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24 tháng 7 2017

a/ \(\frac{1}{2-\sqrt{3}}+\frac{3+\sqrt{3}}{\sqrt{3}}-\frac{4}{\sqrt{3}-1}\)

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

\(=1\)

b/ \(\sqrt{3x+40}-4=x\)

\(\sqrt{3x+40}=x+4\)

Điều kiện: \(\hept{\begin{cases}3x+40\ge0\\x+4\ge0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge-\frac{40}{3}\\x\ge-4\end{cases}}\)

\(\Leftrightarrow x\ge-\frac{40}{3}\)

Ta có: \(3x+40=x^2+8x+16\)

\(\Leftrightarrow x^2+5x-24=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-8\left(l\right)\\x=3\end{cases}}\)

23 tháng 7 2017

a. Ta có \(\frac{1}{2-\sqrt{3}}+\frac{3\sqrt{3}}{\sqrt{3}}-\frac{4}{\sqrt{3}-1}=\frac{2+\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+3-\frac{4\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=\frac{2+\sqrt{3}}{4-3}+3-\frac{4\left(\sqrt{3}+1\right)}{3-1}=2+\sqrt{3}+3-2\sqrt{3}-2=3-\sqrt{3}\)

b. \(\sqrt{3x+40}-4=x\)

ĐK \(3x+40\ge0\Leftrightarrow x\ge-\frac{40}{3}\)

\(\Leftrightarrow\sqrt{3x+40}=x+4\)\(\Leftrightarrow\hept{\begin{cases}x\ge-4\\3x+40=x^2+8x+16\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge-4\\x^2+5x-24=0\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge-4\\\left(x+8\right)\left(x-3\right)=0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge-4\\x=-8;x=3\end{cases}}}\Leftrightarrow x=3\left(tm\right)\)

Vậy x=3

3 tháng 1 2021

1.

\(\sqrt{50}-3\sqrt{8}+\sqrt{32}=5\sqrt{2}-6\sqrt{2}+4\sqrt{2}=3\sqrt{2}\)

2. 

a, ĐK: \(x\in R\)

\(pt\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\)

\(\Leftrightarrow\left|x-2\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

b, ĐK: \(x\ge3\)

\(pt\Leftrightarrow\sqrt{x-3}\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(l\right)\end{matrix}\right.\)

17 tháng 11 2016

b/ Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}.\sqrt{n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Áp dụng vào bài toán ta được

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{99}-\frac{1}{\sqrt{100}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)

Cả 2 câu là n tự nhiên khác 0 hết nhé

17 tháng 11 2016

a/ Ta có: \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)

Áp đụng vào bài toán được

\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{1680}+\sqrt{1681}}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{1681}-\sqrt{1680}\)

\(=\sqrt{1681}-\sqrt{1}=41-1=40\)

8 tháng 9 2020

2. a) \(ĐKXĐ:x\ge\frac{1}{3}\)

 \(\sqrt{3x-1}=4\)\(\Rightarrow\left(\sqrt{3x-1}\right)^2=4^2\)

\(\Leftrightarrow3x-1=16\)\(\Leftrightarrow3x=17\)\(\Leftrightarrow x=\frac{17}{3}\)( thỏa mãn ĐKXĐ )

Vậy \(x=\frac{17}{3}\)

b) \(ĐKXĐ:x\ge1\)

\(\sqrt{x-1}=x-1\)\(\Rightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)

\(\Leftrightarrow x-1=x^2-2x+1\)\(\Leftrightarrow x^2-2x+1-x+1=0\)

\(\Leftrightarrow x^2-3x+2=0\)\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)( thỏa mãn ĐKXĐ )

Vậy \(x=1\)hoặc \(x=2\)

3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)

\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)

Vì \(6>1\)\(\Leftrightarrow\sqrt{6}>\sqrt{1}=1\)\(\Rightarrow\sqrt{6}-1>0\)

\(6>4\)\(\Rightarrow\sqrt{6}>\sqrt{4}=2\)\(\Rightarrow\sqrt{6}-2>0\)

\(\Rightarrow\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|=\left(\sqrt{6}-1\right)-\left(\sqrt{6}-2\right)\)

\(=\sqrt{6}-1-\sqrt{6}+2=1\)

hay \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=1\)

8 tháng 9 2020

2a) \(\sqrt{3x-1}=4\)( ĐKXĐ : \(x\ge\frac{1}{3}\))

Bình phương hai vế

\(\Leftrightarrow\left(\sqrt{3x-1}\right)^2=4^2\)

\(\Leftrightarrow3x-1=16\)

\(\Leftrightarrow3x=17\)

\(\Leftrightarrow x=\frac{17}{3}\)( tmđk )

Vậy phương trình có nghiệm duy nhất là x = 17/3

b) \(\sqrt{x-1}=x-1\)( ĐKXĐ : \(x\ge1\))

Bình phương hai vế 

\(\Leftrightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)

\(\Leftrightarrow x-1=x^2-2x+1\)

\(\Leftrightarrow x^2-2x+1-x+1=0\)

\(\Leftrightarrow x^2-3x+2=0\)

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

\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}\left(tmđk\right)}\)

Vậy phương trình có hai nghiệm là x = 1 hoặc x = 2

3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)

\(=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)

\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}-\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot2+2^2}\)

\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}\)

\(=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)

\(=\sqrt{6}-1-\left(\sqrt{6}-2\right)\)

\(=\sqrt{6}-1-\sqrt{6}+2\)

\(=1\)

26 tháng 10 2020

a) \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-2+\sqrt{3}\)

\(=\frac{\sqrt{3}.\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}.\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-2+\sqrt{3}\)

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

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

b) \(\frac{-3}{2}.\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2.\left(1+\sqrt{5}\right)^2}\)

\(=\frac{-3}{2}.\sqrt{5-4\sqrt{5}+4}+\sqrt{4^2.\left(1+\sqrt{5}\right)^2}\)

\(=\frac{-3}{2}.\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{4^2}.\sqrt{\left(1+\sqrt{5}\right)^2}\)

\(=\frac{-3}{2}.\left|\sqrt{5}-2\right|+4.\left|1+\sqrt{5}\right|\)

\(=\frac{-3}{2}.\left(\sqrt{5}-2\right)+4\left(1+\sqrt{5}\right)\)

\(=\frac{-3\sqrt{5}}{2}+3+4+4\sqrt{5}\)

\(=\frac{-3\sqrt{5}}{2}+4\sqrt{5}+7\)

\(=\frac{-3\sqrt{5}}{2}+\frac{8\sqrt{5}}{2}+\frac{14}{2}\)

\(=\frac{-3\sqrt{5}+8\sqrt{5}+14}{2}=\frac{14+5\sqrt{5}}{2}\)

18 tháng 8 2016

a, = \(\frac{\sqrt{7}-5}{2}-\frac{2\left(3-\sqrt{7}\right)}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\frac{5\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\)

18 tháng 8 2016

a, = \(=\frac{\sqrt{7}-5}{2}-\frac{3-\sqrt{7}}{2}+\frac{6\sqrt{7}+12}{7-4}-\frac{20-5\sqrt{7}}{16-7}=\frac{\sqrt{7}-5-3+\sqrt{7}}{2}+\frac{6\sqrt{7}+12}{3}-\frac{20-5\sqrt{7}}{9}\)

14 tháng 11 2019

Bài 1:

a) \(\frac{4}{\sqrt{5}-\sqrt{3}}-\sqrt{12}\)

\(=\frac{4}{\sqrt{5}-\sqrt{3}}-2\sqrt{3}\)

\(=\frac{4\sqrt{5}+4\sqrt{3}}{\sqrt{5^2}-\sqrt{3^2}}-2\sqrt{3}\)

\(=\frac{4\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-2\sqrt{3}\)

\(=\frac{4\left(\sqrt{5}+\sqrt{3}\right)}{2}-2\sqrt{3}\)

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

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

\(=2\sqrt{5}\)

b) \(\sqrt{\frac{9}{8}}-\sqrt{\frac{49}{2}}+\sqrt{\frac{25}{18}}\)

\(=\frac{3}{2\sqrt{2}}-\frac{7}{\sqrt{2}}+\frac{5}{3\sqrt{2}}\)

\(=\frac{3\sqrt{2}}{2.2}-\frac{7}{\sqrt{2}}+\frac{5\sqrt{2}}{3.2}\)

\(=\frac{3\sqrt{2}}{4}-\frac{7}{\sqrt{2}}+\frac{5\sqrt{2}}{6}\)

\(=-\frac{23\sqrt{2}}{12}\)

14 tháng 11 2019

chung ta den bai 2 :3

a) \(\frac{x}{\sqrt{x}-2}=-1\)

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

\(\Leftrightarrow x-2=-\sqrt{x}\)

bình phương 2 vế ta được:

\(\Leftrightarrow x^2-4x+4=x\)

\(\Leftrightarrow x^2-4x+4-x=0\)

\(\Leftrightarrow x^2-5x+4=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=1\end{cases}}\)

b) \(\sqrt{x-2}=x-4\)

chúng ta lại bình phương hai vế như câu a và chúng ta được:

\(\Leftrightarrow x-2=x^2-8x+16\)

\(\Leftrightarrow x-2-x^2+8x-16=0\)

\(\Leftrightarrow9x-18-x^2=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=3\end{cases}}\)