a, \(\sqrt{\frac{289}{25}}=\frac{\sqrt{289}}{\sqrt{25}}=\frac{17}{5}\)

b, \(\sqrt{2\frac{14}{25}}=\sqrt{\frac{64}{25}}=\frac{8}{5}\)

c, \(\sqrt{\frac{0,25}{9}}=\frac{\sqrt{0,25}}{\sqrt{9}}=\frac{0,5}{3}=\frac{1}{2}.\frac{1}{3}=\frac{1}{6}\)

d, \(\sqrt{\frac{8,1}{16}}\)đề có sai ko cô ? 

a) căn 289 / 225 = 17/15

b) căn 64/ 25 = 8/5

c) căn 0,25 / 9 = 1/6

d) căn 8,1 / 1,6 = 9/4

289225=289225=1715.

1425=2.25+1425=6425=6425=85.

0,259=0,259=0,53=530=16.

8,11,6=8116=8116=94.

a, \(\frac{\sqrt{2}}{\sqrt{18}}=\sqrt{\frac{2}{18}}=\sqrt{\frac{1}{9}}=\frac{1}{3}\)

b, \(\frac{\sqrt{15}}{\sqrt{735}}=\sqrt{\frac{15}{735}}=\sqrt{\frac{1}{49}}=\frac{1}{7}\)

c, \(\frac{\sqrt{12500}}{\sqrt{500}}=\sqrt{\frac{12500}{500}}=\sqrt{\frac{125}{5}}=\sqrt{25}=5\)

d, \(\frac{\sqrt{6^5}}{\sqrt{2^3.3^5}}=\sqrt{\frac{6^5}{2^3.3^5}}=\sqrt{\frac{2^5.3^5}{2^3.3^5}}=\sqrt{2^2}=2\)

a) căn 2 / căn 18 = 1/3

b) căn 15/ căn 735 = 1/7

c) căn 12500 / căn 500 = 5

d) căn 6^5 / 2^3 * 3^5 = 2

218=218=19=19=13.

15735=15735=149=149=17.

12500500=12500500=25=5.

6523.35=6523.35=25.3523.35=22=2.

a, \(\sqrt{\left(2x-1\right)^2}=3\Leftrightarrow\left|2x-1\right|=3\)

Với \(x\ge\frac{1}{2}\)pt có dạng : \(2x-1=3\Leftrightarrow x=2\)( tm )

Với \(x< \frac{1}{2}\)pt có dạng : \(-2x+1=3\Leftrightarrow x=-1\)( tm ) 

Vậy tập nghiệm của pt là S = { -1 ; 2 } 

b, \(\frac{5}{3}\sqrt{15x}-\sqrt{15x}-2=\frac{1}{3}\sqrt{15x}\)ĐK : \(x\ge0\)

\(\Leftrightarrow\frac{2}{3}\sqrt{15x}-2=\frac{1}{3}\sqrt{15x}\Leftrightarrow\frac{1}{3}\sqrt{15x}=2\)

\(\Leftrightarrow\sqrt{15x}=6\)bình phương 2 vế : \(\Leftrightarrow15x=36\Leftrightarrow x=\frac{36}{15}=\frac{12}{5}\)( tm ) 

Vậy tập nghiệm của pt là S = { 12/5 } 

) √ ( 2 x − 1 ) 2 = 3 ⇒ | 2 x − 1 | = 3 ⇔ 2 x − 1 = ± 3 +) TH1: 2 x − 1 = 3 ⇒ 2 x = 4 ⇒ x = 2 +) TH2: 2 x − 1 = − 3 ⇒ 2 x = − 2 ⇒ x = − 1 Vậy x = − 1 ; x = 2 . b) Điều kiện: x ≥ 0 5 3 √ 15 x − √ 15 x − 2 = 1 3 √ 15 x ⇔ 5 3 √ 15 x − √ 15 x − 1 3 √ 15 x = 2 ⇔ ( 5 3 − 1 − 1 3 ) √ 15 x = 2 ⇔ 1 3 √ 15 x = 2 ⇔ √ 15 x = 6 ⇔ 15 x = 36 ⇔ x = 12 5 Vậy x = 12 5 .

a) \sqrt{(2x-1)^{2}}=3(2x−1)2​=3

$\Rightarrow$ $|2x-1|=3$

$\iff$ $2x-1=\pm 3$

+) TH1: $2x-1=3$

$\Rightarrow$ $2x=4$

$\Rightarrow$ $x=2$
+) TH2: $2x-1=-3$

$\Rightarrow$ $2x=-2$
$\Rightarrow$ $x=-1$
Vậy  $x=-1$ ; $x=2$ .
b) Điều kiện: $x\ge 0$

\dfrac{5}{3} \sqrt{15 x}-\sqrt{15x}-2=\dfrac{1}{3} \sqrt{15 x}35​15x​−15x​−2=31​15x​

$\iff$ \dfrac{5}{3} \sqrt{15 x}-\sqrt{15 x}-\dfrac{1}{3} \sqrt{15 x}=235​15x​−15x​−31​15x​=2

$\iff$ \left(\dfrac{5}{3}-1-\dfrac{1}{3}\right) \sqrt{15} x=2(35​−1−31​)15​x=2

$\iff$ \dfrac{1}{3} \sqrt{15 x}=231​15x​=2

$\iff$ \sqrt{15 x}=615x​=6

$\iff$ $15x=36$

$\iff$ x=\dfrac{12}{5}x=512​

Vậy x=\dfrac{12}{5}x=512​ .

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

#Học tốt!!!

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

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

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

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

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

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

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

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

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

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

+ 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.

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

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

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

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

33+1=3(3−1)(3+1)(3−1)=3(3−1)3−1=3(3−1)2.

23−1=2(3+1)(3−1)(3+1)=2(3+1)3−1=3+

2+32−3=(2+3)(2+3)(2−3)(2

b3+b=b(3−b)(3+b)(3−b)=b(3−b)32−(b)2=b(3−b)9−b.

p2.p−1=p(2.p+1)(2.p−1)(2.p+

a) $(\sqrt{a}+1)(b\sqrt{a}+1)$.
b) $(x-y)(\sqrt{x}+\sqrt{y})$.

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

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

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

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

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

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

\(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\dfrac{\sqrt{5}(\sqrt{3}-1)}{1-\sqrt{3}}=-\sqrt{5}\)  

\(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{12}-\sqrt{6}}{2\sqrt{2}-2}=\dfrac{\sqrt{6}(\sqrt{2}-1)}{2(\sqrt{2}-1)}=\dfrac{\sqrt{6}}{2}\)

\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\sqrt{a}(\sqrt{a}-1)}{1-\sqrt{a}}=-\sqrt{a}\)

\(\dfrac{p-2\sqrt{p}}{\sqrt{p}-2}=\dfrac{\sqrt{p}(\sqrt{p}-2)}{\sqrt{p}-2}=\sqrt{p}\)

TRẢ LỜI :

\(=\sqrt{5}+\sqrt{5}+\sqrt{5}=3\sqrt{5}\)

c) √20 - √45 + 3√18 + √72

= √4.5 - √9.5 + 3√9.2 + √36.2

= 2√5 - 3√5 + 9√2 + 6√2

= -√5 + 15√2

a) 3√5                                           b) 9√2 / 2

c) -√5 + 15√2                                d)
3,4√2

 

a) 3 \sqrt{5}35​.

b) \dfrac{9}{\sqrt{2}}2​9​ hay \dfrac{9 \sqrt{2}}{2}292​​.

c) 15 \sqrt{2}-\sqrt{5}152​−5​.

d) 3,4 \cdot \sqrt{2}3,4⋅2​.

Rút gọn ta được:

M=√a−1/√a

Viết M ở dạng M=1−1/√a

suy ra M<1

Với \(x>0;x\ne1\)

\(M=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)

\(=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)

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

\(=1-\frac{1}{\sqrt{a}}< 1\)hay M < 1 

M = 1 - 1/√a < 1

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

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

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

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

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

510=5.1010.10=510(10)2=51010" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">510=5.1010.10=510(10)2=51010

5.105.2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">5.105.2.

525=5.525.5=552.(5.5)=552(5)2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">525=5.525.5=552.(5.5)=552(5)2

552.5=52" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">552.5=52.

1320=1.20320.20=203.(20.20)=203.(20)2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">1320=1.20320.20=203.(20.20)=203.(20)2

203.20=22.560=2560=252.30=530" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">203.20=22.560=2560=252.30=530.

(22+2)5.2=(22+2).252.2=22.2+2.25.(2)2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">(22+2)5.2=(22+2).252.2=22.2+2.25.(2)2

2.2+225.2=2(2+2)5.2=2+25" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">2.2+225.2=2(2+2)5.2=2+25.

y+byby=(y+by).yby.y=yy+by.yb.(y)2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">y+byby=(y+by).yby.y=yy+by.yb.(y)2

yy+b(y)2by=yy+byby" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">yy+b(y)2by=yy+byby

y(y+b)b.y=y+bb" role="presentation" style="border:0px; direction:ltr; display:inline-table; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">y(y+b)b.y=y+bb.

y+byby=(y)2+byby" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">y+byby=(y)2+byby=√y(√y+b)b√y=√y+bb

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

#Ye Chi-Lien

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

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

\(\dfrac{1}{3\cdot\sqrt{20}}=\dfrac{\sqrt{20}}{60}\)   ;  

 

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 \(x\ge0;y\ge0\)

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

b) Ta có :

\(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\)

\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2.2a+2^2.a^2\right)}\)

\(=\frac{2}{2a-1}\sqrt{5a^2\left[1^2-2.1.2a+\left(2a\right)^2\right]}\)

\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2a\right)^2}\)

\(=\frac{2}{2a-1}.\sqrt{5}.\sqrt{a^2}.\sqrt{\left(1-2a\right)^2}\)

\(=\frac{2}{2a-1}\sqrt{5}.\left|a\right|.\left|1-2a\right|\)

Vì a > 0,5 <=> a > 0 <=> | a | = a

Vì a > 0,5 <=> 2a > 2 . 0,5 <=> 2a > 1 hay 1 < 2a

<=> 1 - 2a < 0 <=> | 1 - 2a | = - ( 1 - 2a )

= -1 + 2a = 2a - 1

Thay vào trên ta được : 

\(=\frac{2}{2a-1}\sqrt{5}.\left|a\right|.\left|1-2a\right|=\frac{2}{2a-1}\sqrt{5}.a.\left(2a-1\right)=2a\sqrt{5}\)

Vậy : \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=2a\sqrt{5}\)