\(\frac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=\frac{\left(\sqrt{x}\sqrt{x}\sqrt{y}+\sqrt{y}\sqrt{y}\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)

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

=\(x-y\)

\(\frac{\sqrt{63y^3}}{\sqrt{7y}}=\sqrt{\frac{63y^3}{7y}}=\sqrt{9y^2}=\sqrt{\left(3y\right)^2}=3y\)

a)\(x-4\ne0;x\ge0\)

<=>\(x\ne4;x\ge0\)

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

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

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

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

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

=\(\frac{\sqrt{x}-2}{\sqrt{x}+2}\)

bạn cap cả bài nhìn đau mắt gê :3

a) Thay \(m=-5\) vào PT ta được:

\(x^2-\left(-5\right)x+2.\left(-5\right)-3=0\)

\(\Rightarrow x^2+5x-10-3=0\)

\(\Rightarrow x^2+5x-13=0\)

\(\Delta=5^2-4.1.\left(-13\right)=25+52=77>0\)

PT có 2 nghiệm phân biệt:

\(x_1=-\frac{5+\sqrt{77}}{2}\)

\(x_2=-\frac{5-\sqrt{77}}{2}\)

Vậy với m = -5 thì PT có nghiệm là \(S=\left\{-\frac{5+\sqrt{77}}{2};-\frac{5-\sqrt{77}}{2}\right\}\)

b) PT có nghiệm kép \(\Leftrightarrow\Delta=0\Leftrightarrow\left(-m\right)^2-4.1.\left(2m-3\right)=0\)

\(\Leftrightarrow m^2-8m+12=0\Leftrightarrow\left(m-2\right)\left(m-6\right)=0\)

\(\Leftrightarrow\int^{m-2=0}_{m-6=0}\Leftrightarrow\int^{m=2}_{m=6}\)

Vậy với m = 2 và m = 6 thì PT có nghiệm kép.

c) PT có 2 nghiệm trái dấu \(\Leftrightarrow\int^{\Delta>0}_{2m-36}_{m0\Leftrightarrow m>6\)

ta co    \(\sqrt{x-2}

b,\(GTNN\)

\(\frac{3-4x}{x^2+1}=\frac{\left(x^2-4x+4\right)-\left(x^2+1\right)}{x^2+1}=\frac{\left(x-2\right)^2}{x^2+1}-1\ge-1\)

GTNN của B =-1 tại x=2