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

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

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

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

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

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

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

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

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

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

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

Bạn học tốt nhé

a)0,6.a

b)\(a^2\).(a-3)

c)36.(a-1)

d)\(\dfrac{1.a^2}{a-b}\).(a-b)

a) $\sqrt{0,36a^2}=\sqrt{0,36}.\sqrt{a^2}=0,6.|a|=-0,6a$ (do $a<0$ nên $|a|=-a$).

b) $\sqrt{a^4.(3-a{)}^2}=\sqrt{a^4}.\sqrt{(3-a{)}^2}=a^2.|3-a|=a^2(a-3)$

(do $a>3$ nên $a-3>0$ $\Rightarrow$ $|3-a|=a-3$).

c) $\sqrt{\mathrm{27.48.}(1-a{)}^2}=\sqrt{3^3.3{.4}^2.(1-a{)}^2}=\sqrt{3^4{.4}^2.(1-a{)}^2}$

$=\sqrt{3^4}.\sqrt{4^2}.\sqrt{(1-a{)}^2}=3^2.4.|1-a|$

$=36(a-1)$ (do $a>1$ $\Rightarrow$ $a-1>0$ nên $|1-a|=a-1$).

1a−b.a4.(a−b)2=1a−b.a4.(a−b)2

1a−b.a2.|a−b|

1a−b.a2.(a−b)=a2 (do $a>b$ $\Rightarrow$ $a-b>0$ nên $|a-b|=a-b$

Bài 1:

\(a,ĐK:2+8x\ge0\Leftrightarrow x\ge-\dfrac{1}{4}\\ b,ĐK:-\dfrac{1}{5}x+9\ge0\Leftrightarrow-\dfrac{1}{5}x\ge-9\Leftrightarrow x\le45\\ c,ĐK:11-7x\ge0\Leftrightarrow x\le\dfrac{11}{7}\)

Bài 2:

\(a,=\sqrt{144a^2}-2a=12\left|a\right|-2a=12a-2a=10\\ b,=\sqrt{6}-6\sqrt{6}-\sqrt{6}=-6\sqrt{6}\)

Bài 3:

\(a,\Leftrightarrow\left|2x+3\right|=3\Leftrightarrow\left[{}\begin{matrix}2x+3=3\\2x+3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\\ b,ĐK:x\ge2\\ PT\Leftrightarrow2\sqrt{x-2}-4\sqrt{x-2}+3\sqrt{x-2}=4\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)

(Vì x > 0 nên |x| = x; y2 > 0 với mọi y ≠ 0)

(Vì x2 ≥ 0 với mọi x; và vì y < 0 nên |2y| = – 2y)

(Vì x < 0 nên |5x| = – 5x; y > 0 nên |y3| = y3)

(Vì x2y4 = (xy2)2 > 0 với mọi x ≠ 0, y ≠ 0)

 

a) 1/y 

b) - x^2 y 

c) -25x^2 / y^2

d) 4x/5y

 

yx.x2y4=yx.x2y4=yx.|x||y2|=yx.xy2=1y.

(Do $x>0$ nên $|x|=x$, $y\ne 0$ $\Rightarrow$ $y^2>0$ nên $|y^2|=y^2$)

x44y2=2y2.x44y2=2y2.|x2||2y|=2y2.x2−2y=2y2.x2−2y=−x2y.

(Do $y<0$ nên $|2y|=-2y$ và $x^2\ge 0$ nên $|x^2|=x^2$)

25x2y6=5xy.25x2y6=5xy.|5x||y3|=5xy.−5xy3=5xy.(−5x)y3=−25x2y2.

(Do $x<0$ nên $|5x|=-5x$ và $y>0$ $\Rightarrow$ $y^3>0$ nên $|y^3|=y^3$)

16x4y8=0,2x3y3.16x4y8=210.x3y3.4|x2y4|=15.x3y3.4x2y4=4x3y35x2y4=4x5y.

( Do $x\ne 0$ $\Rightarrow$ $x^2>0$ và $y\ne 0$ $\Rightarrow$ $y^4>0$ nên $|x^2{y}^4|=x^2{y}^4$)

a, \(2\sqrt{a^2}-5a=2\left|a\right|-5a\)do a < 0 

\(=-2a-5a=-7a\)

b, \(\sqrt{25a^2}+3a=\sqrt{\left(5a\right)^2}+3a=\left|5a\right|+3a\)do \(a\le0\)

TH1 : \(-5a+3a=-2a\)với \(a< 0\)

hoặc TH2 : \(5+3=8\)

c, \(\sqrt{9a^4}+3a^2=\sqrt{\left(3a^2\right)^2}+3a^2=\left|3a^2\right|+3a^2\)

\(=3a^2+3a^2=6a^2\)do \(3>0;a^2\ge0\forall a\Rightarrow3a^2\ge0\forall a\)

d, \(5\sqrt{4a^6}-3a^3=5\sqrt{\left(2a^3\right)^2}-3a^3\)

\(=5\left|2a^3\right|-3a^3=-10a^3-3a^3=-13a^3\)do \(a< 0\Rightarrow a^3< 0\)

a) \(2\sqrt{a^2}-5a\)=2\(|a|\)-5a = -2a-5a=-7a

b) \(\sqrt{25a^2}\) +3a = 5\(|a|\) + 3a=5a+3a=8a.

c) \(\sqrt{9a^4}\) + 3\(a^2\)=6\(a^2\)

d) \(5\sqrt{4a^6}\) - 3\(a^3\)=-13\(a^3\)

a, 2|a| - 5a = -2a - 5a = -7a

b, 5|a|+3a = 5a + 3a = 8a

c, |3a^2| + 3a^2 = 3a^2 + 3a^2 = 6a^2

d, 5|2a^3| - 3a^3 = 5. (-2a^3) - 3a^3 = -10a^3- 3a^3=-13a^3

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

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

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

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

\(=3\sqrt{2x}-5\sqrt{4.2x}+7\sqrt{9.2x}+28\)

\(=3\sqrt{2x}-5\sqrt{2^2.2x}+7\sqrt{3^2.2x}+28\)

\(=3\sqrt{2x}-5.2\sqrt{2x}+7.3\sqrt{2x}+28\)

\(=\left(3\sqrt{2x}-5.2\sqrt{2x}+7.3\sqrt{2x}\right)+28\)

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

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

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​.

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

bạn tham khảo nha : https://loigiaihay.com/bai-76-trang-41-sgk-toan-9-tap-1-c44a26988.html

a) a √ a 2 − b 2 − ( 1 + a √ a 2 − b 2 ) : b a − √ a 2 − b 2 = a √ a 2 − b 2 − a + √ a 2 − b 2 √ a 2 − b 2 ⋅ a − √ a 2 − b 2 b = a √ a 2 − b 2 − a 2 − ( √ a 2 − b 2 ) 2 b √ a 2 − b 2 = a √ a 2 − b 2 − a 2 − ( a 2 − b 2 ) b √ a 2 − b 2 = a √ a 2 − b 2 − b 2 b ⋅ √ a 2 − b 2 = a √ a 2 − b 2 − b √ a 2 − b 2 = a − b √ a 2 − b 2 = √ a − b ⋅ √ a − b √ a − b ⋅ √ a + b (do a > b > 0 )$ = √ a − b √ a + b Vậy Q = √ a − b √ a + b . b) Thay a = 3 b vào Q = √ a − b √ a + b , ta được: Q = √ 3 b − b √ 3 b + b = √ 2 b √ 4 b = √ 2 b √ 2 ⋅ √ 2 b = 1 √ 2 = √ 2 2 .

a) \dfrac{a}{\sqrt{a^{2}-b^{2}}}-\left(1+\dfrac{a}{\sqrt{a^{2}-b^{2}}}\right): \dfrac{b}{a-\sqrt{a^{2}-b^{2}}}a2−b2​a​−(1+a2−b2​a​):a−a2−b2​b​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a+\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}-b^{2}}} \cdot \dfrac{a-\sqrt{a^{2}-b^{2}}}{b}=a2−b2​a​−a2−b2​a+a2−b2​​⋅ba−a2−b2​​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a^{2}-\left(\sqrt{a^{2}-b^{2}}\right)^{2}}{b \sqrt{a^{2}-b^{2}}}=a2−b2​a​−ba2−b2​a2−(a2−b2​)2​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{a^{2}-\left(a^{2}-b^{2}\right)}{b \sqrt{a^{2}-b^{2}}}=a2−b2​a​−ba2−b2​a2−(a2−b2)​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{b^{2}}{b \cdot \sqrt{a^{2}-b^{2}}}=a2−b2​a​−b⋅a2−b2​b2​

=\dfrac{a}{\sqrt{a^{2}-b^{2}}}-\dfrac{b}{\sqrt{a^{2}-b^{2}}}=a2−b2​a​−a2−b2​b​

=\dfrac{a-b}{\sqrt{a^{2}-b^{2}}}=a2−b2​a−b​

=\dfrac{\sqrt{a-b} \cdot \sqrt{a-b}}{\sqrt{a-b} \cdot \sqrt{a+b}}=a−b​⋅a+b​a−b​⋅a−b​​ (do $a>b>0$)$

=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}=a+b​a−b​​

Vậy Q=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}Q=a+b​a−b​​.
b) Thay $a=3b$ vào Q=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}Q=a+b​a−b​​, ta được:

Q=\dfrac{\sqrt{3 b-b}}{\sqrt{3 b+b}}=\dfrac{\sqrt{2 b}}{\sqrt{4 b}}Q=3b+b​3b−b​​=4b​2b​​

=\dfrac{\sqrt{2 b}}{\sqrt{2} \cdot \sqrt{2 b}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}=2​⋅2b​2b​​=2​1​=22​​.