+ Ta có:

2√6−√5=2(√6+√5)(√6−√5)(√6+√5)26−5=2(6+5)(6−5)(6+5)

                   =2(√6+√5)(√6)2−(√5)2=2(√6+√5)6−5=2(6+5)(6)2−(5)2=2(6+5)6−5

                   =2(√6+√5)1=2(√6+√5)=2(6+5)1=2(6+5).

+ Ta có:

3√10+√7=3(√10−√7)(√10+√7)(√10−√7)310+7=3(10−7)(10+7)(10−7)

                    =3(√10−√7)(√10)2−(√7)2=3(10−7)(10)2−(7)2=3(√10−√7)10−7=3(10−7)10−7

                    =3(√10−√7)3=√10−√7=3(10−7)3=10−7.

+ Ta có:

1√x−√y=1.(√x+√y)(√x−√y)(√x+√y)1x−y=1.(x+y)(x−y)(x+y)

=√x+√y(√x)2−(√y)2=√x+√yx−y=x+y(x)2−(y)2=x+yx−y

+ Ta có:

2ab√a−√b=2ab(√a+√b)(√a−√b)(√a+√b)2aba−b=2ab(a+b)(a−b)(a+b)

=2ab(√a+√b)(√a)2−(√b)2=2ab(√a+√b)a−b=2ab(a+b)(a)2−(b)2=2ab(a+b)a−b.

\(\frac{2}{\sqrt{6}-\sqrt{5}}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)

\(\frac{3}{\sqrt{10}+\sqrt{7}}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}-\sqrt{7}\right)\left(\sqrt{10}+\sqrt{7}\right)}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\sqrt{10}-\sqrt{7}\)

\(\frac{1}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}+\sqrt{y}}{x-y}\)

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

Nhat Linh bị nhầm câu cuối:

\(\dfrac{y+b\sqrt{y}}{b.\sqrt{y}}=\dfrac{y\sqrt{y}+b.y}{b.y}=\dfrac{\sqrt{y}+b}{b}.\)

\(\frac{5}{\sqrt{10}}=\frac{5\sqrt{10}}{10}=\frac{\sqrt{10}}{2}\)

\(\frac{5}{2\sqrt{5}}=\frac{10\sqrt{5}}{20}=\frac{\sqrt{5}}{2}\)

\(\frac{1}{3\sqrt{20}}=\frac{3\sqrt{20}}{180}=\frac{\sqrt{20}}{60}=\frac{2\sqrt{5}}{60}=\frac{\sqrt{5}}{30}\)

\(\frac{2\sqrt{2}+2}{5\sqrt{2}}=\frac{10\sqrt{2}\left(\sqrt{2}+1\right)}{50}=\frac{20+10\sqrt{2}}{50}=\frac{10\left(2+\sqrt{2}\right)}{50}=\frac{2+\sqrt{2}}{5}\)

\(\frac{y+b\sqrt{y}}{b\sqrt{y}}=\frac{y\left(\sqrt{y}+b\right)}{by}=\frac{\sqrt{y}+b}{b}\)

+ Ta có: 

$\frac{5}{\sqrt{10}}=\frac{5.\sqrt{10}}{\sqrt{10}.\sqrt{10}}=\frac{5\sqrt{10}}{(\sqrt{10}{)}^2}=\frac{5\sqrt{10}}{10}$

$=\frac{5.\sqrt{10}}{5.2}$$=\frac{\sqrt{10}}{2}$.

+ Ta có:

$\frac{5}{2\sqrt{5}}=\frac{5.\sqrt{5}}{2\sqrt{5}.\sqrt{5}}=\frac{5\sqrt{5}}{2.(\sqrt{5}.\sqrt{5})}=\frac{5\sqrt{5}}{2(\sqrt{5}{)}^2}$

$=\frac{5\sqrt{5}}{2.5}=\frac{\sqrt{5}}{2}$.

+ Ta có:

$\frac{1}{3\sqrt{20}}=\frac{1.\sqrt{20}}{3\sqrt{20}.\sqrt{20}}=\frac{\sqrt{20}}{3.(\sqrt{20}.\sqrt{20})}=\frac{\sqrt{20}}{3.(\sqrt{20}{)}^2}$

              $=\frac{\sqrt{20}}{3.20}=\frac{\sqrt{2^2.5}}{60}=\frac{2\sqrt{5}}{60}=\frac{2\sqrt{5}}{2.30}=\frac{\sqrt{5}}{30}$.

+ Ta có:

$\frac{(2\sqrt{2}+2)}{5.\sqrt{2}}=\frac{(2\sqrt{2}+2).\sqrt{2}}{5\sqrt{2}.\sqrt{2}}=\frac{2\sqrt{2}.\sqrt{2}+2.\sqrt{2}}{5.(\sqrt{2}{)}^2}$

                    $=\frac{2.2+2\sqrt{2}}{5.2}=\frac{2(2+\sqrt{2})}{5.2}=\frac{2+\sqrt{2}}{5}$.

+ Ta có:

 $\frac{y+b\sqrt{y}}{b\sqrt{y}}=\frac{(y+b\sqrt{y}).\sqrt{y}}{b\sqrt{y}.\sqrt{y}}=\frac{y\sqrt{y}+b\sqrt{y}.\sqrt{y}}{b.(\sqrt{y}{)}^2}$

                    $=\frac{y\sqrt{y}+b(\sqrt{y}{)}^2}{by}=\frac{y\sqrt{y}+by}{by}$

                    $=\frac{y(\sqrt{y}+b)}{b.y}=\frac{\sqrt{y}+b}{b}$.

Cách khác:

$\frac{y+b\sqrt{y}}{b\sqrt{y}}=\frac{{\left(\sqrt{y}\right)}^2+b\sqrt{y}}{b\sqrt{y}}$$=\frac{\sqrt{y}\left(\sqrt{y}+b\right)}{b\sqrt{y}}=\frac{\sqrt{y}+b}{b}$

Nguồn : Bài 50 trang 30 SGK Toán 9 tập 1 - loigiaihay.com

#Ye Chi-Lien

ĐS:

+ Ta có:

$\frac{3}{\sqrt{3}+1}=\frac{3(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{3\sqrt{3}-3.1}{(\sqrt{3}{)}^2-1^2}$

$=\frac{3\sqrt{3}-3}{3-1}=\frac{3\sqrt{3}-3}{2}$.

+ Ta có:

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

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

+ Ta có:

$\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{(2+\sqrt{3}).(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}=\frac{(2+\sqrt{3}{)}^2}{2^2-(\sqrt{3}{)}^2}$

$=\frac{2^2+\mathrm{2.2.}\sqrt{3}+(\sqrt{3}{)}^2}{4-3}$$=\frac{4+4\sqrt{3}+3}{1}=\frac{(4+3)+4\sqrt{3}}{1}$

$=\frac{7+4\sqrt{3}}{1}=7+4\sqrt{3}$.

+ Ta có:

$\frac{b}{3+\sqrt{b}}=\frac{b(3-\sqrt{b})}{(3+\sqrt{b})(3-\sqrt{b})}$

$=\frac{b(3-\sqrt{b})}{3^2-(\sqrt{b}{)}^2}=\frac{b(3-\sqrt{b})}{9-b};(b\ne 9)$.

+ Ta có:

$\frac{p}{2\sqrt{p}-1}=\frac{p(2\sqrt{p}+1)}{(2\sqrt{p}-1)(2\sqrt{p}+1)}$

$=\frac{p(2\sqrt{p}+1)}{(2\sqrt{p}{)}^2-1^2}=\frac{p(2\sqrt{p}+1)}{4p-1}$$=\frac{2p\sqrt{p}+p}{4p-1}$

$\frac{\mathrm{Bài\; 51\; trang\; 30\; SGK\; Toán\; 9\; tập\; 1\; -\; loigiaihay.com}}{}$

#Ye Chi-Lien

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

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

\(\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3}=\left(2+\sqrt{3}\right)^2=4+4\sqrt{3}+3=7+4\sqrt{3}\)

\(\frac{b}{3+\sqrt{b}}=\frac{b\left(3-\sqrt{b}\right)}{\left(3+\sqrt{b}\right)\left(3-\sqrt{b}\right)}=\frac{b\left(3-\sqrt{b}\right)}{9-b}\)

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

Bài 50:

\(\dfrac{5}{\sqrt{10}}=\dfrac{5\sqrt{10}}{10}=\dfrac{\sqrt{10}}{2}\)

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

\(\dfrac{1}{3\sqrt{20}}=\dfrac{1}{6\sqrt{5}}=\dfrac{\sqrt{5}}{30}\)

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

\(a,=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\\ b,=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\\ c,=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)

(do xy > 0 (gt) nên đưa thừa số xy vào trong căn để khử mẫu)

#Học tốt!!!

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

\(\dfrac{a}{b}\cdot\sqrt{\dfrac{b}{a}}=\dfrac{\sqrt{a\cdot b}}{b}\)

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

\(\sqrt{\dfrac{9\cdot a^3}{36\cdot b}}=\dfrac{\sqrt{a^3\cdot b}}{2\cdot b}\)

\(3\cdot x\cdot y\cdot\sqrt{\dfrac{2}{x\cdot y}}=3\cdot\sqrt{2\cdot x\cdot y}\)

• có nghĩa khi và
  Nếu thì
  Nếu thì
• Tương tự như vậy ta có:
  Nếu thì
  Nếu thì
• Ta có:
  Điều kiện để căn thức có nghĩa là hay Do đó:
  Nếu b>0 thì
  Nếu thì
• Điều kiện để có nghĩa là hay
  Cách 1.
  =
  Cách 2. Biến mẫu thành một bình phương rồi áp dụng quy tắc khai phương một thương:
• Điều kiện để có nghĩa là hay xy>0.
  Do đó