left(\frac{1}{3}-\frac{1}{2007.2009}\right)}{\frac{1}{2}\left(1-\frac{1}{\sqrt{2009}}\right)}\)
Tới đây bài toán đã gọn hơn , bạn tự tính nhé :)
Tính giá trị biểu thức:
\(\text{a) }\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{2010}+\sqrt{2011}}\)
\(\text{b) }\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{121\sqrt{120}+120\sqrt{121}}\)
\(\text{c) }\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...\sqrt{+1+\frac{1}{2010^2}+\frac{1}{2011^2}}\)
Giải phương trình: \(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\)
Thưa bn mk đã làm ra nhưng không biết có đúng không. Xem nhá:
Ta có:
\(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2001}-1}{y-2001}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\Leftrightarrow"\frac{1}{\sqrt{x-2009}}-\frac{1}{2}"^2+\)
\("\frac{1}{\sqrt{y-2010}}-\frac{1}{2}"^2-"\frac{1}{\sqrt{z-2011}}-\frac{1}{2}"^2=0\)
\(\Rightarrow x=2013;y=2014;z=2015\)
P/s: Bn thay Ngoặc Kép thành Ngoặc Đơn nhé
tính tổng
\(S=\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{7}+\sqrt{5}}+...+\frac{1}{\sqrt{2019}+\sqrt{2017}}\)
Bạn ơi cái này mk chỉ ghi cách làm và ct thôi nha
đây dùng hàng đẳng thức (a-b)(a+b)=a^2-b^2
còn kia là công thức toán lớp 6
\(\frac{1}{\sqrt{3}+\sqrt{1}}=\frac{\sqrt{3}-\sqrt{1}}{\left(\sqrt{3}+\sqrt{1}\right)\left(\sqrt{3}-\sqrt{1}\right)}=\frac{\sqrt{3}-\sqrt{1}}{\sqrt{3^2}-\sqrt{1^2}}=\frac{1}{2}\left(\sqrt{3}-\sqrt{1}\right)\)
Tương tự:
\(\frac{1}{\sqrt{5}+\sqrt{3}}=\frac{1}{2}\left(\sqrt{5}-\sqrt{3}\right)\)
.....
\(\frac{1}{\sqrt{2019}+\sqrt{2017}}=\frac{1}{2}\left(\sqrt{2019}-\sqrt{2017}\right)\)
Cộng các vế với nhau ta được:
\(S=\frac{1}{2}\left(\sqrt{2019}-\sqrt{1}\right)=\frac{1}{2}\left(\sqrt{2019}-1\right)\)
Bài này em dùng trục căn thức:
VD: \(\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}-1}{\left(\sqrt{3}\right)^2-1}=\frac{\sqrt{3}-1}{2}\)
Tương tự thì ta có:
\(S=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2}\)
\(=\frac{\sqrt{2019^2}-1}{2}=\frac{2019-1}{2}=1009\)