\(\dfrac{1}{\sqrt{7-\sqrt{24}+1}}\)-\(\dfrac{1}{\sqrt{7+\sqrt{24}-...">
K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

a: \(=\dfrac{1}{\sqrt{6}-1+1}-\dfrac{1}{\sqrt{6}+1-1}\)

\(=\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}\)

=0

b: \(=\dfrac{3+\sqrt{7}-3+\sqrt{7}}{2}=\dfrac{2\sqrt{7}}{2}=\sqrt{7}\)

c: \(=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)

\(=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\)

4 tháng 10 2017

\(a)A=\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\\ =\sqrt{5}+3+\sqrt{3}-\left(\sqrt{5}+3\right)\\ =\sqrt{3}\)

\(b)B=\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\\ =\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}\\ =\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\\ =\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)^2\\ =\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)\\ =2\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)\\ =2\left(25-21\right)=8\)

12 tháng 7 2017

\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+\sqrt{48}}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2-\sqrt{3}\right)^2}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{48-20+10\sqrt{3}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)

\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)

= 5

\(\dfrac{\sqrt{3}-\sqrt{5+\sqrt{24}}+\sqrt{\sqrt{72}+11}}{\sqrt{6+\sqrt{20}}+\sqrt{2}-\sqrt{7+\sqrt{40}}}\)

\(=\dfrac{\sqrt{3}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{2}\right)^2}}{\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{2}-\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}}\)

\(=\dfrac{\sqrt{3}-\sqrt{2}-\sqrt{3}+3+\sqrt{2}}{\sqrt{5}+1+\sqrt{2}-\sqrt{2}-\sqrt{5}}\)

\(=3\)

Câu 1: 

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{2999}{3000}\)

\(\Leftrightarrow1-\dfrac{1}{n+1}=\dfrac{2999}{3000}\)

=>n+1=3000

hay n=2999

AH
Akai Haruma
Giáo viên
24 tháng 6 2018

a)

\(S=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{100}+\sqrt{101}}\)

\(S=\frac{\sqrt{2}-\sqrt{1}}{(\sqrt{2}+\sqrt{1})(\sqrt{2}-\sqrt{1})}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}+....+\frac{\sqrt{101}-\sqrt{100}}{(\sqrt{101}+\sqrt{100})(\sqrt{101}-\sqrt{100})}\)

\(S=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{101}-\sqrt{100}}{101-100}\)

\(S=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{101}-\sqrt{100}\)

\(S=\sqrt{101}-1\)

AH
Akai Haruma
Giáo viên
24 tháng 6 2018

b)

\(S=\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+...+\frac{1}{\sqrt{100}+\sqrt{102}}\)

\(S=\frac{\sqrt{4}-\sqrt{2}}{(\sqrt{4}+\sqrt{2})(\sqrt{4}-\sqrt{2})}+\frac{\sqrt{6}-\sqrt{4}}{(\sqrt{6}+\sqrt{4})(\sqrt{6}-\sqrt{4})}+...+\frac{\sqrt{102}-\sqrt{100}}{(\sqrt{102}+\sqrt{100})(\sqrt{102}-\sqrt{100})}\)

\(S=\frac{\sqrt{4}-\sqrt{2}}{4-2}+\frac{\sqrt{6}-\sqrt{4}}{6-4}+....+\frac{\sqrt{102}-\sqrt{100}}{102-100}\)

\(S=\frac{\sqrt{4}-\sqrt{2}+\sqrt{6}-\sqrt{4}+\sqrt{8}-\sqrt{6}+...+\sqrt{102}-\sqrt{100}}{2}\)

\(S=\frac{\sqrt{102}-\sqrt{2}}{2}\)

11 tháng 10 2018

Mik đăng câu hỏi mà ko thấy ai trả lời hết, với lại h mik giải được rồi nên đăng lên có ai tìm bài này thì có đáp án ha ( mấy CTV đừng hiểu lầm nhé)

a) \(x^2-13x+50=4\sqrt{x-3}\)

ĐKXĐ: \(x\ge3\)

\(\Leftrightarrow x^2-13x+50-4\sqrt{x-3}=0\)

\(\Leftrightarrow x^2-14x+x+49-3-+4-4\sqrt{x-3}=0\)

\(\Leftrightarrow(x^2-14x+49)+(x-3-4\sqrt{x-3}+4)=0\)

\(\Leftrightarrow\left(x-7\right)^2+\left(\sqrt{x-3}-2\right)^2=0\)

\(\Leftrightarrow\left(x-7\right)^2=\left(\sqrt{x-3}-2\right)^2\)

\(\Leftrightarrow x-7=-\sqrt{x-3}+2\)

\(\Leftrightarrow x-9=-\sqrt{x-3}\)

\(\Leftrightarrow x^2-18x+81=x-3\)

\(\Leftrightarrow x^2-19x+84=0\)

\(\Leftrightarrow\left(x+12\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-12=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)

Vậy \(x\in\left\{7;12\right\}\)

11 tháng 10 2018

\(b)\dfrac{4x}{x^2-5x+6}+\dfrac{3x}{x^2-7x+6}=6\)

ĐKXĐ: \(x\ne1,2,3,6\)

Đặt \(t=x^2-6x+6\)

pt \(\Leftrightarrow\dfrac{4x}{t+x}+\dfrac{3x}{t-x}=6\)

\(\Leftrightarrow\dfrac{4x\left(t-x\right)+3x\left(t+x\right)}{\left(t+x\right)\left(t-x\right)}=6\)

\(\Leftrightarrow\dfrac{7tx-x^2}{t^2-x^2}=6\)

\(\Leftrightarrow7tx-x^2=6t^2-6x^2\)

\(\Leftrightarrow-6t^2+7xt+5x^2=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)\left(t-\dfrac{5}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\t-\dfrac{5}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x^2-6x+6-\dfrac{5}{3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x^2-6x+\dfrac{13}{3}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{9+\sqrt{42}}{3}\\x=\dfrac{9-\sqrt{42}}{3}\end{matrix}\right.\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{\dfrac{-1}{2};\dfrac{9\pm\sqrt{42}}{3}\right\}\)