\(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\left(a>0,b>0,a\ne b\right)\\ =\dfrac{\sqrt{ab}.\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\\ =\left(\sqrt{a}+\sqrt{b}\right).\left(\sqrt{a}-\sqrt{b}\right)=a-b\)

\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right).\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(a>=0;a\ne1\right)\\ =\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right).\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\\ =\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)

Để đồ thị hai hàm số là các đường thẳng song song : 

$\{\begin{array}{c}m+1=2\\ -m^2-m\ne -2\end{array}$

$\iff \{\begin{array}{c}m=1\\ -m^2-m+2\ne 0\end{array}$

$\iff \{\begin{array}{c}m=1\left(l\right)\\ m\ne 1\\ m\ne -2\end{array}$

Không tồn tại giá trị của m để hai hàm số..........

b/ Hai đường thẳng trên cắt nhau tại 1 điểm trên trục tung khi

\(\left(m+2\right)x+2m^2+1=3x+3\) với x=0 và \(m+2\ne3\Rightarrow m\ne1\)

\(\Rightarrow2m^2=2\Rightarrow m^2=1\Rightarrow m=-1\)

ĐKXĐ: \(a>0;a\ne4\)

Ta có: \(P=\left(\dfrac{\sqrt{a}+a}{1+\sqrt{a}}\right)\left(\dfrac{a-3\sqrt{a}+2}{\sqrt{a}-2}\right)\)

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

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

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

2/3+(-9/16)

=2/3-9/16

=  (32-27)/48

=5/48

+) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\) ( đk : a > 0 ; b >0 ; a khác b )

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

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

\(=a-b\)

+) \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right).\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\) ( đk : a \(\ge\)0 ; a khác 1 )

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

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

\(=1-a\)

Sửa đề: \(\left(2\sqrt6-4\sqrt3+5\sqrt2+\frac14\sqrt8\right)\cdot3\sqrt6\)

\(=\left(2\sqrt6-4\sqrt3+5\sqrt2+\frac14\cdot2\sqrt2\right)\cdot3\sqrt6\)

\(=\left(2\sqrt6-4\sqrt3+\frac{11}{2}\sqrt2\right)\cdot3\sqrt6\)

\(=2\sqrt6\cdot3\sqrt6-4\sqrt3\cdot3\sqrt6+\frac{11}{2}\sqrt2\cdot3\sqrt6\)

\(=6-12\sqrt{18}+\frac{33}{2}\cdot\sqrt{12}=6-36\sqrt2+\frac{33}{2}\cdot2\sqrt3=6-32\sqrt2+33\sqrt3\)

a: Sửa đề: \(\frac34\sqrt{4x}-\sqrt{4x}+5=\frac14\sqrt{4x}\)

ĐKXĐ: x>=0

Ta có: \(\frac34\sqrt{4x}-\sqrt{4x}+5=\frac14\sqrt{4x}\)

=>\(\frac34\cdot2\sqrt{x}-2\sqrt{x}+5=\frac14\cdot2\sqrt{x}\)

=>\(\frac32\sqrt{x}-2\sqrt{x}-\frac12\sqrt{x}=-5\)

=>\(-\sqrt{x}=-5\)

=>\(\sqrt{x}=5\)

=>x=25(nhận)

b: ĐKXĐ: x<=3

Ta có: \(\sqrt{3-x}-\sqrt{27-9x}+1,25\cdot\sqrt{48-16x}=6\)

=>\(\sqrt{3-x}-3\cdot\sqrt{3-x}+1,25\cdot4\cdot\sqrt{3-x}=6\)

=>\(3\cdot\sqrt{3-x}=6\)

=>\(\sqrt{3-x}=2\)

=>3-x=4

=>x=3-4=-1(nhận)

`\sqrt{9x^2+12x+4}=4`

`<=>\sqrt{(3x+2)^2}=4`

`<=>|3x+2|=4`

`<=>` $\left[\begin{matrix} 3x+2=4\\ 3x+2=-4\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=\dfrac{2}{3}\\ x=-2\end{matrix}\right.$

 anh có thể giái giúp em bài trong trang em được 0 ạ?nếu được em cảm ơn

1

\(T=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{a-4}.\left(\dfrac{a-4}{\sqrt{a}}\right)\)

\(=\dfrac{a-2\sqrt{a}+4-a-2\sqrt{a}-4}{a-4}.\dfrac{a-4}{\sqrt{a}}\)

\(=\dfrac{-4\sqrt{a}}{a-4}.\dfrac{a-4}{\sqrt{a}}=-4\)