+ 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.

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

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

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

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

+ 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

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

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

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

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

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

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

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

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

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

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

+ 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

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

#Học tốt!!!

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

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

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

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

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

\(\sqrt{\dfrac{1}{600}}\)=\(\sqrt{\dfrac{1}{10^2\cdot6}}\)=\(\sqrt{\dfrac{1\cdot6}{10^2\cdot6\cdot6}}\)=\(\dfrac{\sqrt{6}}{60}\)

\(\sqrt{\dfrac{11}{540}}\)=\(\sqrt{\dfrac{11\cdot540}{540\cdot540}}\)=\(\dfrac{\sqrt{5940}}{540}\)=\(\dfrac{\sqrt{165}}{90}\)

\(\sqrt{\dfrac{3}{50}}\)=\(\sqrt{\dfrac{3\cdot50}{50\cdot50}}\)=\(\dfrac{\sqrt{150}}{50}\)=\(\dfrac{\sqrt{6}}{10}\)

\(\sqrt{\dfrac{5}{98}}\)=\(\sqrt{\dfrac{5\cdot98}{98\cdot98}}=\dfrac{\sqrt{490}}{98}=\dfrac{\sqrt{10}}{14}\)

\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{3-\sqrt{3}}{9}\)

\(\sqrt{\dfrac{1}{600}}=\dfrac{\sqrt{6}}{60}\)

\(\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{165}}{90}\)

\(\sqrt{\dfrac{3}{50}}=\dfrac{\sqrt{6}}{10}\)

\(\sqrt{\dfrac{5}{98}}=\dfrac{\sqrt{10}}{14}\)

\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{3-\sqrt{3}}{9}\)

a) Ta có : Vì \(x\ge0\)và \(y\ge0\)nên \(x+y\ge0\)\(\Leftrightarrow\left|x+y\right|=x+y\)

\(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}\)

\(=\frac{2}{x^2-y^2}\sqrt{\frac{3}{2}.\left(x+y\right)^2}\)

\(=\frac{2}{x^2-y^2}.\sqrt{\frac{3}{2}}.\left|x+y\right|\)

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

\(=\frac{2}{x-y}.\sqrt{\frac{3}{2}}\)

\(=\frac{1}{x-y}.2.\sqrt{\frac{3}{2}}\)

\(=\frac{1}{x-y}.\sqrt{\frac{2^2.3}{2}}\)

\(=\frac{1}{x-y}.\sqrt{6}=\frac{\sqrt{6}}{x-y}\)

a, \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{x^2-y^2}\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{2\sqrt{3}\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\sqrt{2}}\)

do \(x\ge0;y\ge0\)

\(=\frac{2\sqrt{3}}{\sqrt{2}\left(x-y\right)}=\frac{2\sqrt{6}}{2\left(x-y\right)}=\frac{\sqrt{6}}{x-y}\)

Rút gọn các biểu thức sau với x≥0x≥0:

a) 2\(\sqrt{3x}\)-4\(\sqrt{3x}\)+27-3\(\sqrt{3x}\)=27-5\(\sqrt{3x}\)

b)3\(\sqrt{2x}\)-5\(\sqrt{8x}\)+7\(\sqrt{18x}\)+28

=3\(\sqrt{2x}\)-10\(\sqrt{2x}\)+21\(\sqrt{2x}\)+28

=14\(\sqrt{2x}\)+28=14(\(\sqrt{2x}\)+2)

a) \(2\sqrt{3x}-4\sqrt{3x}+27-3\sqrt{3x}\)

\(=\left(2\sqrt{3x}-4\sqrt{3x}-3\sqrt{3x}\right)+27\)

\(=-5\sqrt{3x}+27\)

a) 3\(\sqrt{3}\)=\(\sqrt{27}\)>\(\sqrt{12}\)

c) \(\frac{1}{3}\)\(\sqrt{51}\)=\(\sqrt{\frac{51}{9}}\)<\(\frac{1}{5}\)\(\sqrt{150}\)=\(\sqrt{\frac{150}{25}}\)=\(\sqrt{6}\)

b) 3\(\sqrt{5}\)=\(\sqrt{45}\)< 7=\(\sqrt{49}\)

d) \(\frac{1}{2}\sqrt{6}\)=\(\sqrt{\frac{6}{4}}\)=\(\sqrt{\frac{3}{2}}\)< 6\(\sqrt{\frac{1}{2}}\)=\(\sqrt{\frac{36}{2}}\)=\(\sqrt{18}\)

a) Ta có: $3\sqrt{3}=\sqrt{3^2.3}=\sqrt{9.3}=\sqrt{27}$

Vì $\sqrt{27}>\sqrt{12}$ nên $3\sqrt{3}>\sqrt{12}$

Vậy $3\sqrt{3}>\sqrt{12}$.
b) Ta có: $3\sqrt{5}=\sqrt{3^2.5}=\sqrt{45}$

$7=\sqrt{7^2}=\sqrt{49}$

Vì $\sqrt{49}>\sqrt{45}$ nên $7>3\sqrt{5}$

Vậy $7>3\sqrt{5}$.

c) Ta có: $\frac{1}{3}\sqrt{51}=\sqrt{{\left(\frac{1}{3}\right)}^2.51}=\sqrt{\frac{51}{9}}$

$\frac{1}{5}\sqrt{150}=\sqrt{{\left(\frac{1}{5}\right)}^2.150}=\sqrt{\frac{150}{25}}=\sqrt{6}=\sqrt{\frac{6.9}{9}}=\sqrt{\frac{54}{9}}$

Vì $\sqrt{\frac{54}{9}}>\sqrt{\frac{51}{9}}$ nên $\frac{1}{3}\sqrt{51}<\frac{1}{5}\sqrt{150}$

Vậy $\frac{1}{3}\sqrt{51}<\frac{1}{5}\sqrt{150}$.

d) Ta có: $\frac{1}{2}\sqrt{6}=\sqrt{{\left(\frac{1}{2}\right)}^2.6}=\sqrt{\frac{6}{4}}$

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

Vì $\sqrt{3}.\sqrt{\frac{1}{2}}<6\sqrt{\frac{1}{2}}$ nên $\frac{1}{2}.\sqrt{6}<6\sqrt{\frac{1}{2}}$

Vậy $\frac{1}{2}\sqrt{6}<6\sqrt{\frac{1}{2}}$.

Ta có:

+)  3√5=√32.5=√9.5=√45.35=32.5=9.5=45.

+)  −5√2=−√52.2=−√25.2=−√50.−52=−52.2=−25.2=−50.

+) Với xy>0xy>0 thì √xyxy có nghĩa nên ta có:

−23√xy=−√(23)2.xy=−√49xy.−23xy=−(23)2.xy=−49xy.

+) Với x>0x>0 thì √2x2x có nghĩa nên ta có:

x√2x=√x2.2x=√x2.2xx2x=x2.2x=x2.2x=√2x.xx=√2x.

a, \(3\sqrt{5}=\sqrt{9.5}=\sqrt{45}\)

b, \(-5\sqrt{2}=-\sqrt{25.2}=-\sqrt{50}\)

c, \(-\frac{2}{3}\sqrt{xy}=-\sqrt{\frac{4}{9}xy}\)

d, \(x\sqrt{\frac{2}{x}}=\sqrt{\frac{2x^2}{x}}=\sqrt{2x}\)