a) Ta có: \(-7xy\cdot\sqrt{\dfrac{3}{xy}}\)

\(=\dfrac{-7xy\cdot\sqrt{3xy}}{xy}\)

\(=-7\sqrt{3}\cdot\sqrt{xy}\)

b) Ta có: \(ab+b\sqrt{a}+\sqrt{a}+1\)

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

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

$a)-7xy.\sqrt{\dfrac{3}{xy}}$

$=-7.\sqrt{x^2y^2.\dfrac{3}{xy}}(do \,x,y>0a\to xy>0)$

$=-7.\sqrt{\dfrac{xy}{3}}$

$b)ab+b\sqrt{a}+\sqrt{a}+1(a \ge 0)$

$=b\sqrt{a}(\sqrt{a}+1)+\sqrt{a}+1$

$=(\sqrt{a}+1)(b\sqrt{a}+1)$

a) \(-7xy.\sqrt{\dfrac{3}{xy}}=-7xy.\dfrac{\sqrt{3xy}}{xy}=-7\sqrt{3xy}\)

b) \(ab+b\sqrt{a}+\sqrt{a}+1=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)

a: \(-7xy\cdot\sqrt{\dfrac{3}{xy}}=-7xy\cdot\dfrac{\sqrt{3}}{\sqrt{xy}}=-7\sqrt{3xy}\)

b: \(ab+b\sqrt{a}+\sqrt{a}+1\)

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

\(ab+b\sqrt{a}+\sqrt{a}+1=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)

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

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

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

\(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)

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

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

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

\(=\sqrt{a}\left(\sqrt{a}+1\right)+2\sqrt{b}\left(\sqrt{a}+1\right)=\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\sqrt{b}\right)\)

\(a\sqrt{b}+\sqrt{ab}+\sqrt{a}+1\)

\(=\sqrt{ab}\cdot\sqrt{a}+\sqrt{ab}+\sqrt{a}+1\)

\(=\left(\sqrt{ab}\cdot\sqrt{a}+\sqrt{ab}\right)+\left(\sqrt{a}+1\right)\)

\(=\sqrt{ab}\cdot\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)

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

a√b + √(ab) + √a + 1

= [a√b + √(ab)] + (√a + 1)

= √(ab)(√a + 1) + (√a + 1)

= (√a + 1)[√(ab) + 1]

\(A,ĐKXĐ:x;y\ge0\)

\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)

\(=\sqrt{y}\left(\sqrt{x}-2\right)-5\left(\sqrt{x}-2\right)\)

\(=\left(\sqrt{x}-2\right)\left(\sqrt{y}-5\right)\)

\(ĐKXĐ:x;y\ge0\)

\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)

\(=\left(a\sqrt{x}-\sqrt{xy}\right)+\left(b\sqrt{y}-ab\right)\)

\(=\sqrt{x}\left(a-\sqrt{y}\right)+b\left(\sqrt{y}-a\right)\)

\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)

\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)

\(=\left(a-\sqrt{y}\right)\left(\sqrt{x}-b\right)\)

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

a) ĐS: ; b) ĐS: 26; c) ĐS: 12a

d) - = - 6a + 9 -

= - 6a + 9 - = - 6a + 9 - 6│a│.

Khi a ≥ 0 thì │a│= a.

Do đó - = - 6a + 9 -6a = - 12a + 9.

Khi a < 0 thì │a│= a.

Do đó - = - 6a + 9 + 6a = + 9.

1. Ta thấy:

\(\frac{(a-b)^3}{(\sqrt{a}-\sqrt{b})^3}-b\sqrt{b}+2a\sqrt{a}=\frac{(\sqrt{a}-\sqrt{b})^3(\sqrt{a}+\sqrt{b})^3}{(\sqrt{a}-\sqrt{b})^3}-b\sqrt{b}+2a\sqrt{a}\)

\(=(\sqrt{a}+\sqrt{b})^3-b\sqrt{b}+2a\sqrt{a}=a\sqrt{a}+b\sqrt{b}+3\sqrt{ab}(\sqrt{a}+\sqrt{b})-b\sqrt{b}+2a\sqrt{a}\)

\(=3a\sqrt{a}+3\sqrt{ab}(\sqrt{a}+\sqrt{b})=3\sqrt{a}(a+\sqrt{ab}+b)\)

$a\sqrt{a}-b\sqrt{b}=(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)$

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

\(\frac{3a+3\sqrt{ab}}{b-a}=\frac{3\sqrt{a}(\sqrt{a}+\sqrt{b})}{(\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})}=\frac{-3\sqrt{a}}{\sqrt{a}-\sqrt{b}}(2)\)

Từ $(1);(2)$ ta có đpcm.

Câu 2:

Điều kiện đã cho tương đương với:

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

$\Leftrightarrow \frac{(a-b)^2}{a(a+b)(a-b)}+\frac{(a+b)^2}{a(a-b)(a+b)}=\frac{a(3a-b)}{a(a-b)(a+b)}$

$\Leftrightarrow (a-b)^2+(a+b)^2=a(3a-b)$

$\Leftrightarrow 2a^2+2b^2=3a^2-ab$

$\Leftrightarrow a^2-ab-2b^2=0$

$\Leftrightarrow (a+b)(a-2b)=0$

$\Leftrightarrow a=-b$ hoặc $a=2b$

Nếu $a=-b$ thì $|a|=|b|$ (trái giả thiết). Do đó $a=2b$

Khi đó:

$P=\frac{(2b)^3+2(2b)^2.b+3b^3}{2(2b)^3+2b.b^2+b^3}=\frac{19b^3}{19b^3}=1$