K
Khách

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

a: \(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\sqrt{a}\cdot\sqrt{a}-\sqrt{a}}{-\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{-\left(\sqrt{a}-1\right)}=-\sqrt{a}\)

b: \(\dfrac{2+\sqrt{3}}{2-\sqrt{7}}=\dfrac{\left(2+\sqrt{3}\right)\left(2+\sqrt{7}\right)}{4-7}\)

\(=\dfrac{-\left(2+\sqrt{3}\right)\left(2+\sqrt{7}\right)}{3}\)

\(=\dfrac{-4-2\sqrt{7}-2\sqrt{3}-\sqrt{21}}{3}\)

c: \(3xy\cdot\sqrt{\dfrac{2}{xy}}=\dfrac{3xy}{\sqrt{xy}}\cdot\sqrt{2}=3\sqrt{2}\cdot\sqrt{xy}\)

d:

\(\dfrac{3}{\sqrt[3]{3}+\sqrt[3]{2}}=\dfrac{3\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}{3+2}\)

\(=\dfrac{3}{5}\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)\)

e:

\(\dfrac{4}{\sqrt{3}+1}-\dfrac{5}{\sqrt{3}-2}+\dfrac{6}{\sqrt{3}-3}\)

\(=\dfrac{4\left(\sqrt{3}+1\right)}{3-1}-\dfrac{5}{2-\sqrt{3}}-\dfrac{6}{3-\sqrt{3}}\)

\(=2\left(\sqrt{3}+1\right)-\dfrac{5\left(2+\sqrt{3}\right)}{4-3}-\dfrac{6\left(3+\sqrt{3}\right)}{6}\)

\(=2\sqrt{3}+2-10-5\sqrt{3}-3-\sqrt{3}\)

\(=-4\sqrt{3}-11\)

f:

\(\dfrac{1}{1+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{9}}+\dfrac{1}{\sqrt{9}+\sqrt{13}}\)

\(=\dfrac{\sqrt{5}-1}{5-1}+\dfrac{\sqrt{9}-\sqrt{5}}{9-5}+\dfrac{\sqrt{13}-\sqrt{9}}{13-9}\)

\(=\dfrac{-1+\sqrt{5}-\sqrt{5}+\sqrt{9}-\sqrt{9}+\sqrt{13}}{4}=\dfrac{\sqrt{13}-1}{4}\)

1 tháng 11 2023

\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\\ =\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{1-\sqrt{a}}\\ =\dfrac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\\ =-\sqrt{a}\\ \dfrac{2+\sqrt{3}}{2-\sqrt{7}}\\ =\dfrac{\left(2+\sqrt{3}\right)\left(2+\sqrt{7}\right)}{4-7}\\ =\dfrac{4+2\sqrt{7}+2\sqrt{3}+\sqrt{21}}{-3}\\\)

\(3xy\sqrt{\dfrac{2}{xy}}\\ =\sqrt{\dfrac{\left(3xy\right)^2\cdot2}{xy}}\\ =\sqrt{\dfrac{9x^2y^2\cdot2}{xy}}\\ =\sqrt{9xy\cdot2}\\ =\sqrt{18xy}\)

\(\dfrac{4}{\sqrt{3}+1}-\dfrac{5}{\sqrt{3}-2}+\dfrac{6}{\sqrt{3}-3}\\ =\dfrac{4\left(\sqrt{3}+1\right)}{3-1}-\dfrac{5\left(\sqrt{3}+2\right)}{3-4}+\dfrac{6\left(\sqrt{3}+3\right)}{3-9}\\ =\dfrac{4\left(\sqrt{3}+1\right)}{2}-\dfrac{5\left(\sqrt{3}+2\right)}{-1}+\dfrac{6\left(\sqrt{3}+3\right)}{-6}\\ =2\sqrt{3}+2+5\sqrt{3}+10-\sqrt{3}-3\\ =6\sqrt{3}+9\)

\(\dfrac{1}{1+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{9}}+\dfrac{1}{\sqrt{9}+\sqrt{13}}\\ =\dfrac{1-\sqrt{5}}{1-5}+\dfrac{\sqrt{5}-\sqrt{9}}{5-9}+\dfrac{\sqrt{9}-\sqrt{13}}{9-13}\\ =\dfrac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}}{-4}\\ =\dfrac{1-\sqrt{13}}{-4}\)

`# gvy`

bài 1 :Trục căn thức ở mẫu và rút ngọn nếu được. a) \(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\) b) \(\dfrac{26}{5-2\sqrt{3}}\) c) \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\) d) \(\dfrac{2\sqrt{10}-5}{4-\sqrt{10}}\) g) \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1+1}}\) bài 2: tính giá trị các biểu thức sau: a)\(\dfrac{2}{\sqrt{7}-5}-\dfrac{2}{\sqrt{7}+5}\) b)...
Đọc tiếp

bài 1 :Trục căn thức ở mẫu và rút ngọn nếu được.

a) \(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\) b) \(\dfrac{26}{5-2\sqrt{3}}\) c) \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\)

d) \(\dfrac{2\sqrt{10}-5}{4-\sqrt{10}}\) g) \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1+1}}\)

bài 2: tính giá trị các biểu thức sau:

a)\(\dfrac{2}{\sqrt{7}-5}-\dfrac{2}{\sqrt{7}+5}\) b) \(\dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}\)

c) \(\sqrt{12}+\sqrt{48}-\sqrt{(\sqrt{75}-\sqrt{108)}^2}\)

bài 3: thực hiện phép tính.

a) \(\sqrt{(3-2\sqrt{2})^2}+\sqrt{(3+2\sqrt{2})^2}\) b)\(\sqrt{(5-2\sqrt{6})^2}-\sqrt{(5+2\sqrt{6})^2}\)

c) \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\) d) \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

bài 4: thực hiện các phép tính sau.

a) \(\sqrt{125}-4\sqrt{45}+3\sqrt{20}-\sqrt{80}\) b) \(2\sqrt{\dfrac{27}{4}}-\sqrt{\dfrac{48}{9}}\dfrac{2}{5}\sqrt{\dfrac{75}{16}}\)

c) \(\sqrt{8}+\sqrt{72}+\sqrt{98}-5\sqrt{128}\) d) \(2\sqrt{\dfrac{9}{8}}-\sqrt{\dfrac{49}{2}}+\sqrt{\dfrac{25}{18}}\)

bài 5: rút ngọn biểu thức với giả thiết các biểu thức chữ đều có nghĩa.

a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}(x>0;y>0)\)

b) \(\dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}(a;b\ge0)\)

bài 6: giải các phương trình sau:\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

2
7 tháng 8 2018

mn ơi giải giúp mik bài não cũng đc a

mình cảm ơn mn nhiều ạ =))

7 tháng 8 2018

tớ nghĩ tớ giải đc 1-2 bài gì đó nhưng tớ ko bít bấm can lm sao giải cho cậu đc

7 tháng 8 2017

\(\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{5}-3}\)

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

\(=\dfrac{3-\sqrt{5}}{\sqrt{5}-3}\)

= - 1

\(\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{6+2\sqrt{5}}}{2}\)

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

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

\(\dfrac{2+\sqrt{2}}{\sqrt{1,5+\sqrt{2}}}\)

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

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

\(=\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)

= 2

\(\dfrac{\sqrt{20}}{\sqrt{5}}+\dfrac{\sqrt{117}}{\sqrt{13}}+\dfrac{\sqrt{272}}{\sqrt{17}}+\dfrac{\sqrt{105}}{\sqrt{2\dfrac{1}{7}}}\)

\(=4+9+16+49\)

= 78

7 tháng 8 2017

\(\dfrac{x\sqrt{x}-y\sqrt{y}}{x+\sqrt{xy}+y}\)

\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{x+\sqrt{xy}+y}\)

\(=\sqrt{x}-\sqrt{y}\)

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

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

\(\left[-\text{tử}-\right]=\sqrt{2}\left(2+\sqrt{3}\right)-\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^2}+\sqrt{2}\left(2-\sqrt{3}\right)+\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)^2}\)

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

\(\left[-\text{mẫu}-\right]=2-\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}-\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)

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

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

= 3

Ta có:

\(\dfrac{4\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{3}\)

\(=\dfrac{8-\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{3\sqrt{2}}\)

\(=\dfrac{8-\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{3\sqrt{2}}\)

\(=\dfrac{8-\left(\sqrt{3}+1\right)+\left(\sqrt{3}-1\right)}{3\sqrt{2}}=\dfrac{6}{3\sqrt{2}}=\sqrt{2}\)

\(\sqrt{\dfrac{2+a-2\sqrt{2a}}{a+3-2\sqrt{3a}}}\)

\(=\sqrt{\dfrac{\left(\sqrt{a}-\sqrt{2}\right)^2}{\left(\sqrt{a}-\sqrt{3}\right)^2}}\)

\(=\dfrac{\left|\sqrt{a}-\sqrt{2}\right|}{\left|\sqrt{a}-\sqrt{3}\right|}\)

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

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

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

\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)

c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)

23 tháng 6 2017

(bài 1) a) \(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}\) = \(\dfrac{5-2\sqrt{6}-5-2\sqrt{6}}{25-24}\)

= \(\dfrac{-4\sqrt{6}}{1}\) = \(-4\sqrt{6}\)

b) \(\sqrt{6+2\sqrt{5}}-\dfrac{\sqrt{15}-\sqrt{3}}{\sqrt{3}}\) = \(\sqrt{\left(\sqrt{5}+1\right)^2}-\dfrac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{3}}\)

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

c) \(\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\) = \(\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\)

= \(\sqrt{6}.\sqrt{16}\) = \(4\sqrt{6}\)

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

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

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

= \(\dfrac{2\sqrt{3}-3+4-2\sqrt{3}+2\sqrt{2}-\sqrt{6}-1}{2-\sqrt{3}}\)

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

e) \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\) = \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{1+\sqrt{5}}\)

= \(\dfrac{4}{1+\sqrt{3}}-\sqrt{3}\) = \(\dfrac{4-\sqrt{3}-3}{1+\sqrt{3}}\) = \(\dfrac{1-\sqrt{3}}{1+\sqrt{3}}\)

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

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

23 tháng 6 2017

bài 2)

a)\(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(a+b-2\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)

= \(\dfrac{a\sqrt{a}+a\sqrt{b}+b\sqrt{a}+b\sqrt{b}-2a\sqrt{b}-2b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\)

= \(\dfrac{a\sqrt{a}+-a\sqrt{b}+b\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\) = \(\dfrac{a\left(\sqrt{a}-\sqrt{b}\right)-b\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)

= \(\dfrac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\) = \(a-b\)

b) \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)

= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

= \(\dfrac{2\left(a-1\right)}{4\sqrt{a}}.\dfrac{-4a}{a-1}\) = \(-2\)

a: \(=\dfrac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)

\(=\sqrt{7}-4+\dfrac{23}{9}\sqrt{7}+\dfrac{16}{9}\)

\(=\dfrac{32}{9}\sqrt{7}-\dfrac{20}{9}\)

b:\(=\dfrac{2\sqrt{6}+4+2\sqrt{6}-4}{2}+\dfrac{5}{6}\sqrt{6}\)

\(=2\sqrt{6}+\dfrac{5}{6}\sqrt{6}=\dfrac{17}{6}\sqrt{6}\)

c: \(=\dfrac{1}{3}\sqrt{3}+\dfrac{1}{6}\sqrt{2}+\dfrac{1}{\sqrt{3}}\cdot\sqrt{\dfrac{5-2\sqrt{6}}{12}}\)

\(=\dfrac{1}{3}\sqrt{3}+\dfrac{1}{6}\sqrt{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)

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

 

10 tháng 7 2017

bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không

a: \(=\dfrac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)

\(=\sqrt{7}-4+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)

\(=\dfrac{32}{9}\sqrt{7}-\dfrac{20}{9}\)

b: \(=\dfrac{2\sqrt{6}+4+2\sqrt{6}-4}{2}+\dfrac{5\sqrt{6}}{6}\)

\(=2\sqrt{6}+\dfrac{5}{6}\sqrt{6}=\dfrac{17}{6}\sqrt{6}\)

1 tháng 8 2018

\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)

\(\Rightarrow\sqrt{x}+3\)

\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)

\(\Rightarrow\sqrt{y}-1\)

\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)

\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)

\(\Rightarrow\sqrt{xy}\)

1 tháng 8 2018

\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)

\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)

\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)

\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)

\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)

\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)

9 tháng 8 2018

Bài 1 bạn nhóm , trục như thường nhé :D

Bài 2. \(a.A=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)

\(b.B=\sqrt{17-12\sqrt{2}}-\sqrt{9+4\sqrt{2}}=\sqrt{9-2.2\sqrt{2}.3+8}-\sqrt{8+2.2\sqrt{2}+1}=3-2\sqrt{2}-2\sqrt{2}-1=2-4\sqrt{2}\)

\(c.C=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2.\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{43+30\sqrt{2}}=\sqrt{25+2.3\sqrt{2}.5+18}=5+3\sqrt{2}\)

\(d.D=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)

\(D^2=24-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}=24-2\sqrt{81}=24-18=6\)

\(D=-\sqrt{6}\left(do:D< 0\right)\)

9 tháng 8 2018

cảm ơn bn nhé!!! yeu