Question

Rút gọn các biểu thức sau.

a)   \(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}\)

b)    \(\dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}\) với  \(a,b>0\)

c)    \(\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}\) với  \(x\ge0\)

d)     \(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)với  \(x\ge0;x\ne14\)

Answer 1

a, \(\frac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}=\frac{\sqrt{10}+\sqrt{6}}{\sqrt{10.3}+\sqrt{6.3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)

b, Với a;b > 0 

\(\frac{a+\sqrt{ab}}{b+\sqrt{ab}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}=\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{ab}}{b}\)

c, Với x >= 0 

\(\frac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}=\frac{\left(\sqrt{x}-1\right)\left(4\sqrt{x}+7\right)}{4\sqrt{x}+7}=\sqrt{x}-1\)

d, Với x >= 0 ; x khác 14

\(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)

Answer 2

a) \(\frac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}=\frac{\sqrt{10}+\sqrt{6}}{\sqrt{3}\left(\sqrt{10}+\sqrt{6}\right)}=\frac{1}{\sqrt{3}}\)

b) \(\frac{a+\sqrt{ab}}{b+\sqrt{ab}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}=\frac{\sqrt{a}}{\sqrt{b}}\)

c) \(\frac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}=\frac{\left(\sqrt{x}-1\right)\left(4\sqrt{x}+7\right)}{\left(4\sqrt{x}+7\right)}=\sqrt{x}-1\)

d) \(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\frac{x+\sqrt{x}-4\sqrt{x}-4}{x-4\sqrt{x}+3\sqrt{x}-12}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)

Answer 3

a) \(\dfrac{\sqrt{2}}{\sqrt{6}}\)

b) \(\dfrac{\sqrt{a}}{\sqrt{b}}\)

c)\(\sqrt{x}-1\)

d) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

Answer 4

a/ \(\dfrac{1}{\sqrt{3}}\)

b/\(\dfrac{\sqrt{ab}}{b}\)

c/\(\sqrt{x}-1\)

d/\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

Answer 5

a)\(\dfrac{\sqrt{3}}{3}\)

b)\(\dfrac{\sqrt{ab}}{b}\)

c)\(\sqrt{x}-1\)

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

Answer 6

a) \(\sqrt{\dfrac{1}{3}}\)
b) \(\sqrt{\dfrac{a}{b}}\)
c) \(\sqrt{x-1}\)
d) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

Answer 7

a) \(\dfrac{\sqrt{3}}{3}\)

b) \(\dfrac{\sqrt{a}+1}{\sqrt{b}+1}\)

c) \(\sqrt{x}-1\)

d) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

Answer 8

 

Answer 9

a. \(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}=\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{3}\left(\sqrt{10}+\sqrt{6}\right)}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\)

b. \(\dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt[]{b}\right)}=\dfrac{\sqrt{a}}{\sqrt{b}}\)

c. \(\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}=\dfrac{4x-4\sqrt{x}+7\sqrt{x}-7}{4\sqrt{x}+7}=\dfrac{\left(\sqrt{x}-1\right)\left(4\sqrt{x}+7\right)}{4\sqrt{x}+7}=\sqrt{x}-1\)

d. \(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

Answer 10

a) \(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}\) = \(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{10.3}+\sqrt{6.3}}=\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{3}\left(\sqrt{10}+\sqrt{6}\right)}\)= \(\dfrac{1}{\sqrt{3}}\)=\(\dfrac{\sqrt{3}}{3}\)

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

với a,b≥0 thì \(\dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}=\)\(\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}=\dfrac{\sqrt{a}}{\sqrt{b}}\)  =\(\dfrac{\sqrt{ab}}{b}\)

c) \(\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}\)

với x≥0 thì \(\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}\)= \(\dfrac{\left(\sqrt{x}-1\right)\left(4\sqrt{x}+7\right)}{4\sqrt{x}+7}=\sqrt{x}-1\)

d) \(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\dfrac{\left(\sqrt{x}-4.\sqrt{x}+1\right)}{\left(\sqrt{x}-4.\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)( với x≥0; x≠14)

 

Answer 11

Answer 12

a) \(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}\sqrt{18}}=\dfrac{1}{\sqrt{3}}\)

b) \(\dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}=\dfrac{\sqrt{a}}{\sqrt{b}}\)

c) \(\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}=\sqrt{x}-1\)

d) \(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

Answer 13

 

Answer 14

 

Answer 15

 

 

Answer 16

a)   \dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}=\dfrac{\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{\sqrt{2}}{\sqrt{6}}=\dfrac{1}{\sqrt{3}}30​+18​10​+6​​=6​(5​+3​)2​(5​+3​)​=6​2​​=3​1​

b)  Với  $a,b>0$

  \dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}=\dfrac{\left(\sqrt{a}\right)^2+\sqrt{a}.\sqrt{b}}{\left(\sqrt{b}\right)^2+\sqrt{a}.\sqrt{b}}\\ =\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{a}}{\sqrt{b}}b+ab​a+ab​​=(b​)2+a​.b​(a​)2+a​.b​​=b​(a​+b​)a​(a​+b​)​=b​a​​

c)   Với $x\ge 0$

\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}=\dfrac{4x+7\sqrt{x}-4\sqrt{x}-7}{4\sqrt{x}+7}\\ =\dfrac{\sqrt{x}\left(4\sqrt{x}+7\right)-\left(4\sqrt{x}+7\right)}{4\sqrt{x}+7}=\dfrac{\left(4\sqrt{x}+7\right)\left(\sqrt{x}-1\right)}{4\sqrt{x}+7}=\sqrt{x}-1.\\4x​+74x+3x​−7​=4x​+74x+7x​−4x​−7​=4x​+7x​(4x​+7)−(4x​+7)​=4x​+7(4x​+7)(x​−1)​=x​−1.d)     Với    $x\ge 0;x\ne 16$

\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+3\right)-4\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}x−x​−12x−3x​−4​=x+3x​−4x​−12x+x​−4x​−4​=x​(x​+3)−4(x​+3)x​(x​+1)−4(x​+1)​=(x​+3)(x​−4)(x​+1)(x​−4)​=x​+3x​+1​
 

Answer 17

a) \(\dfrac{1}{\sqrt{3}}\)

b) \(\dfrac{\sqrt{a}}{\sqrt{b}}\)

c) \(\dfrac{1}{\sqrt{x}-1}\)

d) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

Answer 18

a) \(\dfrac{1}{\sqrt{3}}\)

b) \(\dfrac{\sqrt{a}}{\sqrt{b}}\)

c) \(\sqrt{x}-1\)

d) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

Answer 19

a,

\(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}=\dfrac{\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{\sqrt{2}}{\sqrt{6}}=\dfrac{1}{\sqrt{3}}\)

b,

với a,b>0

\(\dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}=\dfrac{\left(\sqrt{a}\right)^2+\sqrt{a}.\sqrt{b}}{\left(\sqrt{b}\right)^2+\sqrt{a}.\sqrt{b}}=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{a}}{\sqrt{b}}\)

c, Với $x\ge 0$

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

d,với x≥0,x≠16

\(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+3\right)-4\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

Answer 20

  

Answer 21

\(d.=\dfrac{x+\sqrt{X}-4\sqrt{X}-4}{x+3\sqrt{X}-4\sqrt{X}-12}\\ \\ \\ =\dfrac{\sqrt{X}\left(\sqrt{X}+1\right)-4\left(\sqrt{X}+1\right)}{\sqrt{X}\left(\sqrt{X}+3\right)-4\left(\sqrt{X}+3\right)}=\dfrac{\left(\sqrt{X}+1\right)\left(\sqrt{X}-4\right)}{\left(\sqrt{X}+3\right)\left(\sqrt{X}-4\right)}\\ \\ \\ \\ =\dfrac{\sqrt{X}+1}{\sqrt{X}+3}=\dfrac{\sqrt{X}+1}{\sqrt{X}+3}.\dfrac{\sqrt{X}-3}{\sqrt{X}-3}=\dfrac{X-3\sqrt{X}-\sqrt{X}-3}{x-9}=\dfrac{X-2\sqrt{X}-3}{x-9}\)\(c.\dfrac{4x+7\sqrt{X}-4\sqrt{X}-7}{4\sqrt{X}+7}\\ \\ \\ \\ =\dfrac{\sqrt{x}\left(4\sqrt{X}+7\right)-\left(4\sqrt{X}+7\right)}{4\sqrt{X}+7}\\ =\dfrac{\left(4\sqrt{X}+7\right)\left(\sqrt{X}-1\right)}{4\sqrt{X}+7}\\ \\ \\ =\sqrt{X}-1\)\(b.=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}=\dfrac{\sqrt{a}}{\sqrt{b}}\\ =\dfrac{\sqrt{a}}{\sqrt{b}}.\dfrac{\sqrt{b}}{\sqrt{b}}=\dfrac{\sqrt{ab}}{\sqrt{b}.\sqrt{b}}=\dfrac{\sqrt{ab}}{b}\)\(a.\dfrac{\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{\sqrt{2}}{\sqrt{6}}\\ \\ =\sqrt{\dfrac{2}{6}}=\sqrt{\dfrac{1}{3}}\)

Answer 22

a) \(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}\) = \(\dfrac{\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{5}+\sqrt{3}\right)}\)= \(\dfrac{\sqrt{2}}{\sqrt{6}}\) = \(\dfrac{1}{\sqrt{3}}\)

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

c) \(\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}\)= \(\dfrac{4x+7\sqrt{x}-4\sqrt{x}-7}{4\sqrt{x}+7}\)= \(\dfrac{\left(4\sqrt{x}+7\right)\left(\sqrt{x}-1\right)}{4\sqrt{x}+7}\)= \(\sqrt{x}-1\)

d)\(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)= \(\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)= \(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}\)=\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

Answer 23

  

Answer 24

a)\(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}\)=\(\dfrac{\sqrt{10}+\sqrt{6}\left(\sqrt{30}\right)-\sqrt{18}}{30-18}\)=\(\dfrac{10\sqrt{3}-6\sqrt{3}}{12}\)=\(\dfrac{4\sqrt{3}}{12}\)=\(\dfrac{\sqrt{3}}{3}\)

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

c)\(\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}\)=\(\dfrac{16x\sqrt{x}-16x-49\sqrt{x}+49}{16x-49}\)=\(\dfrac{16x\left(\sqrt{x}-1\right)-49\left(\sqrt{x}-1\right)}{16x-49}\)=\(\dfrac{\left(\sqrt{x}-1\right)\left(16x-49\right)}{16x-49}\)=\(\sqrt{x}\)-1

d)\(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)=\(\dfrac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}\)=\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

Answer 25

A)đkxđ x>=0

-x=1(t/m)

-x=4(t/m)

Vậy pt có 2 nghiệm là S={1;4}

B)đkxđ x>=1

x=2(tm)

Vậy pt có 1 nghiệm là S={2}

C)đkxđ x>=-1/2

x=-1/2(tm)

x=-4/9(tm)

Answer 26

a)=1/√3
b) Với  a,b>0 
=√a/√b
c)voi x>=0
=√x-1
d)Với  x\ge0;x\ne16x≥0;x​=16 
=√x+1/√x+3

Answer 27

Answer 28

  

Answer 29

Answer 30

a) \(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}\)\(=\dfrac{\left(\sqrt{10+\sqrt{6}}\right)\left(\sqrt{30}-\sqrt{18}\right)}{\left(\sqrt{30+\sqrt{18}}\right)\left(\sqrt{30}-\sqrt{18}\right)}\)\(=\dfrac{10\sqrt{3}-6\sqrt{5}+6\sqrt{5}-6\sqrt{3}}{30-6\sqrt{15}+6\sqrt{15}-18}\)\(=\dfrac{4\sqrt{3}}{12}=\dfrac{\sqrt{3}}{3}\)

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

c)\(\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}\)\(=\dfrac{4x-4\sqrt{x}+7\sqrt{x}-7}{4\sqrt{x}+7}\)\(=\dfrac{4\sqrt{x}\left(\sqrt{x}-1\right)-7\left(\sqrt{x}-1\right)}{4\sqrt{x}+7}\)

\(=\dfrac{\left(4\sqrt{x}+7\right)\left(\sqrt{x}-1\right)}{4\sqrt{x}+7}=\sqrt{x}-1\)

d) \(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)\(=\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x-4\sqrt{x}+3\sqrt{x}-12}\)\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-4\right)+3\left(\sqrt{x}-4\right)}\)\(=\dfrac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)