Tính
\(S=\frac{1}{3\sqrt{1}+3\sqrt{3}}+\frac{1}{3\sqrt{3}+3\sqrt{5}}+...+\frac{1}{3\sqrt{2017}+3\sqrt{2019}}\)
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:
\(S=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2}\)
\(=\frac{\sqrt{2019^2}-1}{2}=\frac{2019-1}{2}=1009\)
Tính :
a ) \(S=\frac{1}{3\sqrt{1}+3\sqrt{3}}+\frac{1}{3\sqrt{3}+3\sqrt{5}}+...+\)\(\frac{1}{3\sqrt{2017}+3\sqrt{2019}}\)
b ) \(S=\frac{7}{\sqrt{2.2}+\sqrt{2.3}}+\frac{7}{\sqrt{2.3}+\sqrt{2.4}}\)\(+...+\frac{7}{\sqrt{2.2018}+\sqrt{2.2019}}\)
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 ................
Tính tổng S= \(\frac{1}{1+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{2019^2}+\sqrt{2019^2-2}}\)
\(S=\frac{\sqrt{3}-1}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2019^2-\left(2019^2-2\right)}\)
\(S=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2}\)
\(S=\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+...+\sqrt{2019^2}-\sqrt{2019^2-2}\right)\)
\(S=\frac{1}{2}\left(-1+\sqrt{2019^2}\right)\)
\(S=\frac{\left(2019-1\right)}{2}=1009\)
\(S=\frac{1-\sqrt{3}}{1-3}+\frac{\sqrt{3}-\sqrt{5}}{3-5}+\frac{\sqrt{5}-\sqrt{7}}{5-7}+...+\frac{2019-\sqrt{2019^2-2}}{2019^2-2019^2-2}.\)
\(S=\frac{1-\sqrt{3}}{-2}+\frac{\sqrt{3}-\sqrt{5}}{-2}+\frac{\sqrt{5}-\sqrt{7}}{-2}+...+\frac{2019-\sqrt{2019^2-2}}{-2}.\)
\(-2S=1-\sqrt{3}+\sqrt{3}-\sqrt{5}+\sqrt{5}...+2019-\sqrt{2019^2-2}\)
\(-2S=1-\sqrt{2019^2-2}\Rightarrow S=\frac{\sqrt{2019^2-2}-1}{2}\)
Bài 1. cho \(f\left(x\right)=\left(2x^3-21x-29\right)^{2019}\). Tính f(x) tại \(x=\sqrt[3]{7+\sqrt{\frac{49}{8}}}+\sqrt[3]{7-\sqrt{\frac{49}{8}}}\)
Bài 2. Tìm số tự nhiên n biết rằng: \(\frac{1}{\sqrt{1^3+2^3}}+\frac{1}{\sqrt{1^3+2^3+3^3}}+...+\frac{1}{\sqrt{1^3+2^3+3^3+...+n^3}}=\frac{2015}{2017}\)
Bài 3. Tính \(A=\left(3x^3+8x^2+2\right)\)với \(x=\frac{\sqrt[3]{17\sqrt{5}-38}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)
Bài 4. CMR: \(\sqrt{1}+\sqrt{2}+...+\sqrt{n}\le n.\sqrt{\frac{n+1}{2}}\)
Nhìn cái đề bài đáng sợ kinh, ai giúp tớ vs
1, \(x^3=\left(7+\sqrt{\frac{49}{8}}\right)+\left(7-\sqrt{\frac{49}{8}}\right)+3x\sqrt[3]{\left(7+\sqrt{\frac{49}{8}}\right)\left(7-\sqrt{\frac{49}{8}}\right)}\)
\(=14+3x\cdot\frac{7}{2}=14+\frac{21x}{2}\)
\(\Leftrightarrow x^3-\frac{21}{2}x-14=0\)
Ta có: \(f\left(x\right)=\left(2x^3-21-29\right)^{2019}=\left[2\left(x^3-\frac{21}{2}x-14\right)-1\right]^{2019}=\left(-1\right)^{2019}=-1\)
2, ta có: \(1^3+2^3+...+n^3=\left(1+2+...+n\right)^2=\left[\frac{n\left(n+1\right)}{2}\right]^2\) (bạn tự cm)
Áp dụng công thức trên ta được n=2016
3, \(x=\frac{\sqrt[3]{17\sqrt{5}-38}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}=\frac{\sqrt[3]{\left(\sqrt{5}\right)^3-3.\left(\sqrt{5}\right)^2.2+3\sqrt{5}.2^2-2^3}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{9-2.3\sqrt{5}+5}}\)
\(=\frac{\sqrt[3]{\left(\sqrt{5}-2\right)^3}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}=\frac{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}{\sqrt{5}+3-\sqrt{5}}=\frac{5-4}{3}=\frac{1}{3}\)
Thay x=1/3 vào A ta được;
\(A=3x^3+8x^2+2=3.\left(\frac{1}{3}\right)^3+8.\left(\frac{1}{3}\right)^2+2=3\)
Bài 4
ÁP DỤNG BĐT CAUCHY
là ra
\(\frac{1}{\sqrt{1^3+2^3}}+\frac{1}{\sqrt{1^3+2^3+3^3}}+...+\frac{1}{\sqrt{1^3+2^3+3^3+...+n^3}}=\frac{2015}{2017}\) (1)
Cần CM: \(1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\) quy nạp nhé bn, trên mạng có nhìu
(1) \(\Leftrightarrow\)\(\frac{1}{\sqrt{\left(1+2\right)^2}}+\frac{1}{\sqrt{\left(1+2+3\right)^2}}+...+\frac{1}{\sqrt{\left(1+2+3+...+n\right)^2}}=\frac{2015}{2017}\)
\(\Leftrightarrow\)\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+n}=\frac{2015}{2017}\)
\(\Leftrightarrow\)\(\frac{1}{\frac{2\left(2+1\right)}{2}}+\frac{1}{\frac{3\left(3+1\right)}{2}}+...+\frac{1}{\frac{n\left(n+1\right)}{2}}=\frac{2015}{2017}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{2015}{2017}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{2015}{2017}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{2015}{2017}\)
\(\Leftrightarrow\)\(n=2016\)
tính
\(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}+\frac{1}{\sqrt{4}-\sqrt{5}}+...+\frac{1}{\sqrt{2018}-\sqrt{2019}}\)
\(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}-...-\frac{1}{\sqrt{2018}-\sqrt{2019}}\)
\(=\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{2019}-\sqrt{2018}}{2019-2018}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}\)
\(=\sqrt{2019}-\sqrt{2}\)
rút gọn bt:
\(\frac{\sqrt{2}-\sqrt{6}}{1-\sqrt{3}}-\frac{3+\sqrt{27}}{1+\sqrt{3}}\)
Giải pt:
x=\(\frac{1}{\sqrt{2019}-\sqrt{2018}}\)và y=\(\frac{1}{\sqrt{2018}-\sqrt{2017}}\)
b,So sánh
(giúp mk vs huhu...)
a, \(\frac{\sqrt{2}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\)-\(\frac{3\left(1+\sqrt{3}\right)}{1+\sqrt{3}}\)
=\(\sqrt{2}-3\)
b,X=\(\sqrt{2019}+\sqrt{2018}\)
(Khử mẫu,nhân tử&mẫu vs\(\sqrt{2019}+\sqrt{2018}\))
Y=\(\sqrt{2018}+\sqrt{2017}\)
(Khử mẫu,nhân tử&mẫu vs\(\sqrt{2018}+\sqrt{2017}\))
So sánh:X & Y<=>X-\(\sqrt{2018}\)&Y-\(\sqrt{2018}\)(Trừ hai vế cho \(\sqrt{2018}\)) <=>\(\sqrt{2019}\)&\(\sqrt{2017}\)
Có:2019>2017
=>\(\sqrt{2019}>\sqrt{2017}\)
=>X>Y
Câu b, mk ko bt có lm đúng ko?
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3+3\sqrt{4}}}+...+\frac{1}{2017\sqrt{2016}+2016\sqrt{2017}}\)
Tính giá trị của biểu thức .
\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}=1-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2007}-1}{\sqrt{2007}}\)
Rút gọn \(\frac{1-\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}}+\frac{1-\sqrt{4}+\sqrt{5}}{1+\sqrt{4}+\sqrt{5}}+...+\frac{1-\sqrt{2018}+\sqrt{2019}}{1+\sqrt{2018}+\sqrt{2019}}\)