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=>\(25\cdot\dfrac{\sqrt{a-3}}{5}-7\cdot\dfrac{2}{3}\cdot\sqrt{a-3}-7\sqrt{a^2-9}+18\cdot\dfrac{1}{3}\sqrt{a^2-9}=0\)

=>\(\sqrt{a-3}\cdot\dfrac{1}{3}-\sqrt{a^2-9}=0\)

=>\(\sqrt{a-3}\left(\dfrac{1}{3}-\sqrt{a+3}\right)=0\)

=>a-3=0 hoặc a+3=1/9

=>a=3 hoặc a=-26/9

23 tháng 8 2021

a, ĐK :a >= 3

\(25\sqrt{\frac{a-3}{25}}-7\sqrt{\frac{4a-12}{9}}-7\sqrt{a^2-9}+18\sqrt{\frac{9a^2-81}{81}}=0\)

\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{\left(a-3\right)\left(a+3\right)}+6\sqrt{\left(a-3\right)\left(a+3\right)}=0\)

\(\Leftrightarrow\sqrt{a-3}\left(5-\frac{14}{3}-\sqrt{a+3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{a-3}=0\\\sqrt{a+3}=\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{2}{9}\left(loai\right)\end{cases}}\)

b, \(ĐK:x\ge-\frac{1}{2}\)

\(\Leftrightarrow3\sqrt{2x+1}-2\sqrt{2x+1}+\frac{1}{3}\sqrt{2x+1}=4\)

\(\Leftrightarrow\frac{4}{3}\sqrt{2x+1}=4\)

\(\Leftrightarrow\sqrt{2x+1}=3\)

\(\Leftrightarrow x=4\left(tm\right)\)

23 tháng 8 2021

a) đk: \(a\ge3\)

pt \(\Leftrightarrow25\frac{\sqrt{a-3}}{\sqrt{25}}-7\frac{\sqrt{4\left(a-3\right)}}{\sqrt{9}}-7\sqrt{a^2-9}+18\frac{\sqrt{9\left(a^2-9\right)}}{\sqrt{81}}=0\)

\(\Leftrightarrow5\sqrt{a-3}-\frac{7.2}{3}\sqrt{a-3}-7\sqrt{a^2-9}+\frac{18.3}{9}\sqrt{a^2-9}=0\)

\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{a^2-9}+6\sqrt{a^2-9}=0\)

\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}-\sqrt{a^2-9}=0\)

\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}=\sqrt{a^2-9}\)

\(\Leftrightarrow\frac{1}{9}\left(a-3\right)=a^2-9\)

\(\Leftrightarrow a^2-\frac{1}{9}a-\frac{26}{3}=0\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{26}{9}\left(loại\right)\end{cases}}\)

14 tháng 7 2021

\(25\sqrt{\dfrac{x-3}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\left(x\ge3\right)\)

\(=25\sqrt{\dfrac{1}{25}.\left(x-3\right)}-7\sqrt{\dfrac{4}{9}.\left(x-3\right)}-7\sqrt{x^2-9}+18\sqrt{\dfrac{1}{9}.\left(x^2-9\right)}=0\)

\(=5\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Rightarrow\dfrac{1}{3}\sqrt{x-3}-\sqrt{\left(x-3\right)\left(x+3\right)}=0\Rightarrow\sqrt{x-3}-3\sqrt{\left(x-3\right)\left(x+3\right)}=0\)

\(\Rightarrow\sqrt{x-3}\left(1-3\sqrt{x+3}\right)=0\Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1=3\sqrt{x+3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{26}{9}\left(l\right)\end{matrix}\right.\)

14 tháng 7 2021

cảm ơn nhaa<33

28 tháng 8 2020

\(ĐKXĐ:a\ge3\)

\(25\sqrt{\frac{a-3}{25}}-7\sqrt{\frac{4a-12}{9}}-7\sqrt{a^2-9}+18\sqrt{\frac{9a^2-81}{81}}=0\)

\(\Leftrightarrow25.\sqrt{\frac{1}{25}.\left(a-3\right)}-7\sqrt{\frac{4}{9}.\left(a-3\right)}-7\sqrt{a^2-9}+18\sqrt{\frac{9}{81}.\left(a^2-9\right)}=0\)

\(\Leftrightarrow25.\sqrt{\frac{1}{25}}.\sqrt{a-3}-7.\sqrt{\frac{4}{9}}.\sqrt{a-3}-7\sqrt{a^2-9}+18.\sqrt{\frac{9}{81}}.\sqrt{a^2-9}=0\)

\(\Leftrightarrow25.\frac{1}{5}.\sqrt{a-3}-7.\frac{2}{3}.\sqrt{a-3}-7\sqrt{a^2-9}+18.\frac{1}{3}.\sqrt{a^2-9}=0\)

\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}.\sqrt{a-3}-7\sqrt{a^2-9}+6\sqrt{a^2-9}=0\)

\(\Leftrightarrow\frac{1}{3}.\sqrt{a-3}-\sqrt{a^2-9}=0\)

\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}-\sqrt{\left(a-3\right)\left(a+3\right)}=0\)

\(\Leftrightarrow\sqrt{a-3}.\left(\frac{1}{3}-\sqrt{a+3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{a-3}=0\\\frac{1}{3}-\sqrt{a+3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}a-3=0\\\sqrt{a+3}=\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=3\\a+3=\frac{1}{9}\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=3\\a=\frac{-26}{9}\end{cases}}\)

mà \(a\ge3\)\(\Rightarrow a=\frac{-26}{9}\)không thỏa mãn

Vậy \(a=3\)

28 tháng 8 2020

Bài làm:

đk: \(a\ge3\)

Ta có: \(25\sqrt{\frac{a-3}{25}}-7\sqrt{\frac{4a-12}{9}}-7\sqrt{a^2-9}+18\sqrt{\frac{9a^2-81}{81}}=0\)

\(\Leftrightarrow5\sqrt{a-3}+\frac{14}{3}\sqrt{a-3}-7\sqrt{a^2-9}+6\sqrt{a^2-9}=0\)

\(\Leftrightarrow\sqrt{a^2-9}=\sqrt{a-3}\)

\(\Leftrightarrow\left|a^2-9\right|=\left|a-3\right|\)

\(\Leftrightarrow\orbr{\begin{cases}a^2-9=a-3\\a^2-9=3-a\end{cases}}\Leftrightarrow\orbr{\begin{cases}a^2-a-6=0\\a^2+a-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(a-3\right)\left(a+2\right)=0\\\left(a-3\right)\left(a+4\right)=0\end{cases}}\)

=> \(a\in\left\{-4;-2;3\right\}\)

Mà theo đk thì \(a\ge3\) => a = 3 (thỏa mãn)

Vậy a = 3

29 tháng 7 2018

Ai giúp mình với, mình cần sự giúp đỡ, mai nộp bài rồi

22 tháng 7 2023

\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(ĐK:x\ge5\)

\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

22 tháng 7 2023

\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

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

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)

\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}=0\)    (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))

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

 

15 tháng 7 2017

a) \(\sqrt{\dfrac{25}{81}.\dfrac{16}{49}.\dfrac{196}{9}}=\sqrt{\dfrac{25}{81}}.\sqrt{\dfrac{16}{49}}.\sqrt{\dfrac{196}{9}}=\dfrac{5}{9}.\dfrac{4}{7}.\dfrac{14}{3}=\dfrac{40}{27}\)

b) \(\sqrt{3\dfrac{1}{16}.2\dfrac{14}{25}.2\dfrac{34}{81}}=\sqrt{\dfrac{49}{16}.\dfrac{64}{25}.\dfrac{196}{81}}=\sqrt{\dfrac{49}{16}}.\sqrt{\dfrac{64}{25}}.\sqrt{\dfrac{196}{81}}=\dfrac{7}{4}.\dfrac{8}{5}.\dfrac{14}{9}=\dfrac{196}{45}\)

c) \(\dfrac{\sqrt{640}.\sqrt{34,3}}{\sqrt{567}}=\sqrt{\dfrac{640.34,3}{567}}=\sqrt{\dfrac{64.49}{81}}=\dfrac{\sqrt{64}.\sqrt{49}}{\sqrt{81}}=\dfrac{8.7}{9}=\dfrac{56}{9}\)

d) \(\sqrt{21,6}.\sqrt{810}.\sqrt{11^2-5^2}=\sqrt{21,6.810.\left(11^2-5^2\right)}=\sqrt{216.81.\left(11+5\right)\left(11-5\right)}=\sqrt{36^2.9^2.4^2}=36.9.4=1296\)

a: \(=2\cdot\dfrac{4\sqrt{3}}{5}+3\cdot\dfrac{3\sqrt{3}}{7}-\dfrac{3\sqrt{3}}{2}\)

\(=\dfrac{8\sqrt{3}}{5}+\dfrac{9\sqrt{3}}{7}-\dfrac{3\sqrt{3}}{2}\)

\(=\dfrac{112\sqrt{3}+90\sqrt{3}-105\sqrt{3}}{70}=\dfrac{97\sqrt{3}}{70}\)

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

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

\(=6-2\cdot\sqrt{4}=6-2\cdot2=2\)

c: \(=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}\)

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

23 tháng 4 2017

Áp dụng quy tắc khai phương một thương, hãy tính :

a) 9169 = \(\sqrt{\dfrac{3^2}{13^2}}\) = \(\left|\dfrac{3}{13}\right|\) = \(\dfrac{3}{13}\)

b) 25144 = \(\sqrt{\dfrac{5^2}{12^2}}\) = \(\left|\dfrac{5}{12}\right|\) = \(\dfrac{5}{12}\)

c) 1916 = \(\sqrt{\dfrac{25}{16}}\) = \(\sqrt{\dfrac{5^2}{4^2}}\) = \(\left|\dfrac{5}{4}\right|\) = \(\dfrac{5}{4}\)

d) 2781 = \(\sqrt{\dfrac{169}{81}}\) = \(\sqrt{\dfrac{13^2}{9^2}}\) = \(\left|\dfrac{13}{9}\right|\) = \(\dfrac{13}{9}\)

1. Áp dụng quy tắc khai phương một thương, hãy tính: a, \(\sqrt{\dfrac{36}{121}}\) b, \(\sqrt{\dfrac{9}{16}:\dfrac{25}{36}}\) c, \(\sqrt{0,0169}\) d,\(\dfrac{\sqrt{15}}{\sqrt{735}}\) e, \(\sqrt{\dfrac{81}{8}:\sqrt{3\dfrac{1}{8}}}\) g, \(\dfrac{\sqrt{12,5}}{\sqrt{0,5}}\) 2. Tính: a,\(\sqrt{\dfrac{25}{144}}\) b,\(\sqrt{2\dfrac{7}{81}}\) ...
Đọc tiếp

1. Áp dụng quy tắc khai phương một thương, hãy tính:

a, \(\sqrt{\dfrac{36}{121}}\) b, \(\sqrt{\dfrac{9}{16}:\dfrac{25}{36}}\) c, \(\sqrt{0,0169}\)

d,\(\dfrac{\sqrt{15}}{\sqrt{735}}\) e, \(\sqrt{\dfrac{81}{8}:\sqrt{3\dfrac{1}{8}}}\) g, \(\dfrac{\sqrt{12,5}}{\sqrt{0,5}}\)

2. Tính:

a,\(\sqrt{\dfrac{25}{144}}\) b,\(\sqrt{2\dfrac{7}{81}}\) c,\(\sqrt{\dfrac{2,25}{16}}\) d, \(\sqrt{\dfrac{1,21}{0,49}}\)

3. Áp dụng quy tắc chia hai căn bậc hai, hãy tính:

a, \(\sqrt{18}:\sqrt{2}\) b, \(\sqrt{45}:\sqrt{80}\)

c, (\(\sqrt{20}-\sqrt{45}+\sqrt{5}\) ) : \(\sqrt{5}\) d, \(\dfrac{\sqrt{8^2}}{\sqrt{4^5.2^3}}\)

4. Khẳng định nào sau đây là đúng?

A. \(\sqrt{\dfrac{3}{\left(-5\right)^2}}=-\dfrac{\sqrt{3}}{5}\) B. \(\left(\sqrt{\dfrac{-3}{-5}}\right)^2=\dfrac{3}{5}\)

5. Tính.

a, \(\sqrt{2\dfrac{7}{81}}:\dfrac{\sqrt{6}}{\sqrt{150}}\) b, \(\left(\sqrt{12}+\sqrt{27}-\sqrt{3}\right):\sqrt{3}\)

c, \(\left(\sqrt{\dfrac{1}{5}-\sqrt{\dfrac{9}{5}}+\sqrt{5}}\right):\sqrt{5}\) d, \(\sqrt{\dfrac{2+\sqrt{3}}{\sqrt{2}}}\)

6. So sánh

a, So sánh \(\sqrt{144-49}\)\(\sqrt{144}-\sqrt{49}\);

b, Chứng minh rằng , với hai số a,b thỏa mãn a> b> 0 thì \(\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)

3
13 tháng 11 2018

1

a,\(\sqrt{\dfrac{36}{121}}=\sqrt{\dfrac{6^2}{11^2}}=\dfrac{6}{11}\)

\(\sqrt{\dfrac{9}{16}:\dfrac{25}{36}}=\sqrt{\dfrac{81}{100}}=\sqrt{\dfrac{9^2}{10^2}}=\dfrac{9}{10}\)

13 tháng 11 2018

tương tự lm nốthehe